As you know I have been grappling with this question since days ago, which I copy down here for convenience:
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 $A \in \mathcal B$.
As you know again that this question comes from early chapters of Richard F. Bass's introductory book, therefore any solution should involve no advanced theorems such as Dynkin's. The consensus I got so far is to use the Monotone Class Theorem:
Suppose $\mathscr A_0$ is an algebra, $\mathscr A$ is the smallest $\sigma$-algebra containing $\mathscr A_0$, and $\mathscr M$ is the smallest monotone class containing $\mathscr A_0$, then $\mathscr A = \mathscr M$.
Here are my three questions:
(1) My proof idea (naive, perhaps) is to show that, $$\begin{align} \text{if } A \in \mathcal B &\text{ then } A \in \mathscr A \\ \text{if } (a, b) \in \mathbb R^1 &\text{ then } (a, b) \in \mathscr M. \\ \end{align}$$
Since by the Monotone Theorem $\mathscr A = \mathscr M$, therefore I can arrive at the result. Is this idea so flawed that it is dead on arrival? (Who knows, since breakthroughs sometimes happen when someone asks a really dumb question. :-) )
(2) If my idea is valid, how do you prove that if $A \in \mathcal B \rightarrow A \in \mathscr A$? I mean the mechanic of going from $A \in \mathcal B$ to $ A \in \mathscr A$?
(3) And finally, what is the mechanics of proving that if $(a, b) \in \mathbb R^1 \rightarrow (a, b) \in \mathscr M$?
Thank you for your time.
POST SCRIPT! POST SCRIPT!: I finally came up with solution without any advanced theorems, even the Monotone Theorem, adapted from a solution by @JoshKeneda, who had used Dynkin's Theorem. My solution is posted here at the answer page down below. 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.