A set $E$ with $m(E) < \infty$ is measurable iff for all $\epsilon > 0$ there exists a finite union $U$ of intervals such that $m(E \triangle U) < \epsilon$.
I am able to prove one way, i.e. $E$ measurable implies $m(E \triangle U) < \epsilon$, but have no insight on how to prove the other way.
6005's comment above is right. I'll give an answer to what I believe you're asking. Let $m$ be Lebesgue outer measure and suppose $m(E \triangle U) < \epsilon/2$. It's sufficient to show that for all $A \subset \mathbb{R}$ we have $$m(A) \geq m(A \cap E) + m(A - E).$$
Note that $$A \cap E = (A \cap E \cap U) \cup ((A \cap E) - U)$$ and $$A-E = (A - (E \cup U)) \cup ((A \cap U) - E).$$
Then, using the above, monotonicity, and the fact that $U$ is measurable, we have $$\begin{align} m(A \cap E) + m(A-E) &\leq m(A \cap E \cap U) + m(A \cap E) - U) + m(A - (E \cup U)) + m((A \cap U) - E) \\ &\leq m(A \cap U) + m(E - U) + m(A - U) + m(U - E) \\ &\leq 2m(E \triangle U) + m(A) \\ &\leq \epsilon + m(A). \end{align}$$
But as $\epsilon > 0$ is arbitrary we can conclude.