# Pre-measures and measures – Serlo

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In this article we learn about ${\displaystyle \sigma }$-additivity of volumes and see how it can be used to characterize the continuity of volumes on rings. We call a volume with this property a pre-measure and thus define a notion central to measure theory: measures on ${\displaystyle \sigma }$-algebras.

In the previous article we learned about continuous volumes. Intuitively, we took a volume to be continuous if it allows the volume of a set to be measured by approximation. Based on this reasoning, we came up with a formal definition for the continuity of a volume. The following simpler formulation is equivalent to this, as we have seen:

Definition (Continuous volume)

A volume ${\displaystyle \mu :{\mathcal {R}}\to [0,\infty ]}$ on a ring ${\displaystyle {\mathcal {R}}}$ is called continuous if for every increasing set sequence ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}}$ with limit ${\displaystyle A\in {\mathcal {R}}}$ it holds that: ${\displaystyle \lim _{n\to \infty }\mu (A_{n})=\mu (A).}$

Approximation of simple measurable sets

The advantage of continuity is that one can determine the volume of a complicated set by approximation with sets that are easier to measure. But in order to be able to measure quantities by approximation, one must first know whether the volume is really continuous. And because we have defined continuity above exactly by this approximation property, we would first have to check for all set sequences whether the volumes of the sets really approximate the volume of the limit. Which is what we wanted to use. So we are essentially going around in circles

It would be nice to have a different characterization of continuity. Perhaps we can find one that resembles additivity, which is present for volumes anyway.

### Definition of ${\displaystyle \sigma }$-additivity

In the following, let ${\displaystyle \mu }$ be a volume on a ring ${\displaystyle {\mathcal {R}}}$. We know that for pairwise disjoint sets ${\displaystyle B_{1},\dots ,B_{n}\in {\mathcal {R}}}$ , by additivity we have that

${\displaystyle \mu \left(\biguplus _{i=1}^{n}B_{i}\right)=\sum _{i=1}^{n}\mu (B_{i}).}$

Suppose ${\displaystyle \mu }$ is continuous. An infinite series is simply a limit of a sequence of finite sums, and we guess how additivity can be generalized for continuous volumes: Let ${\displaystyle (B_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}}$ be a sequence of pairwise disjoint sets in ${\displaystyle {\mathcal {R}}}$ such that their union ${\displaystyle A:=\biguplus _{i=1}^{\infty }B_{i}}$ also lies in the ring ${\displaystyle {\mathcal {R}}}$. Then the sets ${\displaystyle A_{n}:=\biguplus _{i=1}^{n}B_{i}}$ form an increasing set sequence with limit ${\displaystyle A\in {\mathcal {R}}}$. From the assumption that ${\displaystyle \mu }$ is continuous, it follows that

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

For a continuous volume ${\displaystyle \mu }$ the additivity is valid also for unions of countably infinitely many disjoint sets. This is of course subject to the union of the infinitely many disjoint sets being again in the domain of definition of ${\displaystyle \mu }$. Volumes satisfying this property are called ${\displaystyle \sigma }$-additive, i.e. "countably additive":

Definition (${\displaystyle \sigma }$-additive volume on a ring)

Let ${\displaystyle {\mathcal {R}}}$ be a ring and ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}}$ a sequence of pairwise disjoint sets whose union ${\displaystyle \biguplus _{n=1}^{\infty }A_{n}}$ lies also within the ring ${\displaystyle {\mathcal {R}}}$. A volume ${\displaystyle \mu :{\mathcal {R}}\to [0,\infty ]}$ is called ${\displaystyle \sigma }$-additive if ${\displaystyle \mu \left(\biguplus _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n}).}$

Hint

It is not excluded that the series on the right side of the equation diverges, i.e. has the value infinity.

### Characterization of continuity (on rings)

We have seen that continuous volumes on rings are ${\displaystyle \sigma }$-additive. Let us recall our original goal: to find an alternative characterization of continuity. We want to investigate whether ${\displaystyle \sigma }$-additivity is suitable as such a characterization.

