# Uniqueness of a continuation – Serlo

Zur Navigation springen Zur Suche springen

Bold text

We derive conditions under which a measure on a ${\displaystyle \sigma }$-algebra is uniquely determined by the values on a generator. On our way, we will learn about Dynkin systems and their relation to ${\displaystyle \sigma }$-algebras. Finally, we will prove the uniqueness theorem for measure continuation.

## Problem

Continuation of functions on sets to measures are usually done in a way that the as many subsets of a basic set ${\displaystyle \Omega }$ enter the ${\displaystyle \sigma }$-algebra, as possible (while sustaining some nice properties). Often, one requires additional conditions, for instance when constructing a geometric volume, that cuboids in ${\displaystyle \mathbb {R} ^{n}}$ get assigned their geometric volume. In general, it is not clear whether such a measure with the desired properties even exists. If it does, we might have several ${\displaystyle \sigma }$-algebras where it can be defined on.

The general construction of a measure is as follows: we start with a small set system ${\displaystyle {\mathcal {C}}}$, such that the function on sets restricted to it fulfills the desired properties. For example, for defining a geometric volume, we choose ${\displaystyle {\mathcal {C}}}$ as the set system of cuboids and ${\displaystyle \mu }$ as the set function that assigns to each cuboid its geometric volume.

Using the existence of a continuation - theorem we know when a function ${\displaystyle \mu \colon {\mathcal {C}}\to [0,\infty ]}$ can be continued to a measure on the ${\displaystyle \sigma }$-algebra ${\displaystyle \sigma ({\mathcal {C}})}$ generated by ${\displaystyle \sigma }$. For the proof of the continuation theorem, one possible continuation has been explicitly stated via outer measures. But can there also be other ways to continue ${\displaystyle \mu }$ to a measure on ${\displaystyle \sigma ({\mathcal {C}})}$? In other words, we are interested in whether a measure ${\displaystyle \mu }$ on the ${\displaystyle \sigma }$-algebra ${\displaystyle \sigma ({\mathcal {C}})}$ is already uniquely determined by its values on the smaller set system ${\displaystyle {\mathcal {C}}}$.

To make them easier to check, uniqueness statements are often re-formulated in mathematics: Suppose, ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are measures on the ${\displaystyle \sigma }$-algebra ${\displaystyle \sigma ({\mathcal {C}})}$ generated by a set system ${\displaystyle {\mathcal {C}}}$. Further, let ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide on ${\displaystyle {\mathcal {C}}}$, which means ${\displaystyle \mu (A)=\nu (A)}$ for all ${\displaystyle A\in {\mathcal {C}}}$. Then uniqueness just means ${\displaystyle \mu =\nu }$ on the whole ${\displaystyle \sigma }$ algebra ${\displaystyle \sigma ({\mathcal {C}})}$.

In the following, we derive conditions for when this is the case.

## The principle of good sets

We will proceed step by step to find conditions on the generator ${\displaystyle {\mathcal {C}}}$ and the two measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ under which uniqueness holds. For this we consider the system of "good sets"

${\displaystyle {\mathcal {G}}:=\{A\in \sigma ({\mathcal {C}})\colon \mu (A)=\nu (A)\}.}$

It contains all sets from ${\displaystyle \sigma ({\mathcal {C}})}$ on which ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide. Uniqueness would mean that all sets in ${\displaystyle \sigma ({\mathcal {C}})}$ are good, i.e. ${\displaystyle {\mathcal {G}}=\sigma ({\mathcal {C}})}$.

Actually, this is equivalent to saying that ${\displaystyle {\mathcal {G}}}$ is a ${\displaystyle \sigma }$-algebra: Since by assumption, ${\displaystyle \mu (C)=\nu (C)}$ is satisfied for all ${\displaystyle C\in {\mathcal {C}}}$, we have ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {G}}}$ holds. But ${\displaystyle {\mathcal {C}}}$ already generates ${\displaystyle \sigma ({\mathcal {C}})}$, i.e. there is no smaller ${\displaystyle \sigma }$-algebra which contains ${\displaystyle {\mathcal {C}}}$. So if ${\displaystyle {\mathcal {G}}}$ is a ${\displaystyle \sigma }$-algebra, then ${\displaystyle {\mathcal {G}}}$ (which was contained in ${\displaystyle \sigma ({\mathcal {C}})}$) must be the entire ${\displaystyle \sigma }$-algebra ${\displaystyle \sigma ({\mathcal {C}})}$ (see: monotonicity and idempotence of the ${\displaystyle \sigma }$-operator).

This type of approach is often used to show that a given property is satisfied for all sets of a set system ${\displaystyle {\mathcal {M}}}$ (like a ${\displaystyle \sigma }$-algebra). It is called the "principle of good sets" and it works like this:

Suppose one can only make statements about a generator of ${\displaystyle {\mathcal {M}}}$, perhaps because ${\displaystyle {\mathcal {M}}}$ can only be characterized in terms of the generator. An example is the Borel ${\displaystyle \sigma }$-algebra, which is extremely large and can only be written down by means of generators and relations. Then it might be smart to proceed indirectly when showing a property of all sets in ${\displaystyle {\mathcal {M}}}$.

We define the set system ${\displaystyle {\mathcal {G}}=\{A\in {\mathcal {M}}\mid A{\text{ has the given property}}\}}$ of "good sets". Then we show:

• ${\displaystyle {\mathcal {G}}}$ is a ${\displaystyle \sigma }$-algebra.
• ${\displaystyle {\mathcal {G}}}$ contains a generator ${\displaystyle {\mathcal {C}}}$ of ${\displaystyle {\mathcal {M}}}$.

It follows that ${\displaystyle {\mathcal {M}}={\mathcal {G}}}$, i.e. all sets in ${\displaystyle {\mathcal {M}}}$ are "good". Hence, we gained control over the extremely tedious set ${\displaystyle {\mathcal {M}}}$ by only using properties of the simple sets in ${\displaystyle {\mathcal {C}}}$.

Theorem (The principle of good sets)

Let ${\displaystyle {\mathcal {M}}}$ be a ${\displaystyle \sigma }$-algebra and let ${\displaystyle {\mathcal {G}}}$ be the system of all sets from ${\displaystyle {\mathcal {M}}}$ for which a given property holds. Assume that

• ${\displaystyle {\mathcal {G}}}$ is a ${\displaystyle \sigma }$-algebra.
• ${\displaystyle {\mathcal {G}}}$ contains a generator ${\displaystyle {\mathcal {C}}}$ of ${\displaystyle {\mathcal {M}}}$.

Then the property holds for all sets from ${\displaystyle {\mathcal {M}}}$, i.e., we have that ${\displaystyle {\mathcal {M}}={\mathcal {G}}}$.

Hint

The principle of good sets is very powerful. It can also be applied to other kinds of set systems (not only for ${\displaystyle \sigma }$-algebras). For instance, it can also be applied to the ring or ${\displaystyle \sigma }$-ring generated by a set system.

In our case, the sets of good sets ${\displaystyle {\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\colon \mu (A)=\nu (A)\}}$ contains all the sets ${\displaystyle A\in \sigma ({\mathcal {C}})}$ on which the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide. The equality of the measures on the generator ${\displaystyle {\mathcal {C}}}$ is known, so ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {G}}}$ holds. Now we are still looking for conditions so that ${\displaystyle {\mathcal {G}}}$ becomes a ${\displaystyle \sigma }$-algebra. So we want to find conditions on the generator ${\displaystyle {\mathcal {C}}}$ and the two measures ${\displaystyle \mu ,\nu }$ such that:

• The basic set ${\displaystyle \Omega }$ is contained in ${\displaystyle {\mathcal {G}}}$.
• ${\displaystyle {\mathcal {G}}}$ is complement stable.
• ${\displaystyle {\mathcal {G}}}$ is union stable with respect to countable unions.

## Existence of an inner approximation

Every ${\displaystyle \sigma }$-algebra contains the basic set ${\displaystyle \Omega }$, so ${\displaystyle \Omega }$ should be in the set of good sets ${\displaystyle {\mathcal {G}}}$. That is, it should hold that ${\displaystyle \mu (\Omega )=\nu (\Omega )}$ for both measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$. In general, this need not be the case even if the two measures agree on ${\displaystyle {\mathcal {C}}}$:

Example (Measure of the basic set is not uniquely determined)

Let ${\displaystyle {\mathcal {C}}}$ be the set system of all one-element subsets of ${\displaystyle \mathbb {R} }$:

${\displaystyle {\mathcal {C}}=\{\{x\}\mid x\in \mathbb {R} \}}$

The ${\displaystyle \sigma }$-algebra generated by this is

${\displaystyle \sigma ({\mathcal {C}})=\{A\subseteq \mathbb {R} :A{\text{ countable or }}A^{\complement }{\text{ countable}}\}}$

(This is proved in the article on generated ${\displaystyle \sigma }$-algebras). We define on ${\displaystyle \sigma ({\mathcal {C}})}$ the two measures ${\displaystyle \mu =0}$ and ${\displaystyle \nu }$ via

${\displaystyle \nu (A):={\begin{cases}0,{\text{ if }}A{\text{ countable}}\\\infty {\text{ else.}}\end{cases}}}$

Then ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are equal on the generator, but not on all of ${\displaystyle \sigma ({\mathcal {C}})}$. Thus, they are not yet uniquely determined by specifying the values on ${\displaystyle {\mathcal {C}}}$. In particular ${\displaystyle \mu (\mathbb {R} )=0\neq \infty =\nu (\mathbb {R} )}$.

