Measure of intervals in the Borel sigma-algebra

This is an exercise from a real analysis book that is supposed to help you with entrance exams. I am trying to teach myself.

Suppose $X$ is a set of real numbers, and $B$ is the Boresl $\sigma$-algebra. $m$ and $n$ are two measures on $(X,B)$ such that $m((a,b))=n((a,b))< \infty$ whenever $-\infty<a<b<\infty$. I want to show that $m(A)=n(A)$ for $A\in B$

I have an idea of letting $S$ be a family of sets where the condition of the measures is satisfied, that is $m((a,b))=n((a,b))$. I think showing that S is a $\sigma$-algebra should be enough.

I am having trouble with the complement part because of the interval. I am pretty comfortable working with sets, but the intervals are throwing me off a bit.

Thanks for any input!

Solutions Collecting From Web of "Measure of intervals in the Borel sigma-algebra"

Let $S$ be the set of measurable sets $A$ such that $m(A\cap I)=n(A\cap I)$ for all bounded intervals $I = (a,b)$. We want to show that $S$ contains the Borel $\sigma$-algebra.

We show that $S$ is a Dynkin system. This means that it has to satisfy three properties:

1.) $X \in S$

2.) $S$ is closed under complements

3.) $S$ is closed under countable disjoint unions.

Let $I = (a,b)$.
Clearly $(1)$ is satisfied, since $m(X\cap I) = m(I) = n(I) = n(X\cap I)$.

Let $A\in S$. Then we have $$m(A^c\cap I) = m(I) – m(A\cap I) = n(I) – n(A\cap I) = n(A^c \cap I).$$ So $A^c \in S$.

Finally, if $A_1, A_2, A_3, …$ are disjoint measurable sets in $S$, then $$m(\bigcup_1^\infty A_i \cap I) = \sum_1^\infty m(A_i \cap I) = \sum_1^\infty n(A_i \cap I) = n(\bigcup_1^\infty A_i \cap I).$$

So $S$ is a Dynkin system. It contains the open intervals, since the intersection of any two bounded open intervals is again a bounded open interval (possibly degenerate) on which $m$ and $n$ agree. This also implies that open intervals form a $\pi$-system. Thus, by the Dynkin $\pi$-$\lambda$ theorem, $S$ contains the Borel $\sigma$-algebra. Thus, for any set $A$ in the Borel $\sigma$-algebra, we have:
$$m(A) = \lim_{j \rightarrow \infty} m(A \cap (-j, j)) = \lim_{j \rightarrow \infty} n(A \cap (-j, j)) = n(A).$$