So now let ${\displaystyle \mu }$ be a ${\displaystyle \sigma }$-additive volume on a ring ${\displaystyle {\mathcal {R}}.}$ Let further ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}}$ be a monotonically increasing set sequence whose limit ${\displaystyle A:=\bigcup _{n=1}^{\infty }A_{n}}$ is again in ${\displaystyle {\mathcal {R}}}$. Let us try to prove the continuity of ${\displaystyle \mu }$, i.e., the property

${\displaystyle \lim _{n\to \infty }\mu (A_{n})=\mu (A).}$

In order to exploit the ${\displaystyle \sigma }$-additivity, we need to transform the sequence of (not necessarily pairwise disjoint) ${\displaystyle A_{n}}$ into a sequence of pairwise disjoint sets whose union is also equal to ${\displaystyle A}$. To do this, we take each ${\displaystyle A_{n}}$ of the sequence and cut out the part already contained in the previous sequence members: define the sets

${\displaystyle B_{n}:=A_{n}\setminus \bigcup _{i=1}^{n-1}A_{i}=A_{n}\setminus A_{n-1},\quad B_{1}=\emptyset .}$

Since rings are stable under taking set differences, the sequence of pairwise disjoint ${\displaystyle B_{n}}$ is also in ${\displaystyle {\mathcal {R}}}$. Further, ${\displaystyle A_{n}=\biguplus _{i=1}^{n}B_{i}}$ and hence ${\displaystyle \biguplus _{i=1}^{\infty }B_{i}=\bigcup _{i=1}^{\infty }A_{i}=A}$ holds. So we have that

${\displaystyle \lim _{n\to \infty }\mu (A_{n})=\lim _{n\to \infty }\mu \left(\biguplus _{i=1}^{n}B_{i}\right)=\lim _{n\to \infty }\sum _{i=1}^{n}\mu (B_{i})=\sum _{i=1}^{\infty }\mu (B_{i}){\overset {(*)}{=}}\mu \left(\biguplus _{i=1}^{\infty }B_{i}\right)=\mu \left(\bigcup _{i=1}^{\infty }A_{i}\right)=\mu (A),}$

where in ${\displaystyle (*)}$ we have exploited the assumption that ${\displaystyle \mu }$ is a ${\displaystyle \sigma }$-additive volume.

Overall, our considerations show that for volumes on rings continuity and ${\displaystyle \sigma }$-additivity are equivalent. We have thus found an alternative characterization of continuity closely related to additivity:

Theorem (Equivalence of continuity and ${\displaystyle \sigma }$-additivity on rings)

For a volume ${\displaystyle \mu }$ on a ring ${\displaystyle {\mathcal {R}}}$ , the following two statements are equivalent:

1. ${\displaystyle \mu }$ is continuous,
2. ${\displaystyle \mu }$ is ${\displaystyle \sigma }$-additive.

Proof (Equivalence of continuity and ${\displaystyle \sigma }$-additivity on rings)

${\displaystyle 1\Longrightarrow 2}$: Let ${\displaystyle \mu }$ be continuous and let ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}}$ be a sequence of pairwise disjoint sets with ${\displaystyle \biguplus _{n=1}^{\infty }A_{n}\in {\mathcal {R}}}$. The sets ${\displaystyle B_{n}:=\biguplus _{i=1}^{n}A_{i}}$ lie in ${\displaystyle {\mathcal {R}}}$ and form a monotonically growing set sequence. With the continuity of ${\displaystyle \mu }$ , we get

${\displaystyle \mu \left(\biguplus _{n=1}^{\infty }A_{n}\right)=\mu \left(\bigcup _{n=1}^{\infty }B_{n}\right)=\mu (\lim _{n\to \infty }B_{n})=\lim _{n\to \infty }\mu (B_{n})=\lim _{n\to \infty }\mu \left(\biguplus _{i=1}^{n}A_{i}\right)=\lim _{n\to \infty }\sum _{i=1}^{n}\mu (A_{i})=\sum _{i=1}^{\infty }\mu (A_{i}).}$

