Proving two measures of Borel sigma-algebra are equal

I am working on this problem on measure theory like this:

Let $X$ be set of $\mathbb R$, and let $\mathcal B$ be its Borel $\sigma$-algebra, and finally let $\mu_1$ and $\mu_2$ be the two measures on $(X,\mathcal B)$ such that $\mu_1((a,b))= \mu_2((a,b)) < \infty$ whenever $−\infty < a < b < \infty$. Show that $\mu_1(A) = \mu_2(A)$ whenever $A \in \mathcal B$.​

Here is what I was at first thinking: Since $a,b \in \mathbb R$ and since $A$ is an arbitrary subset of $\mathcal B$, so if only I can prove that $(a,b) \in \mathcal B$, then I am done. But I was told by a responder to my posting at Physics Forum here that this reasoning is wrong, since not all sets in $\mathcal B$ are open. I am hitting a deadend again.

Therefore I am posting this question here looking for help, thanks for your time and effort.

POST SCRIPT - 1: I should have mentioned this: This problem comes from the 3rd. chapter of an introductory text by Richard F. Bass here, therefore any solution shouldn't involve any advanced theorems such as Dynkin's, etc. Sorry for this belated info, thanks though to all who have taken time to help.

POST SCRIPT - 2: I finally came up with solution without any advanced theorems, adapted from a solution by @JoshKeneda, who used Dynkin's Theorem. I have submitted this work to my professor, he ok'd it except for (5) because it is true only when the $A_i$'s are pairwise disjoint. Feel free to drop me a message if you have ideas to improve (5). Thanks to all and especially to @JoshKeneda.

DISCLOSURE: This question is very similar to an old MSE posting here, which was put on hold due to being incomplete. My posting has all the correction to the first posting. Always conscientious of community rule and guideline, I have tried avoiding duplication by posting this question elsewhere here and here, but I did not receive any meaningful helps $-$ understandably, as those two outside forums are not specialized in math. Thank you for your understanding.

Solution 1:

By Dynkin's $\pi$-$\lambda$ Theorem two finite measures of the same mass that agree on a $\pi$-system agree on the $\sigma$-algebra generated by that $\pi$-system.

To prove this using Dynkin, consider $\Lambda:=\{ B \in \mathcal{B} : \mu_1(B \cap (a,b)) = \mu_2(B \cap (a,b))\}$ is a Dynkin system containing the $\pi$-system of open intervals. Thus $\Lambda$ contains the $\sigma$-algebra generated by open intervals (which is $\mathcal{B}$) and hence $\Lambda = \mathcal{B}$. Thus the restriction of $\mu_1$ and the restriction of $\mu_2$ are identical on any finite interval. An arbitrary set $B$ can be written as the increasing union of $B\cap (-n,n)$ for larger and larger $n$, so the result follows.

Solution 2:

Let $\mathcal{H}$ be the vector space of bounded measurable functions $h$ such that $\int h d\mu_1 = \int h d\mu_2$. (Check this is a vector space, easy). Clearly $1 \in \mathcal{H}$ (though it is not immediate). Suppose that $h_n \geq 0$ and $h_n$ is a sequence in $\mathcal{H}$ such that $h_n$ increases monotonically to the bounded function $h$. Thus $$\int h d\mu_1 = \lim_n \int h_n d\mu_1 = \lim_n\int h_n d\mu_2 = \int h d\mu_2$$ so $h \in \mathcal{H}.$ Since $\mathcal{H}$ contains the indicators of every set in the $\pi$-system $\Pi$ of finite open intervals, the monotone class theorem implies $\mathcal{H}$ contains all bounded $\sigma(\Pi) = \mathcal{B}$ measurable functions, in particular it contains the indicators $1_B$ for all $B \in \mathcal{B}$. Thus $\mu_1(B)=\mu_2(B)$ for all $B \in \mathcal{B}$.

Solution 3:

(1) Let $I := (a, b)$, let $S := \{A \in \mathcal B \mid m(A \cap I) = n(A \cap I)$}. We need to prove that $S$ is in the Borel $\sigma$-algebra.

(2) Let $A = X \in \mathcal B$, $$\begin{align} m(X \cap I ) &= m(I) \\ n(X \cap I ) &= n(I) \\ \text{therefore, } m(X \cap I ) &= n(X \cap I ) \\ \end{align}$$

(3) Let $A = \emptyset \in \mathcal B$, $$\begin{align} m(\emptyset \cap I) &= m(\emptyset) = 0 \\ n(\emptyset \cap I) &= n(\emptyset) = 0 \\ \text{therefore, } m(\emptyset \cap I) &= n(\emptyset \cap I). \end{align}$$

(4) For $\forall A \in \mathcal B, A^c \in \mathcal B$, $$\begin{align} m(A^c \cap I) &= m((X \setminus A) \cap I)\\ &= m((X \cap I) \setminus (A \cap I))\\ &= m(X \cap I) - m(A \cap I)\\ &= m(I) - m(A \cap I)\\ \text{similarly, } n(A^c \cap I) &= n(I) - n(A \cap I)\\ \text{therefore, } m(A^c \cap I) &= n(A^c \cap I).\\ \end{align}$$

(5) Let $(A_i)_{i \in \mathbb N} \in A$, $$\begin{align} m(\bigcup _{i = 1}^{\infty} (A_i \cap I)) &= \sum_{i = 1}^{\infty} m (A_i \cap I)\\ &= \sum_{i = 1}^{\infty} n (A_i \cap I)\\ &= n (\bigcup _{i = 1}^{\infty} (A_i \cap I))\\ \end{align}$$

(6) Similarly, let $(A_i)_{i \in \mathbb N} \in A$, $$\begin{align} m (\bigcap _{i = 1}^{\infty} (A_i \cap I)) &= m ((\bigcap_{i=1}^{\infty}A_i) \cap I)\\ &= n ((\bigcap_{i=1}^{\infty} A_i) \cap I)\\ &= n (\bigcap _{i = 1}^{\infty} (A_i \cap I))\\ \end{align}$$

(7) Having proven that $m(A \cap I) = n(A \cap I)$ is in Borel $\sigma$-algebra, now we need to prove that $m(A) = n(A)$:

$$\begin{align} \lim _{k \to \infty} (m(A \cap (-k, k)) &= \lim _{k \to \infty} (m (A)) \\ &= m(A) \\ \text{similarly, } \lim_{k \to \infty} (n(A \cap (-k, k)) &= \lim _{k \to \infty} (n (A)) \\ &= n (A)\\ \text{hence, } n(A) &= m(B). \qquad \blacksquare \\ \end{align}$$