# Criteria of finite Lebesgue measurability2

### The problem:

Let $E$ be a Lebesgue measurable subset of $\mathbb{R}^d$ with finite measure. To show that $\forall \epsilon>0$, $\exists$ a compact set $V$ such that $m(E \setminus V) \leq \epsilon$.

### My approach:

$E$ is Lebesgue measurable. Fix $\epsilon > 0$. Now we can employ the inner approximation by closed set criterion for Lebesgue measurability to assert that $\exists$ a closed set $F$, contained in $E$, such that $m(E \setminus F) \leq \frac{\epsilon}{3}$.

Now if $E$ is bounded, so is $F$ and we are done. So let us assume that $E$ is unbounded, so that $F$ is not necessarily bounded. By monotonicity, $m(F) \leq m(E) < \infty$.