${\displaystyle 1\Longleftarrow 2}$: Let ${\displaystyle \mu }$ be ${\displaystyle \sigma }$-additive and let ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}}$ be a monotonically growing set sequence with limit ${\displaystyle A:=\bigcup _{n\in \mathbb {N} }A_{n}\in {\mathcal {R}}}$. Define ${\displaystyle B_{n}:=A_{n}\setminus \bigcup _{i=1}^{n-1}A_{i}\in {\mathcal {R}}}$ for ${\displaystyle n\geq 2}$ and ${\displaystyle B_{1}=\emptyset }$. We have that ${\displaystyle A_{n}=\biguplus _{i=1}^{n}B_{i}}$ and therefore also ${\displaystyle \biguplus _{i=1}^{\infty }B_{i}=\bigcup _{i=1}^{\infty }A_{i}=A\in {\mathcal {R}}}$. It follows, using the ${\displaystyle \sigma }$ additivity of ${\displaystyle \mu }$, that

${\displaystyle \lim _{n\to \infty }\mu (A_{n})=\lim _{n\to \infty }\mu \left(\biguplus _{i=1}^{n}B_{i}\right)=\lim _{n\to \infty }\sum _{i=1}^{n}\mu (B_{i})=\sum _{i=1}^{\infty }\mu (B_{i})=\mu \left(\biguplus _{i=1}^{\infty }B_{i}\right)=\mu \left(\bigcup _{i=1}^{\infty }A_{i}\right)=\mu (A).}$

So ${\displaystyle \mu }$ is continuous.

Warning

Sometimes volumes are considered on domains of definition other than rings (such as so-called semi-rings). But for the equivalence of continuity and ${\displaystyle \sigma }$-additivity it is important that the volume is really defined on a ring: In the proof it is needed that the domain of definition is closed under set differences and finite unions.

## Examples

Wir erinnern zunächst an ein Beispiel aus dem Artikel über stetige Inhalte. Dort betrachten wir die Grundmenge ${\displaystyle \mathbb {N} }$ und den Inhalt ${\displaystyle \mu :{\mathcal {P}}(\mathbb {N} )\to [0,\infty ]}$, der von einer beliebigen Teilmenge der natürlichen Zahlen bestimmt, ob sie endlich oder unendlich ist:

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

Der Inhalt wurde als unstetig erkannt, da die Bedingung der Stetigkeit für die aufsteigende Mengenfolge der Mengen ${\displaystyle A_{n}=\{1,\dots ,n\}}$ mit Grenzwert ${\displaystyle A=\mathbb {N} }$ nicht erfüllt ist. Tatsächlich ist er auch nicht ${\displaystyle \sigma }$-additiv. Ein Gegenbeispiel sind die paarweise disjunkten Mengen ${\displaystyle B_{n}=\{n\}}$, die man wie oben durch Bilden der Differenzen ${\displaystyle A_{n}\setminus A_{n-1}}$ aus den ${\displaystyle A_{n}}$ gewinnen kann. Für diese gilt

${\displaystyle \sum _{n=1}^{\infty }\mu (B_{n})=\sum _{n=1}^{\infty }\mu (\{n\})=0\neq \infty =\mu \left(\biguplus _{n=1}^{\infty }\{n\}\right)=\mu (\mathbb {N} ).}$

Ein Beispiel für einen ${\displaystyle \sigma }$-additiven (und also stetigen) Inhalt auf einem Ring ist dagegen der Inhalt mit ${\displaystyle \mu (A)=|A|}$, ebenfalls auf der Potenzmenge ${\displaystyle {\mathcal {P}}(\mathbb {N} )}$ definiert, der die Anzahl der Elemente einer Teilmenge von ${\displaystyle \mathbb {N} }$ bestimmt. (Dieser wurde hier genauer behandelt.) Es ist offenkundig, dass dieser Inhalt ${\displaystyle \sigma }$-additiv ist: Sind ${\displaystyle B_{1},B_{2},\dots \subseteq \mathbb {N} }$ paarweise disjunkt, so gilt natürlich ${\displaystyle \left\vert \biguplus _{n=1}^{\infty }B_{n}\right\vert =\sum _{n=1}^{\infty }|B_{n}|.}$

Genauso ist natürlich jeder stetige Inhalt ${\displaystyle \sigma }$-additiv, wie unsere Überlegungen im vorherigen Abschnitt gezeigt haben. Beispiele für stetige Inhalte haben wir im Artikel zu stetigen Inhalten gesehen.