We can still generalize this example: Let ${\displaystyle \Omega \neq \emptyset }$ be a set and let ${\displaystyle {\mathcal {C}}=\{\emptyset \}}$. The ${\displaystyle \sigma }$-algebra generated by ${\displaystyle {\mathcal {C}}}$ is ${\displaystyle \sigma ({\mathcal {C}})=\{\emptyset ,\Omega \}}$. Let ${\displaystyle \mu =0}$ be the zero measure on it, and for a ${\displaystyle r>0}$ let the measure ${\displaystyle \nu }$ be defined by ${\displaystyle \nu (\emptyset )=0,\nu (\Omega )=r}$. Then, we have ${\displaystyle \mu (\emptyset )=\nu (\emptyset )}$, so ${\displaystyle \mu }$ and ${\displaystyle \nu }$ agree on ${\displaystyle {\mathcal {C}}}$. However, ${\displaystyle \mu (\Omega )=0\neq r=\nu (\Omega )}$.

So with which conditions on the generator ${\displaystyle {\mathcal {C}}}$ or the two measures ${\displaystyle \mu ,\nu }$ can we enforce ${\displaystyle \mu (\Omega )=\nu (\Omega )}$?

### Idea and definition of an inner approximation

We know that the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide on the sets in ${\displaystyle {\mathcal {C}}}$. The idea is to cover the basic set with at most countably many sets from ${\displaystyle E_{1},E_{2},\ldots \in {\mathcal {C}}}$, i.e. their union should be ${\displaystyle \bigcup _{n=1}^{\infty }E_{n}=\Omega }$.

Example (Coverings of ${\displaystyle \Omega =[0,\infty )}$)

• The sets ${\displaystyle E_{n}=[n,n+2]}$ are neither pairwise disjoint nor contained in each other.
• The sets ${\displaystyle E_{n}=[n,n+1)}$ are pairwise disjoint.
• The sets ${\displaystyle E_{n}=[0,n)}$ are contained in each other in ascending order.

Now we want to infer from ${\displaystyle \mu (E_{n})=\nu (E_{n})}$ for all ${\displaystyle n\in \mathbb {N} }$ (which holds since these sets are from ${\displaystyle {\mathcal {C}}}$), that ${\displaystyle \mu (\Omega )=\nu (\Omega )}$ holds as well. For this the ${\displaystyle E_{n}}$ should be either pairwise disjoint or contained in each other in ascending order:

• If the ${\displaystyle E_{n}}$ are pairwise disjoint, we can use the ${\displaystyle \sigma }$-additivity of measures: ${\displaystyle \mu (\Omega )=\mu \left(\biguplus _{i=1}^{\infty }E_{i}\right)=\sum _{n=1}^{\infty }\mu (E_{i})=\sum _{n=1}^{\infty }\nu (E_{i})=\nu \left(\biguplus _{i=1}^{\infty }E_{i}\right)=\nu (\Omega )}$
• If they are contained in each other in ascending order (i.e. ${\displaystyle E_{1}\subseteq E_{2}\subseteq \ldots }$), they form a monotonic set sequence with limit ${\displaystyle \bigcup _{i=1}^{\infty }E_{i}=\Omega }$. Then we can use the continuity of the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$: ${\displaystyle \mu (\Omega )=\mu (\lim _{n\to \infty }E_{n})=\lim _{n\to \infty }\mu (E_{i})=\lim _{n\to \infty }\nu (E_{i})=\nu (\lim _{n\to \infty }E_{n})=\nu (\Omega )}$

If neither is the case, then we can not necessarily conclude ${\displaystyle \mu (\Omega )=\nu (\Omega )}$ from ${\displaystyle \mu (E_{n})=\nu (E_{n})}$ for all ${\displaystyle n}$. We consider the case where the ${\displaystyle E_{n}}$ are contained in each other in ascending order and define the notion of exhaustion:

Definition (Exhaustion)

Let ${\displaystyle \Omega }$ be a set and let ${\displaystyle E_{n}\subseteq \Omega }$ be subsets for each ${\displaystyle n\in \mathbb {N} }$ with ${\displaystyle E_{n}\subseteq E_{n+1}}$. If ${\displaystyle \Omega =\bigcup _{n\in \mathbb {N} }E_{n}}$, then the sequence ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ is called an exhaustion of ${\displaystyle \Omega }$.

Example (Exhaustion)

For ${\displaystyle \Omega :=\mathbb {R} }$ and ${\displaystyle {\cal {{C}:=\{(a,b]:a,b\in \mathbb {R} \}}}}$ there exists an exhaustion ${\displaystyle (E_{n})_{n}\subseteq {\mathcal {C}}}$ with ${\displaystyle E_{n}:=(-n,n]}$.

Since measures are continuous, íf ${\displaystyle {\mathcal {C}}}$ contains an exhaustion ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ of ${\displaystyle \Omega }$, then we have ${\displaystyle \mu (\Omega )=\lim _{n\to \infty }\mu (E_{n})=\lim _{n\to \infty }\nu (E_{n})=\nu (\Omega )}$.

### Intermediate result

To conclude, w have found the following first condition on the set system ${\displaystyle {\mathcal {C}}}$:

To ensure that the basic set is in the "set of good sets" ${\displaystyle {\mathcal {G}}}$, that is, that ${\displaystyle \mu (\Omega )=\nu (\Omega )}$ holds, we require that ${\displaystyle {\mathcal {C}}}$ contains an exhaustion of ${\displaystyle \Omega }$.

Hint

The condition is automatically satisfied if ${\displaystyle \Omega \in {\mathcal {C}}}$ holds, e.g. if ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are probability measures: Then one can assume a priori ${\displaystyle \Omega \in {\mathcal {C}}}$.

## The inner approximating sets have finite measure

By requiring that ${\displaystyle \Omega }$ has an exhaustion by sets from ${\displaystyle {\mathcal {C}}}$, we ensured that ${\displaystyle \Omega }$ lies in the "set of good sets" ${\displaystyle {\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\colon \mu (A)=\nu (A)\}}$. It remains to investigate under which conditions ${\displaystyle {\mathcal {G}}}$ is closed under formation of differences and countable unions. For this purpose let us first examine under which operations ${\displaystyle {\mathcal {G}}}$ is already closed with our previous assumptions:

### Unions

Let ${\displaystyle A}$ and ${\displaystyle B}$ be two sets from the set of good sets ${\displaystyle {\mathcal {G}}}$. Thus we have ${\displaystyle \mu (A)=\nu (A)}$ and the same for ${\displaystyle B}$. Now, we take their union. Things look good if ${\displaystyle B\subseteq A}$: in this case ${\displaystyle A\cup B=A}$ , so ${\displaystyle \mu (A\cup B)=\mu (A)=\nu (A)=\nu (A\cup B)}$ is definitely also in the set of good sets ${\displaystyle {\mathcal {G}}}$. The same happens for ${\displaystyle A\subseteq B}$

Similarly, the measure value of the union of ${\displaystyle A}$ and ${\displaystyle B}$ is uniquely determined if the two sets are disjoint: In that case, from additivity of the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ it follows that

${\displaystyle \mu (A\uplus B)=\mu (A)+\mu (B)=\nu (A)+\nu (B)=\nu (A\uplus B)}$

So the disjoint union is also uniquely measurable and lies again in ${\displaystyle {\mathcal {G}}}$.

If we additionally exploit the ${\displaystyle \sigma }$-additivity of measures, then we even know the measure of countably infinite unions of disjoint sets. Given a sequence of pairwise disjoint sets ${\displaystyle A_{1},A_{2},\dots \in {\mathcal {G}}}$, the ${\displaystyle \sigma }$-additivity of measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ implies

${\displaystyle \mu \left(\biguplus _{i=1}^{\infty }A_{i}\right)=\sum _{i=1}^{\infty }\mu (A_{i})=\sum _{i=1}^{\infty }\nu (A_{i})=\nu \left(\biguplus _{i=1}^{\infty }A_{i}\right).}$

### Set differences

Let again ${\displaystyle A}$ and ${\displaystyle B}$ be two sets from ${\displaystyle {\mathcal {G}}}$, i.e. let ${\displaystyle \mu (A)=\nu (A)}$, ${\displaystyle \mu (B)=\nu (B)}$ hold. As in the case of the union, the difference of the two sets is again in ${\displaystyle {\mathcal {G}}}$ if ${\displaystyle A}$ and ${\displaystyle B}$ are disjoint. In that case ${\displaystyle A\setminus B=A}$ and we have that ${\displaystyle \mu (A\setminus B)=\mu (A)=\nu (A)=\nu (A\setminus B)}$. (Likewise with the roles on ${\displaystyle A}$ and ${\displaystyle B}$ exchanged).

