Proof using Monotone Class Theorem

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 $$(a, b) \in \mathbb R^1 \rightarrow (a, b) \in \mathscr M$$?

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.

You don't even need the monotone class theorem (that is at least as advanced as Dynkin-stuff).

First extend the property of $\mu_1$ and $\mu_2$ coinciding, to half-open intervals $(a,b]$ as follows:

Pick $c>b$. Then use the fact that $\mu_1(a,c) = \mu_2(a,c)$ and $\mu_1(b,c) = \mu_2(b,c)$, together with additivity of measures to get the result. Then do this with the other kind of half-open intervals

Using continuity of measures, you can also prove the same holds for open and closed rays, since a ray is just an increasing union of intervals.

Now make an algebra $\mathscr A_0$ (closed under unions and complements) out of your choice of basic intervals, which generates the Borel $\sigma$-algebra. In other words, we are applying the standard argument for showing that all Borel sets satisfy some property:

1) Show that the collection $\mathscr{B}_+$ of Borel sets on which $\mu_1=\mu_2$ is a $\sigma$-algebra.

2) You showed that $\mathscr A_0$ belongs to $\mathscr{B}_+$, hence $$\mathscr{B} = \sigma(\mathscr{A}_0) \subset \mathscr{B}_+$$

• Thank you, let me take a look. Thanks again. Feb 9, 2015 at 16:57
• FYI after I posted this question twice, you are the only one responding with a simple solution without advanced theorems. I am very, very interested in understanding your solution and eventually adopt it into my assignment. Is there anyway for you to elaborate your solution into a step-by-step analysis, as I am just learning the rope in this course? Thank you for your time. Feb 10, 2015 at 2:07