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Let $\mu_1$ and $\mu_2$ be two finite measures defined on $\sigma(\mathcal{F})$ such that, $\forall A \in \mathcal{F}$, $\mu_1(A)=\mu_2(A)$. Show that they must agree on $\sigma(\mathcal{F})$.

I attempted proof of this as follows:

let $\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) | \mu_1(A)=\mu_2(A) \rbrace$. If we show $\mathcal{M}$ is a monotone class, we are done by the Monotone Class Theorem.

To that end, let $\lbrace A_n \rbrace$ be a sequence of non-decreasing sets from $\mathcal{M}$. We "disjointify" the sets : Let $B_k=A_k- \bigcup_{j=1}^{k-1}A_j$. Note that $\lbrace B_i \rbrace$ are pairwise disjoint and $\bigcup_{j=1}^{\infty}A_j=\bigcup_{j=1}^{\infty}B_j$. Now, since $\mu_1$ and $\mu_2$ are countably additive, $$\mu_1(\bigcup\limits_{j=1}^{\infty}B_j)= \sum\limits_{j=1}^{\infty}\mu_1(B_j),$$ $$\mu_2(\bigcup\limits_{j=1}^{\infty}B_j)= \sum\limits_{j=1}^{\infty}\mu_2(B_j).$$ Because $B_i$ is in $\mathcal{F}$, we have $\mu_1(B_i)=\mu_2(B_i), \forall i \in \mathbb{N}$, and hence $$\mu_1(\bigcup\limits_{j=1}^{\infty}B_j)=\mu_2(\bigcup\limits_{j=1}^{\infty}B_j)$$ We conclude that $\mathcal{M}$ is closed under increasing union. Now, to show $\mathcal{M}$ is closed under decreasing limits of sets, Let $\lbrace C_n \rbrace$ be non-increasing sets from $\mathcal{M}.$ We want to show $\lim C_n = \bigcap_{n=1}^{\infty}C_n$ is in $\mathcal{M}$. For this, define $S_k=\mu_1(C_k)$ and $T_k=\mu_2(C_k)$ as sequences of real numbers. Then, $\lim_{k \rightarrow \infty}S_k=\lim_{k \rightarrow \infty}\mu_1(S_k)=\lim_{k \rightarrow \infty}\mu_1(T_k)=\mu_1(\bigcap_{n=1}^{\infty}C_n)=\mu_2(\bigcap_{n=1}^{\infty}C_n)$, using continuity from above of a finite measure. Thus, $\mathcal{M}$ is a monotone class of $\mathcal{F}$ and by monotone class theorem, $\mathcal{M}(\mathcal{F})=\sigma(\mathcal{F})$.

Does this look alright?

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  • $\begingroup$ Check your definition of $\mathcal{M}$. $\endgroup$ – d.k.o. Dec 16 '20 at 9:08
  • $\begingroup$ @d.k.o. I've used the technique of defining so called "good sets" that satisfy equality of measure on $\mathcal{F}$. $\endgroup$ – Avijit Dikey Dec 16 '20 at 9:42
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    $\begingroup$ Using your definition, $\mathcal{M}$ cannot be equal $\sigma(\mathcal{F})$. $\mathcal{M}$ should be a monotone class containing $\mathcal{F}$. $\endgroup$ – d.k.o. Dec 16 '20 at 11:24
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    $\begingroup$ $\mathcal{M}$ is a family of sets in $\sigma(F)$ for which the conclusion holds. $\endgroup$ – d.k.o. Dec 16 '20 at 11:38
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    $\begingroup$ Ah ofcourse! I understand now. Ive changed the definition of $\mathcal{M}$, does the proof hold now? $\endgroup$ – Avijit Dikey Dec 16 '20 at 11:42
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Your proof is almost correct, but it is not OK. You defined $\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) | \mu_1(A)=\mu_2(A) \rbrace$. Then you take $\lbrace A_n \rbrace$ be a sequence of non-decreasing sets from $\mathcal{M}$. But when you "disjointify" the sets, taking
$B_k=A_k- \bigcup_{j=1}^{k-1}A_j$, all you can say is that $B_i$ is in $\sigma(\mathcal{F})$. You can not say $B_i$ is in $\mathcal{F}$ nor $B_i$ is in $\mathcal{M}$.