In the case ${\displaystyle A\subseteq B}$, the difference of ${\displaystyle A}$ and ${\displaystyle B}$ is equal to ${\displaystyle A\setminus B=\emptyset }$. Since ${\displaystyle \mu (\emptyset )=0=\nu (\emptyset )}$ , it lies again in ${\displaystyle {\mathcal {G}}}$. Moreover, because of the additivity of the measure ${\displaystyle \mu }$, we have that

${\displaystyle \mu (B)=\mu ((B\setminus A)\uplus A)=\mu (B\setminus A)+\mu (A).}$

In the first equation, we used that ${\displaystyle A}$ is a subset of ${\displaystyle B}$. The same is true for the measure ${\displaystyle \nu }$ instead of ${\displaystyle \mu }$. Rearranging the above formula together with ${\displaystyle \mu (A)=\nu (A)}$ and ${\displaystyle \mu (B)=\nu (B)}$ yields

${\displaystyle \mu (B\setminus A)=\mu (B)-\mu (A)=\nu (B)-\nu (A)=\nu (B\setminus A).}$

This equation is dangerous! If ${\displaystyle A}$ and ${\displaystyle B}$ have infinite measure, we get the ill-defined expression "${\displaystyle \infty -\infty }$", which cannot be sensibly defined:

Example (${\displaystyle \infty -\infty }$ cannot be defined)

We consider the measure ${\displaystyle \lambda }$ defined by the geometric length in ${\displaystyle \mathbb {R} }$. (We know that such a measure exists by the continuation theorem.) For the two sets ${\displaystyle A=\mathbb {R} }$ and ${\displaystyle B=[0,\infty )}$, we have that ${\displaystyle A\setminus B=(-\infty ,0)}$. So it follows ${\displaystyle \lambda (A\setminus B)=\lambda ((-\infty ,0))=\infty }$, and in this case, "${\displaystyle \infty -\infty =\infty }$".

By contrast, for the sets ${\displaystyle A=B=\mathbb {R} }$, we have that ${\displaystyle A\setminus B=\emptyset }$. So ${\displaystyle \lambda (A\setminus B)=\lambda (\emptyset )=0}$ and we get "${\displaystyle \infty -\infty =0}$".

This shows that in general we cannot make sense of the expression ${\displaystyle \infty -\infty }$, without further assumptions.

### Differences of sets with infinite measure

A way out of this problem is to approximate ${\displaystyle A}$ and ${\displaystyle B}$ by ascending sequences ${\displaystyle (A_{n})_{n\in \mathbb {N} },(B_{n})_{n\in \mathbb {N} }}$ of sets of finite measure and take a limit. For this, the sets of the set sequence should also be good sets. We can then calculate the measure of the differences ${\displaystyle A_{n}\setminus B_{n}}$ as above, since ${\displaystyle \mu (A_{n})}$ and ${\displaystyle \mu (B_{n})}$ are both finite. But we have to be careful: For this to work, the subset relation ${\displaystyle A_{n}\subseteq B_{n}}$ must also hold for the set sequences ${\displaystyle (A_{n})_{n\in \mathbb {N} },(B_{n})_{n\in \mathbb {N} }}$ for all ${\displaystyle n\in \mathbb {N} }$. So we cannot just choose the sequences ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (B_{n})_{n\in \mathbb {N} }}$ in an arbitrary manner. They need to grow "equally fast" and at the same time approximate ${\displaystyle A}$ and ${\displaystyle B}$ equally fast.

Recall: we assumed that there is an exhaustion ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ of the basic set ${\displaystyle \Omega }$ with sets from the set system ${\displaystyle {\mathcal {C}}}$, i.e., a monotonically growing set sequence in ${\displaystyle {\mathcal {C}}}$ with limit ${\displaystyle \Omega }$. (This was to guarantee ${\displaystyle \mu (\Omega )=\nu (\Omega )}$.)

Then the sets ${\displaystyle A_{n}:=A\cap E_{n}}$ also form an increasing set sequence with limit

${\displaystyle \bigcup _{n=1}^{\infty }A_{n}=\bigcup _{n=1}^{\infty }(A\cap E_{n})=A\cap \bigcup _{n=1}^{\infty }E_{n}=A\cap \Omega =A}$.

The same is true for the sequence ${\displaystyle B_{n}:=B\cap E_{n}}$. Moreover, because ${\displaystyle B\subseteq A}$, we also have ${\displaystyle B_{n}=B\cap E_{n}\subseteq A\cap E_{n}=A_{n}}$, so the subset relation is satisfied for every member of the sequence. Because of ${\displaystyle B_{n}\setminus A_{n}=(B\cap E_{n})\setminus (A\cap A_{n})=(B\setminus A)\cap E_{n}}$ the sequence of ${\displaystyle (B_{n}\setminus A_{n})_{n\in \mathbb {N} }}$ is also monotonically increasing.

Let us now use the same calculation as for the differences of sets with finite measure

${\displaystyle \mu (B_{n}\setminus A_{n})=\mu (B_{n})-\mu (A_{n})=\nu (B_{n})-\nu (A_{n})=\nu (B_{n}\setminus A_{n})}$

and then turn to the limit ${\displaystyle n\to \infty }$. For this we need:

• ${\displaystyle \mu (A_{n})=\nu (A_{n})}$ and ${\displaystyle \mu (B_{n})=\nu (B_{n})}$ for all ${\displaystyle n}$. That is, intersections ${\displaystyle C\cap E_{n}}$ of sets ${\displaystyle C\in {\mathcal {C}}}$ with the ${\displaystyle E_{n}}$ are said to lie in ${\displaystyle {\mathcal {G}}}$.
• Each set of the exhaustion has finite measure, i.e., ${\displaystyle \mu (E_{n})<\infty }$ for all ${\displaystyle n\in \mathbb {N} }$. Only then, because of monotonicity, we can be sure that ${\displaystyle \mu (A_{n})=\mu (A\cap E_{n})\leq \mu (E_{n})<\infty }$ and ${\displaystyle \mu (B_{n})=\mu (B\cap E_{n})\leq \mu (E_{n})<\infty }$ also holds, ´which is our goal.

Hint

Note that the "finiteness" of an exhaustion depends on the considered measure. For example, the exhaustion ${\displaystyle ((-n,n])_{n\in \mathbb {N} }}$ is finite with respect to ${\displaystyle \mu \equiv 0}$, but not with respect to the measure ${\displaystyle {\tilde {\mu }}}$ that assigns the value ${\displaystyle \infty }$ to every set except ${\displaystyle \emptyset }$.

If ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (B_{n})_{n\in \mathbb {N} }}$ satisfy these conditions, we can calculate the original difference ${\displaystyle B\setminus A}$ (${\displaystyle A\subseteq B}$):

${\displaystyle \mu (B\setminus A)=\lim _{n\to \infty }\mu (B_{n}\setminus A_{n})=\lim _{n\to \infty }(\mu (B_{n})-\mu (A_{n}))=\lim _{n\to \infty }(\nu (B_{n})-\nu (A_{n}))=\lim _{n\to \infty }\nu (B_{n}\setminus A_{n})=\nu (B\setminus A).}$

Note that we were only able to swap difference and measure because the sets ${\displaystyle A_{n},B_{n}}$ had finite measure. We could do that for the sets ${\displaystyle A,B}$ with infinite measure. For this we had to construct ${\displaystyle (A_{n})_{n\in \mathbb {N} }=(A\cap E_{n})_{n\in \mathbb {N} },(B_{n})_{n\in \mathbb {N} }=(B\cap E_{n})_{n\in \mathbb {N} }}$.

The following examples shows that the sets ${\displaystyle E_{n}}$ of the exhaustion indeed need to have finite measure in order to get uniqueness.

Example

We consider the set system of half-open intervals in ${\displaystyle \mathbb {R} }$

${\displaystyle {\mathcal {C}}=\{(a,b]:-\infty \leq a\leq b<\infty \}.}$

It generates a ${\displaystyle \sigma }$-algebra ${\displaystyle \sigma ({\mathcal {C}})}$ (the so-called Borel ${\displaystyle \sigma }$-algebra), which also contains all one-element subsets of ${\displaystyle \mathbb {R} }$. Now consider the two measures ${\displaystyle \sigma ({\mathcal {C}})}$ defined on ${\displaystyle \mu }$ and ${\displaystyle \nu }$ by

${\displaystyle \mu (A)={\begin{cases}0,\;{\text{ if }}A=\emptyset ,\\\infty ,\;{\text{ else.}}\end{cases}}}$

and

${\displaystyle \nu (A)=\#(\mathbb {Q} \cap A)}$

for all ${\displaystyle A\in \sigma ({\mathcal {C}})}$. (${\displaystyle \mu }$ is thus the trivial measure that takes only the values ${\displaystyle 0}$ and ${\displaystyle \infty }$, while ${\displaystyle \nu }$ counts the number of rational numbers contained in ${\displaystyle A}$). Because ${\displaystyle \mathbb {Q} }$ is dense in ${\displaystyle \mathbb {R} }$, the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide on the set system ${\displaystyle {\mathcal {C}}}$. Furthermore, the exhaustion ${\displaystyle (E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}},E_{n}:=(-n,n]}$ of ${\displaystyle \mathbb {R} }$ exists with ${\displaystyle \mu (E_{n})=\infty =\nu (E_{n})}$ for all ${\displaystyle n\in \mathbb {N} }$.

