# Extending measure on Field to $\sigma$-Field

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?

• Check your definition of $\mathcal{M}$. – d.k.o. Dec 16 '20 at 9:08
• @d.k.o. I've used the technique of defining so called "good sets" that satisfy equality of measure on $\mathcal{F}$. – Avijit Dikey Dec 16 '20 at 9:42
• Using your definition, $\mathcal{M}$ cannot be equal $\sigma(\mathcal{F})$. $\mathcal{M}$ should be a monotone class containing $\mathcal{F}$. – d.k.o. Dec 16 '20 at 11:24
• $\mathcal{M}$ is a family of sets in $\sigma(F)$ for which the conclusion holds. – d.k.o. Dec 16 '20 at 11:38
• Ah ofcourse! I understand now. Ive changed the definition of $\mathcal{M}$, does the proof hold now? – Avijit Dikey Dec 16 '20 at 11:42

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 $$\mathcal{M} \supseteq \sigma(\mathcal{F})$$.

• 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 – Avijit Dikey Dec 17 '20 at 2:26
• @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 $\mu_2$. – Ramiro Dec 17 '20 at 3:55