Prove that the two measures are the same

Question

Suppose $\mu$ and $\nu$ are positive measures on the Borel $\sigma$-algebra on $[0, 1]$ such that

$\int fd\mu=\int fd\nu$

whenever $f$ is real-valued and continuous on $[0, 1]$. Prove that $\mu=\nu$.

I have no idea how to deal with this question. How should I do?

Answer 1

As you probabely know we have that $$\mu(A) = \int 1_{A} d\mu$$ For any measure $\mu$ and for all $A$. This almost proves it, exept for the fact that this indicator function might not be continuous. However, if A is an open interval you can immediatly see that the indicator is the monotone limit of continuous functions. Using the the monotone convergence theorem we now have the result for every element of the semi ring of half open intervals, since the integral is the same for half open or open intervals. The value of the measure on the semi ring uniquely determines it on the sigma algebra it generates, which is the borel algebra so we are done.

Edit: We do need the measures to by $\sigma$-finite for the uniqueness theorem to apply

Answer 2

Using increasing limits of bump functions that tend to the indicator function of intervals, prove that the measures coincide on intervals.

Once they coincide on intervals they coincide on open sets by $\sigma$-additivity.

The Borel $\sigma$-algebra is generated by the open sets.