The two measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are not equal on the ${\displaystyle \sigma }$-algebra generated by ${\displaystyle {\mathcal {C}}}$: For all one-element sets ${\displaystyle \{x\}\subseteq \mathbb {R} }$, we have that ${\displaystyle \mu (\{x\})=\infty }$ holds, but ${\displaystyle \nu (\{x\})=1}$ or ${\displaystyle =0}$. So the measures ${\displaystyle \mu }$ or ${\displaystyle \nu }$ are not yet uniquely determined by their values on ${\displaystyle {\mathcal {C}}}$.

### Intermediate result

We conclude what has been found out:

• The "set of good sets" ${\displaystyle {\mathcal {G}}}$ is already closed under (at most countably infinitely many) disjoint unions; unions of sets which are subsets of each other; differences of disjoint sets; and differences of sets which are subsets of each other and have finite measure.
• To ensure that the basic set is in ${\displaystyle {\mathcal {G}}}$, that is, ${\displaystyle \mu (\Omega )=\nu (\Omega )}$, we require that ${\displaystyle {\mathcal {C}}}$ contains an exhaustion ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ of ${\displaystyle \Omega }$.
• In order for differences of sets of infinite measure, which are subsets of each other, are in ${\displaystyle {\mathcal {G}}}$, we require, that the exhaustion sets ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ all have finite measure and that for all ${\displaystyle A\in {\mathcal {G}}}$ the intersections ${\displaystyle A\cap E_{n}}$ lie again in ${\displaystyle {\mathcal {G}}}$.

Apart from these conditions and ${\displaystyle \mu |_{\mathcal {C}}=\nu |_{\mathcal {C}}}$, we make no requirements on ${\displaystyle \mu }$, ${\displaystyle \nu }$, and ${\displaystyle {\mathcal {C}}}$.

The last condition is somewhat unsatisfactory, because it involves ${\displaystyle {\mathcal {G}}}$ (which may include complicated sets). But we want to find conditions that refer only to the a priori given measures ${\displaystyle \mu ,\nu }$ and the generator ${\displaystyle {\mathcal {C}}}$, respectively. We will still work on this and weaken the condition later.

Hint

In particular, the exhaustion requirements are directly satisfied if ${\displaystyle \mu (\Omega )=\nu (\Omega )<\infty }$ holds This is always the case if ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are probability measures. Then ${\displaystyle (\Omega )_{n\in \mathbb {N} }}$ is a monotonically increasing sequence of sets with finite measure, which obviously converges to ${\displaystyle \Omega }$.

## Dynkin systems

As before, let ${\displaystyle {\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\mid \mu (A)=\nu (A)\}}$ be the system of good sets on which the two measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide. Assuming that the conditions from the previous intermediate result are satisfied, we now know that the two measures are equal on the following sets:

• The basic set ${\displaystyle \Omega }$: this is guaranteed by the exhaustion ${\displaystyle (E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}}$.
• Unions of finitely or countably infinitely many pairwise disjoint sets in ${\displaystyle {\mathcal {G}}}$: this holds because of ${\displaystyle \sigma }$-additivity and continuity of the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$.
• Differences of sets from ${\displaystyle {\mathcal {G}}}$, where one is contained in the other: this is guaranteed by finiteness of the exhaustion and by the condition that intersections of sets from ${\displaystyle {\mathcal {G}}}$ with sets of the exhaustion are again in ${\displaystyle {\mathcal {C}}}$.

Thus we can already characterize the set system ${\displaystyle {\mathcal {G}}}$ of "good sets" more precisely. It contains the basic set and is closed under the operations "disjoint union" and "difference of sets contained in each other".

A set system with these properties is called a "Dynkin system".

### Definition

Definition (Dynkin system)

A set system ${\displaystyle {\mathcal {D}}\subseteq {\mathcal {P}}(\Omega )}$ is called a Dynkin system if it holds that:

1. ${\displaystyle \Omega \in {\mathcal {D}}}$
2. for every two sets ${\displaystyle A,B\in {\mathcal {D}}}$ with ${\displaystyle A\subseteq B}$ we also have ${\displaystyle B\setminus A\in {\mathcal {D}}}$.
3. for each countably many pairwise disjoint sets ${\displaystyle A_{1},A_{2},\dots \in {\mathcal {D}}}$ we also have ${\displaystyle \biguplus _{n\in \mathbb {N} }A_{n}\in {\mathcal {D}}}$.

Hint

It follows directly from the definition that every ${\displaystyle \sigma }$-algebra is a Dynkin system. The converse is not true: A Dynkin system is not always a ${\displaystyle \sigma }$-algebra, because it lacks closure under non-disjoint countable unions.

An equivalent characterization of a Dynkin system is the following.

Theorem

A set system ${\displaystyle {\mathcal {D}}\subseteq {\mathcal {P}}(\Omega )}$ is a Dynkin system if and only if:

1. ${\displaystyle \Omega \in {\mathcal {D}}}$
2. for every countably many ${\displaystyle A\in {\mathcal {D}}}$, we also have ${\displaystyle A^{\complement }\in {\mathcal {D}}}$.
3. for each countably many pairwise disjoint sets ${\displaystyle A_{1},A_{2},\dots \in {\mathcal {D}}}$ we also have ${\displaystyle \biguplus _{n\in \mathbb {N} }A_{n}\in {\mathcal {D}}}$.

Proof

Let ${\displaystyle {\mathcal {D}}}$ be a Dynkin system in the sense of the definition above. Points 1. and 3. in the theorem are identical to the definition. We only need to show point 2. So let ${\displaystyle A\in {\mathcal {D}}}$ be arbitrary. We have that ${\displaystyle \Omega \in {\mathcal {D}}}$ and ${\displaystyle A\subseteq \Omega }$, so it follows from property 2. of the definition that also ${\displaystyle A^{\complement }=\Omega \setminus A\in {\mathcal {D}}}$.

Conversely, let the three properties from the theorem be satisfied. We show that then ${\displaystyle {\mathcal {D}}}$ is a Dynkin system in the sense of the definition above. Again, we only need to show point 2. So let ${\displaystyle A,B\in {\mathcal {D}}}$ be arbitrary with ${\displaystyle A\subseteq B}$. Let ${\displaystyle B\setminus A=B\cap A^{\complement }=(B^{\complement }\uplus A)^{\complement }}$. The union on the right hand side is disjoint because of ${\displaystyle A\subseteq B}$. By assumptions 2. and 3. the set ${\displaystyle (B^{\complement }\uplus A)^{\complement }=B\setminus A}$ thus indeed lies in ${\displaystyle {\mathcal {D}}}$.

So with the preconditions from the intermediate result of the previous section, ${\displaystyle {\mathcal {G}}}$ is already a Dynkin system. All that is still missing for a ${\displaystyle \sigma }$ algebra is closure under arbitrary countable unions.

Moreover, since the two measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ agree on the generator ${\displaystyle {\mathcal {C}}}$, we have that ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {G}}}$ holds. So the Dynkin system generated by ${\displaystyle {\mathcal {C}}}$ also lies in the set of good sets ${\displaystyle {\mathcal {G}}}$, which is defined analogously to the generated ${\displaystyle \sigma }$-algebra:

Definition (Generated Dynkin system)

Let ${\displaystyle \Omega }$ be a set and ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {P}}(\Omega )}$ be a set system. The Dynkin system

${\displaystyle \delta ({\mathcal {C}}):=\bigcap \{{\mathcal {D}}:{\mathcal {D}}{\text{ is a Dynkin system with }}{\mathcal {C}}\subseteq {\mathcal {D}}\}}$

is then called the Dynkin system generated by ${\displaystyle {\mathcal {C}}}$. In other words, ${\displaystyle \delta ({\mathcal {C}})}$ is the smallest Dynkin system containing ${\displaystyle {\mathcal {C}}}$.

As with the definition of the generated ${\displaystyle \sigma }$-algebra, we need to convince ourselves that ${\displaystyle \delta ({\mathcal {C}})}$ is well-defined. This can be done completely analogously to the proof we already gave for generated ${\displaystyle \sigma }$-algebras.

