Measure is equal to Lebesgue measure

Question

Let $\mathscr{M} \subseteq \mathscr{P}(\mathbb{R})$ be the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$, and suppose that $\mu : \mathscr{M} \rightarrow [0, \infty]$ is a measure on $(\mathbb{R}, \mathscr{M})$ with the property that $\mu(I) = \ell(I)$ for all intervals $I$. Prove that $\mu = m$, where $m$ denotes Lebesgue measure.

The question also provides two useful hints: show separately the inequalities $m \leq \mu$ and $\mu \leq m$, and to use the characterisation of Lebesgue measurable sets proven in an earlier question, namely that $E \subseteq \mathbb{R}$ is measurable if and only if for every $\epsilon>0$ there is an open set $U \supseteq E$ such that $m(U \setminus E) \leq \epsilon$.

I'm kind of stuck and can't seem to make any progress, so any help would be great! Thanks.

Answer 1

Any open set in $\mathbb R$ is a countable disjoint union of open intervals. Hence $\mu (U)=m(U)$ for any open set $U$. Now $\mu (E) \leq \mu (U)=m(U) \leq m(E)+\epsilon$. Let $\epsilon \to 0$ to get $\mu (E) \leq m(E)$. The reverse inequality is proved by a similar argument.

Answer 2

$HINT$

$\mu(I)=\int_Id\mu(x)=\int1_I(x)d\mu(x)=\int 1_I(x)dm(x)=\int_Idm(x)=m(I)$ for every interval $I$

Use the fact that step functions are dense on the class of simple functions and then the density of simple functions to $L^1$ functions of the form $f(x)=1_A$ where $A$ is a measurable subset if the real line.