## Pre-measures

For volumes satisfying the useful ${\displaystyle \sigma }$-additivity, we will use a separate term:

A pre-measure is a ${\displaystyle \sigma }$-additive and hence special continuous volume

Definition (Pre-measure)

A function on sets ${\displaystyle \mu :{\mathcal {R}}\to [0,\infty ]}$ on a ${\displaystyle \sigma }$-ring ${\displaystyle {\mathcal {R}}}$ is called a premeasure if for all sequences of pairwise disjoint sets ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ in ${\displaystyle {\mathcal {R}}}$ it holds that:

1. ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=\sum _{n\in \mathbb {N} }\mu (A_{n})}$
2. ${\displaystyle \mu (\emptyset )=0}$.

Hint

One can also define the notion of a pre-measure in general on rings, then one simply requires that ${\displaystyle \mu }$ is a ${\displaystyle \sigma }$-additive volume. That is one requires ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=\sum _{n\in \mathbb {N} }\mu (A_{n})}$ only for sequences of pairwise disjoint sets whose union is contained in ${\displaystyle {\mathcal {R}}}$ (on ${\displaystyle \sigma }$-rings this is always the case).

Every pre-measure is also a volume. The non-negativity as well as ${\displaystyle \mu (\emptyset )=0}$ holds by definition, the finite additivity we get from the ${\displaystyle \sigma }$-additivity by choosing all ${\displaystyle A_{n}=\emptyset }$ starting from a certain index.

For volumes, as shown in the sigma-additivity section, the equivalence between continuity and ${\displaystyle \sigma }$-additivity holds. Because ${\displaystyle \sigma }$-additive volumes are even pre-measures, a volume is continuous if and only if it is a pre-measure.

## Sigma-algebras and measures

We defined what a pre-measure is and thus characterized (on rings) continuity of volumes alternatively. As a natural domain of definition of a continuous volume we had learned ${\displaystyle \sigma }$-rings, since they are rings which additionally contain the limits of monotone set sequences.

### Definition: ${\displaystyle \sigma }$-algebra

Let ${\displaystyle {\mathcal {R}}}$ be a ${\displaystyle \sigma }$-ring. It is useful to require that the basic set be "measurable", i.e., ${\displaystyle \Omega \in {\mathcal {R}}}$. This is important, for example, in probability theory, where ${\displaystyle \Omega }$ is the certain event. Moreover, with ${\displaystyle \Omega \in {\mathcal {R}}}$ we obtain directly the complement stability via the difference stability of rings, which is often useful (e.g. counter-events in probability theory).

A ${\displaystyle \sigma }$-algebra is a very special set system

Definition (${\displaystyle \sigma }$-algebra)

A ${\displaystyle \sigma }$-ring ${\displaystyle {\mathcal {A}}}$ with ${\displaystyle \Omega \in {\mathcal {A}}}$ is called ${\displaystyle \sigma }$-algebra. That means, a ${\displaystyle \sigma }$-algebra must satisfy

1. ${\displaystyle \Omega \in {\mathcal {A}}}$
2. ${\displaystyle A,B\in {\mathcal {A}}\implies A\setminus B\in {\mathcal {A}}}$
3. ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ : sequence of sets in ${\displaystyle {\mathcal {A}}}$ ${\displaystyle \implies }$ ${\displaystyle \bigcup _{n\in \mathbb {N} }A_{n}\in {\mathcal {A}}}$

So basically, a ${\displaystyle \sigma }$-algebra is a kind of "larger version of a ${\displaystyle \sigma }$-ring", as a ${\displaystyle \sigma }$-algebra must always contain the larges possible set ${\displaystyle \Omega }$, while a ${\displaystyle \sigma }$-ring not necessarily has to.

There is another common and equivalent definition of a ${\displaystyle \sigma }$-algebra, which is often easier to verify in practice.