Theorem

The intersection in the above definition is not empty. Furthermore, the intersection of any number of Dynkin systems is again a Dynkin system.

Hint

Just as for generated ${\displaystyle \sigma }$-algebras, the following properties hold for the Dynkin system generated by a set system ${\displaystyle {\mathcal {C}}}$:

1. Extensivity: ${\displaystyle {\mathcal {C}}\subseteq \delta ({\mathcal {C}})}$
2. Minimality: ${\displaystyle \delta ({\mathcal {C}})}$ is the smallest Dynkin system containing ${\displaystyle {\mathcal {C}}}$. If ${\displaystyle {\mathcal {C}}}$ is a Dynkin system, then ${\displaystyle \delta ({\mathcal {C}})={\mathcal {C}}}$.
3. Idempotence: ${\displaystyle \delta ({\mathcal {C}}))=\delta ({\mathcal {C}})}$.
4. Monotonicity: ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {E}}\implies \delta ({\mathcal {C}})\subseteq \delta ({\mathcal {E}})}$

### Examples

Here are a few examples of Dynkin systems:

Example (${\displaystyle \sigma }$-algebras)

It follows directly from the definition that any ${\displaystyle \sigma }$-algebra is also a Dynkin system.

However, not every Dynkin system is a ${\displaystyle \sigma }$-algebra:

Example (Subsets with even cardinality)

Let ${\displaystyle n>2}$ be an even number and ${\displaystyle \Omega }$ be a set with exactly ${\displaystyle n}$ elements. The set system

${\displaystyle {\mathcal {D}}:=\{A\subseteq \Omega \mid {\text{number of elements in }}A{\text{ is even}}\}}$

is a Dynkin system:

Since ${\displaystyle |\Omega |=n}$ is even, we have that ${\displaystyle \Omega \in {\mathcal {D}}}$.

If ${\displaystyle A,B\in {\mathcal {D}}}$ are even with ${\displaystyle A\subseteq B}$, then ${\displaystyle |B\setminus A|=|B|-|A|}$ is even since ${\displaystyle |A|,|B|}$ are even.

Since ${\displaystyle \Omega }$ is finite, we need only consider unions of finitely many disjoint sets for the third condition. Let ${\displaystyle A_{1},\dots ,A_{k}\in {\mathcal {D}}}$ be pairwise disjoint sets, each containing an even number of elements. Then, because of disjointness, the union ${\displaystyle \biguplus _{i=1}^{k}A_{i}}$ also contains an even number of elements and so lies in ${\displaystyle {\mathcal {D}}}$.

However, ${\displaystyle {\mathcal {D}}}$ is not a ${\displaystyle \sigma }$-algebra: Let ${\displaystyle x,y,z\in \Omega }$ be three distinct elements (these exist, since by assumption ${\displaystyle |\Omega |=n>2}$). Then ${\displaystyle \{x,y\},\{y,z\}\in {\mathcal {D}}}$ holds, but their union ${\displaystyle \{x,y,z\}}$ contains an odd number of elements, so it is not contained in ${\displaystyle {\mathcal {D}}}$.

## Motivation for cut-stability

We considered conditions under which the set system ${\displaystyle {\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\mid \mu (A)=\nu (A)\}}$ of "good" sets is a Dynkin system, i.e.

• it contains the basic set,
• it is closed under taking complements,
• it is closed under taking (countable) disjoint unions.

Since we assume that the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide on the generator ${\displaystyle {\mathcal {C}}}$, ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {G}}}$ holds. Since ${\displaystyle {\mathcal {G}}}$ is itself a Dynkin system, therefore the Dynkin system ${\displaystyle \delta ({\mathcal {C}})}$ generated by ${\displaystyle {\mathcal {C}}}$ is also contained in ${\displaystyle {\mathcal {G}}}$.

We would like to have ${\displaystyle \sigma ({\mathcal {C}})\subseteq {\mathcal {G}}}$ , such that ${\displaystyle \mu }$ and ${\displaystyle \nu }$ agree on their whole domain of definition. So ${\displaystyle {\mathcal {G}}}$ should not just a Dynkin system, but a ${\displaystyle \sigma }$-algebra. So what about closedness under non-disjoint finite/countably infinite unions? Let us first look at finite unions. Let ${\displaystyle A,B\in {\mathcal {G}}}$ be good sets (i.e., ${\displaystyle \mu (A)=\nu (A)}$ and ${\displaystyle \mu (B)=\nu (B)}$), non-disjoint, and neither ${\displaystyle A\subseteq B}$ nor ${\displaystyle B\subseteq A}$.

Initially, we have ${\displaystyle \mu (A)=\nu (A)}$ and ${\displaystyle \mu (B)=\nu (B)}$, but it is not yet clear whether also ${\displaystyle \mu (A\cup B)=\nu (A\cup B)}$ . Actually every value may appear on the lift-hand side, as long as we do not violate the monotonicity of ${\displaystyle \mu }$, i.e. as long as ${\displaystyle \mu (A\cup B)\geq \mu (A)}$ and ${\displaystyle \mu (A\cup B)\geq \mu (B)}$.

What other conditions must be satisfied for ${\displaystyle \mu (A\cup B)=\nu (A\cup B)}$ to hold? It suffices if the intersection ${\displaystyle A\cap B}$ is again a good set, i.e. ${\displaystyle \mu (A\cap B)=\nu (A\cap B)}$: If ${\displaystyle A}$ and ${\displaystyle B}$ both have finite measure, then we have

${\displaystyle \mu (A\cup B)=\mu (A)+\mu (B)-\mu (A\cap B)=\nu (A)+\nu (B)-\nu (A\cap B)=\nu (A\cup B).}$

If either set has infinite measure, the equality ${\displaystyle \mu (A\cup B)=\infty =\nu (A\cup B)}$ holds anyway. Intersections of good sets should therefore be good sets again.

A set system, which is not left, when taking arbitrary intersections between its sets is called "cut-stable":

Definition (Cut stability of a set system)

A set system ${\displaystyle {\mathcal {M}}}$ is called cut stable, if

${\displaystyle A,B\in {\mathcal {M}}\implies A\cap B\in {\mathcal {M}}}$

Hint

By induction, the set system is then closed under intersection of any finite number of sets (not only 2), See below.

Is the cut-stability of the system of good sets ${\displaystyle {\mathcal {G}}}$ in addition to our previous conditions already enough for it to be an ${\displaystyle \sigma }$-algebra? Suppose ${\displaystyle {\mathcal {G}}}$ is cut-stable. The previous reasoning for two sets can be extended by induction to any finite union of good sets: Let ${\displaystyle A_{1},A_{2},\dots ,A_{n}\in \sigma ({\mathcal {C}})}$ be good sets, i.e., ${\displaystyle \mu (A_{i})=\nu (A_{i})}$ for all ${\displaystyle i=1,\dots ,n}$. We also assume again that all ${\displaystyle A_{i}}$ have finite measure (otherwise the equality holds anyway). Then

{\displaystyle {\begin{aligned}\mu \left(\bigcup _{i=1}^{n}A_{i}\right)=\mu \left(A_{n}\cup \bigcup _{i=1}^{n-1}A_{i}\right)&=\mu \left(A_{n}\right)+\mu \left(\bigcup _{i=1}^{n-1}A_{i}\right)-\mu \left(A_{n}\cap \bigcup _{i=1}^{n-1}A_{i}\right)\\&{\overset {\text{i.a.}}{=}}\nu \left(A_{n}\right)+\nu \left(\bigcup _{i=1}^{n-1}A_{i}\right)-\mu \left(\bigcup _{i=1}^{n-1}\underbrace {(A_{n}\cap A_{i})} _{\in {\mathcal {G}}}\right)\\&{\overset {\text{i.a.}}{=}}\nu \left(A_{n}\right)+\nu \left(\bigcup _{i=1}^{n-1}A_{i}\right)-\nu \left(\bigcup _{i=1}^{n-1}(A_{n}\cap A_{i})\right)=\nu \left(\bigcup _{i=1}^{n}A_{i}\right).\end{aligned}}}

(with "i.a." meaning "induction assumption") Making use of the cut-stability, we have that ${\displaystyle {\mathcal {G}}}$ is closed under arbitrary finite unions.

Countably infinite unions ${\displaystyle \bigcup _{n\in \mathbb {N} }A_{n}}$ can be made "artificially" disjoint by cutting the preceding ones out of each set: Define

${\displaystyle B_{n}:=A_{n}\setminus (A_{1}\cup \dots \cup A_{n-1})=A_{n}\cap (A_{1}\cup \dots \cup A_{n-1})^{\complement },}$

Then ${\displaystyle \bigcup _{n\in \mathbb {N} }A_{n}=\biguplus _{n\in \mathbb {N} }B_{n}}$ is a disjoint union of sets. If the ${\displaystyle A_{n}}$ are from ${\displaystyle {\mathcal {G}}}$, then so are the ${\displaystyle B_{n}}$, provided that ${\displaystyle {\mathcal {G}}}$ is cut-stable: According to the preceding, the finite union ${\displaystyle A_{1}\cup \dots \cup A_{n-1}}$ lies in ${\displaystyle {\mathcal {G}}}$, and so does the complement of this set (since ${\displaystyle {\mathcal {G}}}$ is a Dynkin system). And by cut-stability also the intersection of it with ${\displaystyle A_{n}}$ lies in ${\displaystyle {\mathcal {G}}}$. Now, ${\displaystyle {\mathcal {G}}}$ is closed under countable disjoint unions (Dynkin system), so also ${\displaystyle \biguplus _{n\in \mathbb {N} }B_{n}=\bigcup _{n\in \mathbb {N} }A_{n}}$ is a good set.

