Serlo: EN: Uniqueness of a continuation

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We derive conditions under which a measure on a $\sigma$-algebra is uniquely determined by the values on a generator. On our way, we will learn about Dynkin systems and their relation to $\sigma$-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 $\mathrm{\Omega }$ enter the $\sigma$-algebra, as possible (while sustaining some nice properties). Often, one requires additional conditions, for instance when constructing a geometric volume, that cuboids in ${ℝ}^n$ 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 $\sigma$-algebras where it can be defined on.

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

Using the existence of a continuation - theorem we know when a function $\mu :𝒞\to [0,\mathrm{\infty }]$ can be continued to a measure on the $\sigma$-algebra $\sigma (𝒞)$ generated by $\sigma$. 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 $\mu$ to a measure on $\sigma (𝒞)$? In other words, we are interested in whether a measure $\mu$ on the $\sigma$-algebra $\sigma (𝒞)$ is already uniquely determined by its values on the smaller set system $𝒞$.

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

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 $𝒞$ and the two measures $\mu$ and $\nu$ under which uniqueness holds. For this we consider the system of "good sets"

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

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

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

Suppose one can only make statements about a generator of $ℳ$, perhaps because $ℳ$ can only be characterized in terms of the generator. An example is the Borel $\sigma$-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 $ℳ$.

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

• $𝒢$ is a $\sigma$-algebra.
• $𝒢$ contains a generator $𝒞$ of $ℳ$.

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

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

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

Existence of an inner approximation

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

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

Idea and definition of an inner approximation

We know that the measures $\mu$ and $\nu$ coincide on the sets in $𝒞$. The idea is to cover the basic set with at most countably many sets from $E_1,E_2,\dots \in 𝒞$, i.e. their union should be $\bigcup _{n=1}^{\mathrm{\infty }}{E}_n=\mathrm{\Omega }$.

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

• If the $E_n$ are pairwise disjoint, we can use the $\sigma$-additivity of measures: $\mu (\mathrm{\Omega })=\mu \left(\underset{i=1}{\overset{\mathrm{\infty }}{⨄}}{E}_i\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (E_i)=\sum _{n=1}^{\mathrm{\infty }}\nu (E_i)=\nu \left(\underset{i=1}{\overset{\mathrm{\infty }}{⨄}}{E}_i\right)=\nu (\mathrm{\Omega })$
• If they are contained in each other in ascending order (i.e. $E_1\subseteq E_2\subseteq \dots$), they form a monotonic set sequence with limit $\bigcup _{i=1}^{\mathrm{\infty }}{E}_i=\mathrm{\Omega }$. Then we can use the continuity of the measures $\mu$ and $\nu$: $\mu (\mathrm{\Omega })=\mu (\underset{n\to \mathrm{\infty }}{\lim }{E}_n)=\underset{n\to \mathrm{\infty }}{\lim }\mu (E_i)=\underset{n\to \mathrm{\infty }}{\lim }\nu (E_i)=\nu (\underset{n\to \mathrm{\infty }}{\lim }{E}_n)=\nu (\mathrm{\Omega })$

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

Mathe für Nicht-Freaks: Vorlage:Definition Mathe für Nicht-Freaks: Vorlage:Beispiel Since measures are continuous, íf $𝒞$ contains an exhaustion $(E_n{)}_{n\in \mathbb{N}}$ of $\mathrm{\Omega }$, then we have $\mu (\mathrm{\Omega })=\underset{n\to \mathrm{\infty }}{\lim }\mu (E_n)=\underset{n\to \mathrm{\infty }}{\lim }\nu (E_n)=\nu (\mathrm{\Omega })$.

Intermediate result

To conclude, w have found the following first condition on the set system $𝒞$:

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

The inner approximating sets have finite measure

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

Unions

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

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

So the disjoint union is also uniquely measurable and lies again in $𝒢$.

If we additionally exploit the $\sigma$-additivity of measures, then we even know the measure of countably infinite unions of disjoint sets. Given a sequence of pairwise disjoint sets $A_1,A_2,\dots \in 𝒢$, the $\sigma$-additivity of measures $\mu$ and $\nu$ implies

Set differences

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

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

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

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

Differences of sets with infinite measure

A way out of this problem is to approximate $A$ and $B$ by ascending sequences $(A_n{)}_{n\in \mathbb{N}},(B_n{)}_{n\in \mathbb{N}}$ 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 $A_n\setminus B_n$ as above, since $\mu (A_n)$ and $\mu (B_n)$ are both finite. But we have to be careful: For this to work, the subset relation $A_n\subseteq B_n$ must also hold for the set sequences $(A_n{)}_{n\in \mathbb{N}},(B_n{)}_{n\in \mathbb{N}}$ for all $n\in \mathbb{N}$. So we cannot just choose the sequences $(A_n{)}_{n\in \mathbb{N}}$ and $(B_n{)}_{n\in \mathbb{N}}$ in an arbitrary manner. They need to grow "equally fast" and at the same time approximate $A$ and $B$ equally fast.