Definition (${\displaystyle \sigma }$-algebra)

A set system ${\displaystyle {\mathcal {A}}\subseteq {\mathcal {P}}(\Omega )}$ with

1. ${\displaystyle \Omega \in {\mathcal {A}}}$
2. ${\displaystyle A\in {\mathcal {A}}\implies A^{\complement }\in {\mathcal {A}}}$
3. ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ : sequence of sets in ${\displaystyle {\mathcal {A}}}$ ${\displaystyle \implies }$ ${\displaystyle \bigcup _{n\in \mathbb {N} }A_{n}\in {\mathcal {A}}}$

is called a ${\displaystyle \sigma }$-algebra.

Theorem (Both definitions of a ${\displaystyle \sigma }$-algebra are equivalent)

The definitions agree in two out of three points. We now show the equivalence of the third points.

1. ${\displaystyle \Longrightarrow :}$ Let ${\displaystyle {\mathcal {A}}}$ be a ${\displaystyle \sigma }$-algebra according to the first definition. Then by complement stability and by ${\displaystyle \Omega \in {\mathcal {A}}}$ we have for all ${\displaystyle A\in {\mathcal {A}}}$ that ${\displaystyle A^{\complement }=\Omega \setminus A\in {\mathcal {A}}}$. Thus ${\displaystyle {\mathcal {A}}}$ is a ${\displaystyle \sigma }$-algebra in the sense of the second definition.
2. ${\displaystyle \Longleftarrow :}$ Let ${\displaystyle {\mathcal {A}}}$ be a ${\displaystyle \sigma }$-algebra according to the second definition. Let ${\displaystyle A,B\in {\mathcal {A}}}$. Then, because of complement stability and union stability, ${\displaystyle A\setminus B=(B\cup A^{\complement })^{\complement }\in {\mathcal {A}}}$. Thus ${\displaystyle {\mathcal {A}}}$ is also a ${\displaystyle \sigma }$-algebra in the sense of the first definition.

So the definitions are equivalent.

### Definition: measure, measurable space, measure space

The crucial property of a pre-measure is its additivity with respect to countable disjoint unions as long as the countable disjoint union is again contained in the ring. For ${\displaystyle \sigma }$-algebras this is always the case. So we don't have to check the property, which makes our life a lot easier. hence, pre-measures on ${\displaystyle \sigma }$-algebras are easy to handle and appear a lot in mathematics. They have an own name: "measures".

The notion of a "measure" is actually the dominant one in mathematics. "Pre-measure" can rather be seen as generalizations of it, which are defined on "smaller" rings or ${\displaystyle \sigma }$-rings and are then extended to "larger" ${\displaystyle \sigma }$-algebras.

A measure is a special kind of pre-measure - and at the same time that function on sets with the most desirable properties.

Definition (Measure)

A pre-measure ${\displaystyle \mu :{\mathcal {A}}\to \mathbb {R} }$ is called measure if ${\displaystyle {\mathcal {A}}}$ is a ${\displaystyle \sigma }$-algebra.

Definition (measurable space and measure space)

If ${\displaystyle {\mathcal {A}}}$ is a ${\displaystyle \sigma }$-algebra with basic set ${\displaystyle \Omega }$, we call ${\displaystyle (\Omega ,{\mathcal {A}})}$ a measurable space.

Moreover, if ${\displaystyle \mu }$ is a measure on ${\displaystyle {\mathcal {A}}}$, we call ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ a measure space.

A special case of measures are the so-called probability measures. For a set of elementary results ${\displaystyle \Omega }$, they assign to each event ${\displaystyle A\subseteq \Omega }$ the probability that an outcome of a random experiment lies in ${\displaystyle A}$. In this notion, the certain event (${\displaystyle x\in \Omega }$) should have probability ${\displaystyle 1}$. So we say that a measure ${\displaystyle \mu }$ is a probability measure, if and only if ${\displaystyle \mu (\Omega )=1}$:

Definition (Probability space and probability measure)

A measure space ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ with ${\displaystyle \mu (\Omega )=1}$ is called a probability space. ${\displaystyle \mu }$ is then called probability measure. The elements of the ${\displaystyle \sigma }$-algebra ${\displaystyle {\mathcal {A}}}$ are called events.