These considerations show: If the Dynkin system ${\displaystyle {\mathcal {G}}}$ is additionally cut stable, it is also closed under arbitrary, at most countable unions, i.e., it is a sigma-algebra. We summarize this in a theorem:

Theorem

Cut stable Dynkin systems are ${\displaystyle \sigma }$-algebras.

Proof

Let ${\displaystyle {\mathcal {D}}\subseteq {\mathcal {P}}(\Omega )}$ be a cut stable Dynkin system. Then ${\displaystyle \Omega \in {\mathcal {D}}}$ and complement stability is given, as we have a Dynkin system. The remaining ${\displaystyle \sigma }$-algebra property is the union stability with respect to countable unions. So far, this is only given for pairwise disjoint unions. We first note that because ${\displaystyle A\setminus B=A\cap B^{\complement }}$, difference stability follows from complement stability and intersection stability of ${\displaystyle {\mathcal {D}}}$.

Now let ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ be any sequence of sets in ${\displaystyle {\mathcal {D}}}$. We define ${\displaystyle (B_{n})_{n\in \mathbb {N} }}$ by ${\displaystyle B_{1}=A_{1}}$, ${\displaystyle B_{n+1}=A_{n+1}\setminus \bigcup _{i=1}^{n}A_{i}}$ for all ${\displaystyle n\in \mathbb {N} }$. Since ${\displaystyle A_{i}\subseteq B_{i}}$ for all ${\displaystyle i\in \mathbb {N} }$ we have that

${\displaystyle \bigcup _{i=1}^{n}B_{i}=\bigcup _{i=1}^{n-1}B_{i}\cup \left(A_{n}\setminus \left(\bigcup _{i=1}^{n-1}A_{i}\right)\right)=\bigcup _{i=1}^{n-1}B_{i}\cup A_{n}.}$

With a simple induction argument it follows that ${\displaystyle \bigcup _{i=1}^{n}B_{i}=\bigcup _{i=1}^{n}A_{i}}$ and finally ${\displaystyle B_{n+1}=A_{n+1}\setminus \left(\bigcup _{i=1}^{n}B_{i}\right)}$.

In particular, it follows from the construction of the sequence ${\displaystyle (B_{n})_{n\in \mathbb {N} }}$ that elements of this sequence are pairwise disjoint. By definition, ${\displaystyle B_{1}=A_{1}\in {\mathcal {D}}}$. Suppose now that ${\displaystyle B_{1},\dots ,B_{n}\in {\mathcal {D}}}$ holds (induction assumption). Then with stability with respect to disjoint unions it follows that ${\displaystyle \bigcup _{i=1}^{n}B_{i}\in {\mathcal {D}}}$. With difference stability, we further have ${\displaystyle B_{n+1}=A_{n+1}\setminus \left(\bigcup _{i=1}^{n}B_{i}\right)\in {\mathcal {D}}}$. Then, by induction, it also holds that ${\displaystyle B_{i}\in {\mathcal {D}}}$ for all ${\displaystyle i\in \mathbb {N} }$.

But since the ${\displaystyle B_{i}}$ were additionally pairwise disjoint, it follows from the disjoint countable union stability that ${\displaystyle \bigcup _{n\in \mathbb {N} }B_{n}\in {\mathcal {D}}}$.

From this we finally get

${\displaystyle \bigcup _{n\in \mathbb {N} }A_{n}=\bigcup _{n\in \mathbb {N} }\bigcup _{i=1}^{n}A_{i}=\bigcup _{n\in \mathbb {N} }\bigcup _{i=1}^{n}B_{i}=\bigcup _{n\in \mathbb {N} }B_{n}\in {\mathcal {D}}.}$

### Intermediate result

We conclude our so far obtained results:

• To ensure that the basic set lies in the set of good sets ${\displaystyle {\mathcal {G}}}$, that is ${\displaystyle \mu (\Omega )=\nu (\Omega )}$, we require that ${\displaystyle {\mathcal {C}}}$ contains an exhaustion ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ of ${\displaystyle \Omega }$.
• To ensure that differences of sets of infinite measure which are subsets of each other, are again in ${\displaystyle {\mathcal {G}}}$, we require, that the exhaustion sets ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ all have finite measure and that for all ${\displaystyle A\in {\mathcal {G}}}$ cuts ${\displaystyle A\cap E_{n}}$ lie again in ${\displaystyle {\mathcal {G}}}$.
• With these two conditions, ${\displaystyle {\mathcal {G}}}$ is already a Dynkin system. For it to be a ${\displaystyle \sigma }$-algebra, ${\displaystyle {\mathcal {G}}}$ should be cut stable, that is, cuts of good sets should be good again.

Except for these conditions and ${\displaystyle \mu |_{\mathcal {C}}=\nu |_{\mathcal {C}}}$, we make no further requirements on ${\displaystyle \mu }$, ${\displaystyle \nu }$, and ${\displaystyle {\mathcal {C}}}$.

Hint

While we had previously required in the second point that for all ${\displaystyle A\in {\mathcal {G}}}$ cuts ${\displaystyle A\cap E_{n}}$ must also be in ${\displaystyle {\mathcal {G}}}$, this is no longer necessary because of the third condition.

Next, we will answer the question, which additional conditions on ${\displaystyle \mu }$, ${\displaystyle \nu }$, or ${\displaystyle {\mathcal {C}}}$ will make ${\displaystyle {\mathcal {G}}}$ cut-stable.

## Cut-stability of the generator

We have found conditions on the measures ${\displaystyle \mu ,\nu }$ and the generator ${\displaystyle {\mathcal {C}}}$ by which the set of good sets ${\displaystyle {\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\mid \mu (A)=\nu (A)\}}$ is a Dynkin system. In particular, ${\displaystyle {\mathcal {G}}}$ thus contains the Dynkin system ${\displaystyle \delta ({\mathcal {C}})}$ generated by ${\displaystyle {\mathcal {C}}}$, since ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {G}}}$. We want ${\displaystyle \sigma ({\mathcal {C}})\subseteq {\mathcal {G}}}$ to hold as well. For this it suffices to find additional conditions under which ${\displaystyle \delta ({\mathcal {C}})}$ is cut-stable: Since every cut-stable Dynkin system is a ${\displaystyle \sigma }$-algebra, it then follows that ${\displaystyle \sigma ({\mathcal {C}})=\delta ({\mathcal {C}})\subseteq {\mathcal {G}}\subseteq \sigma ({\mathcal {C}})}$, and we are done.

So under what conditions is the Dynkin system generated by ${\displaystyle {\mathcal {C}}}$ cut-stable? This apparently depends only on the properties of the set system ${\displaystyle {\mathcal {C}}}$, not on the measures ${\displaystyle \mu }$ or ${\displaystyle \nu }$. In fact, it is sufficient if ${\displaystyle {\mathcal {C}}}$ is cut-stable. This has to do with the fact that the cut operation is compatible with the union and complement operations of a Dynkin system, and thus the cut-stability is inherited from the generator to the generated Dynkin system. We show this in the following theorem.

Theorem

Any Dynkin system generated by a cut-stable set system ${\displaystyle {\mathcal {C}}}$ is itself cut-stable.

Proof

For ${\displaystyle B\in \delta ({\mathcal {C}})}$ we define ${\displaystyle {\mathcal {D}}(B)=\{D\in \delta ({\mathcal {C}})|D\cap B\in \delta ({\mathcal {C}})\}}$, i.e., the set of sets in our Dynkin system that are cut-stable "with respect to ${\displaystyle B}$". By definition, ${\displaystyle {\mathcal {D}}(B)\subseteq \delta ({\mathcal {C}})}$ holds for any ${\displaystyle B\in \delta ({\mathcal {C}})}$. Now we show two things:

1. ${\displaystyle {\mathcal {D}}(B)}$ is a Dynkin system for every ${\displaystyle B\in \delta ({\mathcal {C}})}$.
2. ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {D}}(B)}$ for all ${\displaystyle B\in \delta ({\mathcal {C}})}$.

