Show that $m(A)=n(A)$ whenever $A\in\mathcal B$? I'm trying complete this problem from Bass' Real Analysis book:

Let $\mathcal B$ be the Borel $\sigma$-algebra on $\Bbb{R}$, and suppose that $m,n$ are measures on $(\Bbb{R},\cal B)$ such that $m(a,b)=n(a,b)<\infty$ whenever $-\infty<a<b<\infty$. Show that $m(A)=n(A)$ for all $A\in\cal B$.

I'm not really sure what the best way to proceed is. Obviously the result is true if $A$ is open, because then it's just a countable union of disjoint open intervals. However, I'm not sure how to classify what "every" element of $\cal B$ looks like; of course, it includes all closed intervals, and intervals of the form $[a,\infty)$, and so on... a proof by cases seems like it would make sense but I don't know how to be sure I've gotten every case.
Just to be clear, I'd prefer an insightful hint rather than somebody giving me the full solution.
 A: Your question is Exercise 3.9 in Chatper 3 of Bass's Real Analysis. In Chapter 2 of his book (Proposition 2.8), it is shown that $\mathcal{B}$ is generated by the collection
$$
\mathcal{C}=\{(a,b)\mid a,b\in\mathbb{R}\}.
$$
What you need is the monotone class theorem in the very last section of Chapter 2.
Hints: 

  
*
  
*Show that $\mathcal{M}:=\{m(A)=n(A)\mid A\in\mathcal{B}\}$ is a monotone class. 
  
*[Added:] Let $\mathcal{A}$ be the algebra generated by $\mathcal{C}$. Show that $m$ and $n$ agree on $\mathcal{A}$ also and thus $\mathcal{A}\subset\mathcal{M}$.  


[Added later:] It is worth noting that this is a common trick to deal with $\sigma$-algebras: instead of showing directly some statement (in this question $m(A)=n(A)$)  is true on a $\sigma$-algebra, one defines a collection of subsets where the statement holds. Then one shows that this collection coincides with the $\sigma$-algebra.

For the second bullete point of the hints, one can apply the similar idea. Instead of showing directly that $m$ and $n$ agree on the $\mathcal{A}$, one can show that
$\mathcal{M}$
 is an algebra. Then we must have $\mathcal{A}\subset\mathcal{M}$ by the choice of $\mathcal{A}$. 
A: It's hard to show something about every Borel subset by fixing $A\in\mathcal{B}$ because like you noted, an arbitrary Borel set might be quite complicated and we can't just write it down as some convenient union. Instead, it is often more fruitful to consider the collection with the desired property and show that it contains the Borel sets, for which it is sufficient to show that such a collection contains the open sets and is a $\sigma$-algebra (then it must contain the Borel $\sigma$-algebra, which is defined as the smallest $\sigma$-algebra containing all of the open sets). So in this case, fix $\mathcal{A}=\{A\subset\mathbb{R}\mid m(A)=n(A)\}$ and try to show that such a set is a $\sigma$-algebra that contains the open sets. As Jack noted, the monotone class theorem is often useful when you're trying to prove a collection is a $\sigma$-algebra, and you should think about using it here.