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

Then the sets $A_n:=A\cap E_n$ also form an increasing set sequence with limit

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

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

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

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

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

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

The following examples shows that the sets $E_n$ of the exhaustion indeed need to have finite measure in order to get uniqueness.

Intermediate result

We conclude what has been found out:

• The "set of good sets" $𝒢$ 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 $𝒢$, that is, $\mu (\mathrm{\Omega })=\nu (\mathrm{\Omega })$, we require that $𝒞$ contains an exhaustion $(E_n{)}_{n\in \mathbb{N}}$ of $\mathrm{\Omega }$.
• In order for differences of sets of infinite measure, which are subsets of each other, are in $𝒢$, we require, that the exhaustion sets $(E_n{)}_{n\in \mathbb{N}}$ all have finite measure and that for all $A\in 𝒢$ the intersections $A\cap E_n$ lie again in $𝒢$.

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

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

Dynkin systems

As before, let $𝒢=\{A\in \sigma (𝒞)\mid \mu (A)=\nu (A)\}$ be the system of good sets on which the two measures $\mu$ and $\nu$ 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 $\mathrm{\Omega }$: this is guaranteed by the exhaustion $(E_n{)}_{n\in \mathbb{N}}\subseteq 𝒞$.
• Unions of finitely or countably infinitely many pairwise disjoint sets in $𝒢$: this holds because of $\sigma$-additivity and continuity of the measures $\mu$ and $\nu$.
• Differences of sets from $𝒢$, where one is contained in the other: this is guaranteed by finiteness of the exhaustion and by the condition that intersections of sets from $𝒢$ with sets of the exhaustion are again in $𝒞$.

Thus we can already characterize the set system $𝒢$ 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

Relationship between different set systems.

Mathe für Nicht-Freaks: Vorlage:Definition

An equivalent characterization of a Dynkin system is the following.

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

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

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

Examples

Here are a few examples of Dynkin systems:

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

Motivation for cut-stability

We considered conditions under which the set system $𝒢=\{A\in \sigma (𝒞)\mid \mu (A)=\nu (A)\}$ 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 $\mu$ and $\nu$ coincide on the generator $𝒞$, $𝒞\subseteq 𝒢$ holds. Since $𝒢$ is itself a Dynkin system, therefore the Dynkin system $\delta (𝒞)$ generated by $𝒞$ is also contained in $𝒢$.

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

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

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

If either set has infinite measure, the equality $\mu (A\cup B)=\mathrm{\infty }=\nu (A\cup B)$ 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":

Is the cut-stability of the system of good sets $𝒢$ in addition to our previous conditions already enough for it to be an $\sigma$-algebra? Suppose $𝒢$ is cut-stable. The previous reasoning for two sets can be extended by induction to any finite union of good sets: Let $A_1,A_2,\dots ,A_n\in \sigma (𝒞)$ be good sets, i.e., $\mu (A_i)=\nu (A_i)$ for all $i=1,\dots ,n$. We also assume again that all $A_i$ have finite measure (otherwise the equality holds anyway). Then

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

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

Then $\bigcup _{n\in \mathbb{N}}{A}_n=\underset{n\in \mathbb{N}}{⨄}{B}_n$ is a disjoint union of sets. If the $A_n$ are from $𝒢$, then so are the $B_n$, provided that $𝒢$ is cut-stable: According to the preceding, the finite union $A_1\cup \dots \cup A_{n-1}$ lies in $𝒢$, and so does the complement of this set (since $𝒢$ is a Dynkin system). And by cut-stability also the intersection of it with $A_n$ lies in $𝒢$. Now, $𝒢$ is closed under countable disjoint unions (Dynkin system), so also $\underset{n\in \mathbb{N}}{⨄}{B}_n=\bigcup _{n\in \mathbb{N}}{A}_n$ is a good set.

These considerations show: If the Dynkin system $𝒢$ 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:

Intermediate result

We conclude our so far obtained results:

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

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

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

Cut-stability of the generator

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

So under what conditions is the Dynkin system generated by $𝒞$ cut-stable? This apparently depends only on the properties of the set system $𝒞$, not on the measures $\mu$ or $\nu$. In fact, it is sufficient if $𝒞$ 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.

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

This relation between Dynkin systems and $\sigma$-algebras is very useful and simplifies many proofs about measures. This is because for Dynkin systems one can exploit the $\sigma$-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 $\sigma (𝒞)=𝒢$, that is, equality of $\mu$ and $\nu$ on all of $\sigma (𝒞)$:

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

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

Uniqueness of measure continuations

We can now formulate and prove the uniqueness theorem.

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

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

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

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