For once we have shown these statements, it follows that ${\displaystyle {\mathcal {D}}(B)\subseteq \delta ({\mathcal {C}})\subseteq \delta ({\mathcal {D}}(B))={\mathcal {D}}(B)}$, so ${\displaystyle {\mathcal {D}}(B)=\delta ({\mathcal {C}})}$. Here we used monotonicity in the second step and idempotence of the ${\displaystyle \delta }$ operator in the third step. Now if ${\displaystyle A,B\in \delta ({\mathcal {C}})}$ are arbitrary, then because of ${\displaystyle {\mathcal {D}}(B)=\delta ({\mathcal {C}})}$ we know in particular that ${\displaystyle A\in {\mathcal {D}}(B)}$ holds, so ${\displaystyle A\cap B\in \delta ({\mathcal {C}})}$. This shows the desired cut-stability.

So let us now show the necessary statements.

Proof step: ${\displaystyle {\mathcal {D}}(B)}$ is a Dynkin system for all ${\displaystyle B\in \delta ({\mathcal {C}})}$.

Let ${\displaystyle B\in \delta ({\mathcal {C}})}$ be arbitrary. Obviously ${\displaystyle \Omega \cap B=B\in \delta ({\mathcal {C}})}$, so ${\displaystyle \Omega \in {\mathcal {D}}(B)}$.

Let ${\displaystyle A\in {\mathcal {D}}(B)}$ be arbitrary. Then ${\displaystyle A\cap B\in \delta ({\mathcal {C}})}$ holds. Furthermore, because of ${\displaystyle B\in \delta ({\mathcal {C}})}$, of course ${\displaystyle B^{\complement }\in \delta ({\mathcal {C}})}$. From this we conclude

${\displaystyle A^{\complement }\cap B=(A^{\complement }\cup B^{\complement })\cap B=(A\cap B)^{\complement }\cap B=((A\cap B)\uplus B^{\complement })^{\complement }.}$

This, as the complement of a disjoint union of two elements of ${\displaystyle \delta ({\mathcal {C}})}$, is also an element of ${\displaystyle \delta ({\mathcal {C}})}$ and so ${\displaystyle A^{\complement }\in {\mathcal {D}}(B)}$, i.e. ${\displaystyle {\mathcal {D}}(B)}$ is complement stable.

Now let ${\displaystyle (A_{k})_{k\in \mathbb {N} }}$ be a sequence of pairwise disjoint sets in ${\displaystyle {\mathcal {B}}}$. Then for all ${\displaystyle k\in \mathbb {N} }$, it holds that ${\displaystyle A_{k}\cap B\in \delta ({\mathcal {C}})}$. The sequence ${\displaystyle (A_{k}\cap B)_{k\in \mathbb {N} }}$ is also a sequence of pairwise disjoint sets in ${\displaystyle \delta ({\mathcal {C}})}$. From the stability of ${\displaystyle \delta ({\mathcal {C}})}$ under countable disjoint unions, it follows that

${\displaystyle \left(\biguplus _{k\in \mathbb {N} }A_{k}\right)\cap B=\biguplus _{k\in \mathbb {N} }(A_{k}\cap B)\in \delta ({\mathcal {C}}).}$

So ${\displaystyle \bigcup _{k\in \mathbb {N} }A_{k}\in {\mathcal {D}}(B)}$. Thus the three properties of a Dynkin system are satisfied and we are done.

Proof step: ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {D}}(B)}$ for all ${\displaystyle B\in \delta ({\mathcal {C}})}$.

Let ${\displaystyle B\in \delta ({\mathcal {C}})}$ and ${\displaystyle E\in {\mathcal {C}}}$ be arbitrary. Then for all ${\displaystyle D\in {\mathcal {C}}}$, due to the cut-stability of ${\displaystyle {\mathcal {C}}}$, it holds that ${\displaystyle E\cap D\in {\mathcal {C}}\subseteq \delta ({\mathcal {C}})}$. So, in particular, ${\displaystyle D\in {\mathcal {D}}(E)}$. Since ${\displaystyle D\in {\mathcal {C}}}$ was arbitrary, we also have ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {D}}(E)}$. As ${\displaystyle {\mathcal {D}}(E)}$ is a Dynkin system (as shown before), we get ${\displaystyle \delta ({\mathcal {C}})\subseteq \delta ({\mathcal {D}}(E))={\mathcal {D}}(E)}$. Further, ${\displaystyle B\in \delta ({\mathcal {C}})\subseteq {\mathcal {D}}(E)}$ holds. That is, ${\displaystyle B\cap E\in \delta ({\mathcal {C}})}$ and so ${\displaystyle E\in {\mathcal {D}}(B)}$. Since ${\displaystyle E\in {\mathcal {C}}}$ was chosen arbitrary, it finally follows that ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {D}}(B)}$.

Since every average-stable Dynkin system is a ${\displaystyle \sigma }$ algebra, it follows directly:

Theorem

If ${\displaystyle {\mathcal {C}}}$ is a cut-stable set system, then ${\displaystyle \delta ({\mathcal {C}})=\sigma ({\mathcal {C}})}$ holds.

This relation between Dynkin systems and ${\displaystyle \sigma }$-algebras is very useful and simplifies many proofs about measures. This is because for Dynkin systems one can exploit the ${\displaystyle \sigma }$-additivity of the measure, since only disjoint unions need to be considered. In the proof of the uniqueness theorem we will see in a moment a first example where this enables to perform a proof.

### Intermediate result

We summarize the conditions we found to get ${\displaystyle \sigma ({\mathcal {C}})={\mathcal {G}}}$, that is, equality of ${\displaystyle \mu }$ and ${\displaystyle \nu }$ on all of ${\displaystyle \sigma ({\mathcal {C}})}$:

• To ensure that the basic set is in the set of good sets ${\displaystyle {\mathcal {G}}}$, that is, ${\displaystyle \mu (\Omega )=\nu (\Omega )}$, we require that ${\displaystyle {\mathcal {C}}}$ contains an exhaustion ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ of ${\displaystyle \Omega }$.
• To ensure that differences of sets of infinite measure, which are subsets of each other, are again in ${\displaystyle {\mathcal {G}}}$, we require that the exhaustion sets ${\displaystyle (E_{n})_{n\in \mathbb {N} }}$ all have finite measure.
• To ensure that the Dynkin system ${\displaystyle \delta ({\mathcal {C}})}$ generated by ${\displaystyle {\mathcal {C}}}$ is cut-stable, i.e., it is a ${\displaystyle \sigma }$-algebra, we require that the generator ${\displaystyle {\mathcal {C}}}$ must be cut-stable.

In general, one cannot drop the cut- stability of ${\displaystyle {\mathcal {C}}}$ if the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ shall also agree on ${\displaystyle \sigma ({\mathcal {C}})}$, as the following example shows.

Example

Consider the (rather small as it is finite) basic set ${\displaystyle \Omega =\{1,2,3,4\}}$ and the set system ${\displaystyle {\mathcal {C}}=\{\emptyset ,\{1,2\},\{2,3\},\Omega \}}$. Then ${\displaystyle {\mathcal {C}}}$ is not cut-stable, because ${\displaystyle \{2\}=\{1,2\}\cap \{2,3\}\notin {\mathcal {C}}}$. Consider the two (probability) measures

${\displaystyle \mu :={\frac {1}{2}}(\delta _{1}+\delta _{2}),\quad \nu :={\frac {1}{2}}(\delta _{2}+\delta _{4}).}$

Here ${\displaystyle \delta _{x}}$ denotes the Dirac measure, which is defined by.

${\displaystyle \delta _{x}(A):={\begin{cases}1,{\text{ if }}x\in A,\\0{\text{ else.}}\end{cases}}}$

Then ${\displaystyle \mu }$ and ${\displaystyle \nu }$ agree on ${\displaystyle {\mathcal {C}}}$. Moreover, because of ${\displaystyle \Omega \in {\mathcal {C}}}$, there is a finite exhaustion of the basic set. Nevertheless, ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are not equal on the ${\displaystyle \sigma }$-algebra generated by ${\displaystyle {\mathcal {C}}}$ : This ${\displaystyle \sigma }$-algebra contains the set ${\displaystyle \{1,2,3\}=\{1,2\}\cup \{2,3\}}$, and we have ${\displaystyle \mu (\{1,2,3\})=2\neq 1=\nu (\{1,2,3\})}$.

## Uniqueness of measure continuations

We can now formulate and prove the uniqueness theorem.

Theorem (Uniqueness of measure continuations)

Let ${\displaystyle \mu }$ and ${\displaystyle \nu }$ be two measures on the ${\displaystyle \sigma }$-algebra ${\displaystyle {\mathcal {A}}}$ over the basic set ${\displaystyle \Omega }$. Let there be a generator ${\displaystyle {\mathcal {C}}}$ of ${\displaystyle {\mathcal {A}}}$ with the following properties:

• ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide on ${\displaystyle {\mathcal {C}}}$, i.e. ${\displaystyle \mu (C)=\nu (C)}$ for all ${\displaystyle C\in {\mathcal {C}}}$,
• There exists in ${\displaystyle {\mathcal {C}}}$ an exhaustion of ${\displaystyle \Omega }$ with sets of finite measure: a monotonically increasing sequence ${\displaystyle (E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}}$ with limit ${\displaystyle \Omega }$ and ${\displaystyle \mu (E_{n})=\nu (E_{n})<\infty }$,
• ${\displaystyle {\mathcal {C}}}$ is cut-stable, i.e. ${\displaystyle A,B\in {\mathcal {C}}\implies A\cap B\in {\mathcal {C}}}$.