The way to correct your proof is simple. Here it is:

Let $\mu_1$ and $\mu_2$ be two finite measures defined on $\sigma(\mathcal{F})$ such that, $\forall A \in \mathcal{F}$, $\mu_1(A)=\mu_2(A)$. Show that they must agree on $\sigma(\mathcal{F})$.

Let $\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) \,| \, \mu_1(A)=\mu_2(A) \rbrace$. Let us show that $\mathcal{M}$ is a monotone class.

First, let $\lbrace A_n \rbrace$ be a monotone non-decreasing sequence of sets from $\mathcal{M}$.

So, for all $n$, $\mu_1(A_n) =\mu_2(A_n)$.

Since $\lbrace A_n \rbrace$ is be a sequence of non-decreasing sets in $ \sigma(\mathcal{F})$, for any measure $\nu$ defined on $ \sigma(\mathcal{F})$, we have $$\nu\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\nu(A_n)$$ So applying this to $\mu_1$ and $\mu_2$, we get $$\mu_1\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\mu_1(A_n)=\lim_{n \to \infty}\mu_2(A_n) =\mu_2\left (\bigcup\limits_{n=1}^{\infty}A_n \right) $$ So $\bigcup\limits_{n=1}^{\infty}A_n \in \mathcal{M}$, and we conclude that $\mathcal{M}$ is closed under monotone non-decreasing union.

Now, to show $\mathcal{M}$ is closed under decreasing limits of sets.

Let $\lbrace C_n \rbrace$ be a monotone non-increasing sequence of sets from $\mathcal{M}.$

So, for all $n$, $\mu_1(C_n) =\mu_2(C_n)$.

Using continuity from above of a finite measure, we have $$\mu_1\left (\bigcap\limits_{n=1}^{\infty}C_n \right) = \lim_{n \to \infty}\mu_1(C_n)=\lim_{n \to \infty}\mu_2(C_n) =\mu_2\left (\bigcap\limits_{n=1}^{\infty}C_n \right) $$ So $\bigcap\limits_{n=1}^{\infty}C_n \in \mathcal{M}$, and we conclude that $\mathcal{M}$ is closed under monotone non-increasing intersection.

Thus, $\mathcal{M}$ is a monotone class containing $\mathcal{F}$ and by monotone class theorem, $\mathcal{M} \supseteq \sigma(\mathcal{F})$. So, for all $A\in\sigma(\mathcal{F})$, we have $\mu_1(A)=\mu_2(A)$.

Remark: As defined, $\mathcal{M}$ may be bigger than $\sigma(\mathcal{F})$, but all we need to prove the result is $\mathcal{M} \supseteq \sigma(\mathcal{F})$.

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  • $\begingroup$ I'm not sure why we can conclude $\mu_1\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\mu_1(A_n)=\lim_{n \to \infty}\mu_2(A_n) =\mu_2\left (\bigcup\limits_{n=1}^{\infty}A_n \right)$ without disjointing them first $\endgroup$ – Avijit Dikey Dec 17 '20 at 2:26
  • $\begingroup$ @AvijitDikey , It is a general property that if $\lbrace A_n \rbrace$ is be a sequence of non-decreasing sets in $ \sigma(\mathcal{F})$, for any measure $\nu$ defined on $ \sigma(\mathcal{F})$, we have $$\nu\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\nu(A_n)$$ To prove this general property, you may "disjoint" the $A_n$, but having only one measure in consideration. What you can not do is to "disjoint" the $A_n$ while comparing $\mu_1$ to $\mu_2$. $\endgroup$ – Ramiro Dec 17 '20 at 3:55

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