At this point, we see why it is crucial to consider measures in Probability theory and not just pre-measures: A pre-measure is only defined on a ring or a ${\displaystyle \sigma }$-ring, and this ring does not contain ${\displaystyle \Omega }$. So the statement ${\displaystyle \mu (\Omega )=1}$ might make no sense, if we are given only a "pre-measure"!

## Examples for measures

We now consider a few examples of measures on ${\displaystyle \sigma }$-algebras.

The first three examples are more or less trivial. Here, let ${\displaystyle \Omega }$ be an arbitrary basic set and ${\displaystyle {\mathcal {A}}}$ be a ${\displaystyle \sigma }$-algebra over ${\displaystyle \Omega }$.

Example (Zero measure)

Let ${\displaystyle \mu \colon {\mathcal {A}}\to [0,\infty ]}$ with ${\displaystyle \mu (A)=0}$ for all ${\displaystyle A\in {\mathcal {A}}}$. Then ${\displaystyle \mu }$ is obviously a measure and ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ is a measure space. We call ${\displaystyle \mu }$ the zero measure.

Example

Another trivial example is given by

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

Again, it is clear that this is a measure: By definition, ${\displaystyle \mu (\emptyset )=0}$. Let ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {A}}}$ be a sequence of disjoint sets. Then ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=\sum _{n\in \mathbb {N} }\mu (A_{n})}$: If ${\displaystyle A_{n}=\emptyset }$ holds for each ${\displaystyle n}$, then ${\displaystyle \mu (\emptyset )=0}$ is on the left and a sum of zeros is on the right, so equality holds. If at least one of the ${\displaystyle A_{n}}$ is nonempty, then ${\displaystyle \infty }$ stands on both sides of the equation.

The next example can also be considered for any basic set ${\displaystyle \Omega }$, but is only of interest if it is overcountable.

Example

Let ${\displaystyle \mu \colon {\mathcal {A}}\to [0,\infty ]}$ be defined by

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

It is easy to see, as in the previous example, that ${\displaystyle \mu }$ is a measure: obviously ${\displaystyle \mu (\emptyset )=0}$ holds. Let ${\displaystyle (A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {A}}}$ be a sequence of disjoint sets. Then ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=\sum _{n\in \mathbb {N} }\mu (A_{n})}$: If ${\displaystyle A_{n}}$ is countable for every ${\displaystyle n}$, then ${\displaystyle \biguplus _{n\in \mathbb {N} }A_{n}}$ as a countable union of countable sets is itself countable. Consequently, ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=0}$ is on the left and a sum of zeros is on the right, so equality holds. If at least one of the ${\displaystyle A_{n}}$ is overcountable, then we have ${\displaystyle \infty }$ on both sides of the equation.

The following examples are a bit less trivial:

Example (A volume on a finite basic set always generates a measure space)

Let ${\displaystyle \mu :{\mathcal {P}}(\Omega )\to [0,\infty ]}$ be a volume over a finite basic set ${\displaystyle \Omega }$ defined on the whole power set ${\displaystyle {\mathcal {P}}(\Omega )}$. Then ${\displaystyle (\Omega ,{\mathcal {P}}(\Omega ),\mu )}$ is a measure space: It is clear that ${\displaystyle {\mathcal {P}}(\Omega )}$ is a ${\displaystyle \sigma }$-algebra. We show that a volume ${\displaystyle \mu }$ over a finite basic set is already a measure. What we have to prove for this is ${\displaystyle \sigma }$-additivity: let ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ be a sequence of pairwise disjoint sets in ${\displaystyle {\mathcal {P}}(\Omega )}$. From the finiteness of ${\displaystyle \Omega }$ it follows that ${\displaystyle A_{n}=\emptyset }$ for all but finitely many ${\displaystyle n\in \mathbb {N} }$.