Then ${\displaystyle \mu =\nu }$ holds on all of ${\displaystyle {\mathcal {A}}}$. So in particular a measure ${\displaystyle \mu }$ is then already uniquely determined by the values on ${\displaystyle {\mathcal {C}}}$.

Proof (Uniqueness of measure continuations)

We perform the proof with the "principle of good sets" and define the set system ${\displaystyle {\mathcal {G}}=\{A\in {\mathcal {A}}\mid \mu (A)=\nu (A)\}\subseteq {\mathcal {A}}}$ as containing "good sets". It contains those sets from ${\displaystyle {\mathcal {A}}}$ on which ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide. By assumption, ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {G}}}$ holds. We still have to show that ${\displaystyle {\mathcal {G}}}$ is a ${\displaystyle \sigma }$-algebra. It is enough to show that ${\displaystyle {\mathcal {C}}}$ is a Dynkin system since its generator is cut-stability . Now

${\displaystyle {\mathcal {A}}=\sigma ({\mathcal {C}})=\delta ({\mathcal {C}})\subseteq \delta ({\mathcal {G}})={\mathcal {G}}}$

where we have exploited in the second equality that the Dynkin system generated by a cut-stable set system is already a ${\displaystyle \sigma }$-algebra. We now establish the properties of a Dynkin system for ${\displaystyle {\mathcal {G}}}$ in two steps: First assuming that ${\displaystyle \mu ,\nu }$ are finite measures (i.e. ${\displaystyle \mu (A)<\infty ,\nu (A)<\infty }$ for all ${\displaystyle A\in {\mathcal {A}}}$), then generalizing to the infinite measure case.

Proof step: Proof for finite ${\displaystyle \mu ,\nu }$

We have ${\displaystyle \Omega \in {\mathcal {G}}}$: Let ${\displaystyle (E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}}$ be the exhaustion of ${\displaystyle \Omega }$ from the assumption. For this exhaustion, we have equality ${\displaystyle \mu (E_{n})=\nu (E_{n})}$ for all ${\displaystyle n\in \mathbb {N} }$. Then from the continuity of the two measures ${\displaystyle \mu ,\nu }$ we get ${\displaystyle \mu (\Omega )=\lim _{n\to \infty }\mu (E_{n})=\lim _{n\to \infty }\nu (E_{n})=\nu (\Omega )}$.

${\displaystyle {\mathcal {G}}}$ is complement stable: let ${\displaystyle A\in {\mathcal {G}}}$. Because of ${\displaystyle \mu (\Omega )=\nu (\Omega )<\infty }$ for finite measures, we can exploit the subtractivity and obtain

${\displaystyle \mu (A^{\complement })=\mu (\Omega \setminus A)=\mu (\Omega )-\mu (A)=\nu (\Omega )-\nu (A)=\nu (\Omega \setminus A)=\nu (A^{\complement }).}$

${\displaystyle {\mathcal {G}}}$ is also stable under disjoint unions: Let ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {G}}}$ be a sequence of pairwise disjoint sets in ${\displaystyle {\mathcal {G}}}$. Taking advantage of the ${\displaystyle \sigma }$-additivity of ${\displaystyle \mu }$ and ${\displaystyle \nu }$ we obtain

${\displaystyle \mu \left(\biguplus _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n})=\sum _{n=1}^{\infty }\nu (A_{n})=\nu \left(\biguplus _{n=1}^{\infty }A_{n}\right).}$

Proof step: Proof in the general case

Now, we turn to non-finite measures.

We define for ${\displaystyle n\in \mathbb {N} }$ the dimensions ${\displaystyle \mu _{n},\nu _{n}\colon {\mathcal {A}}\to [0,\infty ]}$ with ${\displaystyle \mu _{n}(A):=\mu (A\cap E_{n})}$ and ${\displaystyle \nu _{n}(A):=\nu (A\cap E_{n})}$, where ${\displaystyle E_{n}\in {\mathcal {C}}}$ are the exhaustion sets from the assumption.

Since by assumption ${\displaystyle {\mathcal {C}}}$ is cut-stable and ${\displaystyle \mu =\nu }$ holds on ${\displaystyle {\mathcal {C}}}$, the measures ${\displaystyle \mu _{n}}$ and ${\displaystyle \nu _{n}}$ also coincide on ${\displaystyle {\mathcal {C}}}$. Furthermore, because of ${\displaystyle \mu _{n}(A)=\mu (A\cap E_{n})\leq \mu (E_{n})<\infty }$ and an analogous statement for ${\displaystyle \nu _{n}}$, the two measures ${\displaystyle \mu _{n}}$ , ${\displaystyle \nu _{n}}$ are finite. So we can apply the statement already proved for the finite case and obtain that ${\displaystyle \mu _{n}=\nu _{n}}$ holds on all ${\displaystyle {\mathcal {A}}}$ for all ${\displaystyle n\in \mathbb {N} }$. The limit transition ${\displaystyle n\to \infty }$ gives that also ${\displaystyle \mu }$ and ${\displaystyle \nu }$ on all ${\displaystyle {\mathcal {A}}}$ are equal.

Hint

This theorem is a good example of the usefulness of Dynkin systems: Because of the disjointness of all unions, we could conveniently exploit the ${\displaystyle \sigma }$-additivity of the measures ${\displaystyle \mu }$ and ${\displaystyle \nu }$ in the proof.

Because of the cut-stability of ${\displaystyle {\mathcal {C}}}$ the sequence of sets ${\displaystyle (E_{n})_{n}}$ exhausting ${\displaystyle \Omega }$ need not be monotonically increasing. It is sufficient to require that there exists a sequence ${\displaystyle (A_{n})_{n}\subseteq {\mathcal {C}}}$ such that ${\displaystyle \Omega =\bigcup _{n=1}^{\infty }A_{n}}$ and ${\displaystyle \mu (A_{n})=\nu (A_{n})<\infty }$ holds: If there exists such a sequence, we can define ${\displaystyle E_{n}=\bigcup _{i=1}^{n}A_{i}}$ and obtain a monotonically increasing sequence with limit ${\displaystyle \Omega }$. Moreover, these sets also satisfy ${\displaystyle \mu (E_{n})=\nu (E_{n})<\infty }$, as we saw in the section on cut-stability: Cut-stability ensures that ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide even on finite (possibly non-disjoint) unions.

One therefore sometimes finds the following formulation of the uniqueness theorem:

Theorem (Uniqueness theorem (alternative version))

Let ${\displaystyle \mu }$ and ${\displaystyle \nu }$ be two measures on the ${\displaystyle \sigma }$-algebra ${\displaystyle {\mathcal {A}}}$ over the basic set ${\displaystyle \Omega }$. Let there be a generator ${\displaystyle {\mathcal {C}}}$ of ${\displaystyle {\mathcal {A}}}$ with the following properties:

• ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide on ${\displaystyle {\mathcal {C}}}$, i.e. ${\displaystyle \mu (C)=\nu (C)}$ for all ${\displaystyle C\in {\mathcal {C}}}$,
• There is a sequence ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}}$ with ${\displaystyle \Omega =\bigcup _{n=1}^{\infty }A_{n}}$ and ${\displaystyle \mu (A_{n})=\nu (A_{n})<\infty }$,
• ${\displaystyle {\mathcal {C}}}$ is cut-stable, i.e. ${\displaystyle A,B\in {\mathcal {C}}\implies A\cap B\in {\mathcal {C}}}$.

Then ${\displaystyle \mu =\nu }$ holds on all of ${\displaystyle {\mathcal {A}}}$.

If ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are probability measures, then the second condition is always automatically satisfied: Because of ${\displaystyle \mu (\Omega )=\nu (\Omega )=1<\infty }$ one can assume without restriction ${\displaystyle \Omega \in {\mathcal {C}}}$ and choose the constant sequence ${\displaystyle (\Omega )_{n\in \mathbb {N} }}$. In probability theory, therefore, one often finds the following version of the uniqueness theorem:

Theorem (Uniqueness theorem for probability measures)

Let ${\displaystyle \mu }$ and ${\displaystyle \nu }$ be two probability measures on the ${\displaystyle \sigma }$-algebra ${\displaystyle {\mathcal {A}}}$ over the basic set ${\displaystyle \Omega }$. Let there be a cut-stable generator ${\displaystyle {\mathcal {C}}}$ of ${\displaystyle {\mathcal {A}}}$ on which ${\displaystyle \mu }$ and ${\displaystyle \nu }$ coincide. Then ${\displaystyle \mu =\nu }$ holds on all of ${\displaystyle {\mathcal {A}}}$.