Let ${\displaystyle k\in \mathbb {N} }$ be ${\displaystyle A_{n}=\emptyset }$ for all ${\displaystyle n\geq k}$. Then ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=\mu \left(\biguplus _{n=1}^{k}A_{n}\right)=\sum _{n=1}^{k}\mu (A_{n})=\sum _{n\in \mathbb {N} }\mu (A_{n})}$. Thus ${\displaystyle \mu }$ is a ${\displaystyle \sigma }$-additive volume on a ${\displaystyle \sigma }$-algebra, that is, a measure.

Example (Counting-measure over ${\displaystyle \mathbb {N} }$)

Consider the measurable space ${\displaystyle (\mathbb {N} ,{\mathcal {P}}(N))}$ and ${\displaystyle \mu :{\mathcal {P}}(N)\to \mathbb {R} \cup \{\infty \}}$, ${\displaystyle \mu (A)=|A|}$. Then ${\displaystyle \mu }$ is a measure.

Obviously, ${\displaystyle \mu (\emptyset )=0}$. What remains to be shown is the ${\displaystyle \sigma }$-additivity. Let ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ be a sequence of pairwise disjoint sets in ${\displaystyle {\mathcal {P}}(\mathbb {N} )}$. We distinguish two cases:

1. We finitely often have ${\displaystyle A_{n}\neq \emptyset }$. In that case it suffices to see that ${\displaystyle \mu }$ is a volume on ${\displaystyle {\mathcal {P}}(\mathbb {N} )}$, as shown in this example.
2. We infinitely often have ${\displaystyle A_{n}\neq \emptyset }$. In that case, it follows from the disjointness of ${\displaystyle A_{n}}$ that ${\displaystyle \left\vert \biguplus _{n\in \mathbb {N} }A_{n}\right\vert =\infty }$ holds. Therefore, we have ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=\infty =\sum _{n\in \mathbb {N} ,A_{n}\neq \emptyset }1\leq \sum _{n\in \mathbb {N} ,A_{n}\neq \emptyset }\mu (A_{n})\leq \sum _{n\in \mathbb {N} }\mu (A_{n})}$, so ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=\infty =\sum _{n\in \mathbb {N} }\mu (A_{n})}$.

Thus the ${\displaystyle \sigma }$ additivity holds in both cases, so ${\displaystyle \mu }$ is indeed a measure.

Example (Dirac measure)

Let ${\displaystyle (\Omega ,{\mathcal {A}})}$ be a measurable space, ${\displaystyle x\in \Omega }$. Then ${\displaystyle \mu :\Omega \to \mathbb {R} ,\mu (A)={\begin{cases}1{\text{, if }}x\in A\\0{\text{ if }}x\notin A\end{cases}}}$is called the Dirac measure. It is easy to see that ${\displaystyle \mu }$ is a measure. Obviously ${\displaystyle \mu (\emptyset )=0}$. Now, let ${\displaystyle (A_{n})_{n\in \mathbb {N} }}$ be any sequence of pairwise disjoint sets in ${\displaystyle {\mathcal {A}}}$, then there are two possibilities.

1. ${\displaystyle x}$ lies in exactly one set ${\displaystyle A_{k}}$. In that case ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=1=1+0=\mu (A_{k})+\sum _{n\in \mathbb {N} ,n\neq k}\mu (A_{n})=\sum _{n\in \mathbb {N} }\mu (A_{n})}$.
2. ${\displaystyle x\notin A_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$. Then ${\displaystyle \mu \left(\biguplus _{n\in \mathbb {N} }A_{n}\right)=0=\sum _{n\in \mathbb {N} }0=\sum _{n\in \mathbb {N} }\mu (A_{n}).}$

Thus the ${\displaystyle \sigma }$ additivity holds and ${\displaystyle \mu }$ is indeed a measure.

We will get to know more interesting examples when we have taken a closer look at the construction of measures. For now, we don't even know if there is a ${\displaystyle \sigma }$-algebra over ${\displaystyle \mathbb {R} }$ containing the intervals ${\displaystyle [a,b]}$ and on which the elementary geometric length ${\displaystyle \lambda ([a,b])=b-a}$ is a measure.