Let $E$ be a Lebesgue measurable subset of $\mathbb R^n$ and $m$ be the Lebesgue measure (defined via Carathéodory's approach). Show that $$m(E) = \sup\{m(K): K\subseteq E, \mbox{ compact}\}$$
The usual measure is defined $$m(E)=\inf\{\sum l(U_n): E \subseteq \cup U_n\}$$
and so it suffices to show that $$m(E) \leq \sup\{m(K): K\subseteq E, \mbox{ compact}\} \leq m(E)$$
It seems clear to me that $$\sup\{m(K): K\subseteq E, \mbox{ compact}\} \leq m(E)$$ and so the other inequality is all that needs to be proved. However, I am unsure of I would accomplish this as it seems this could (potentially) always force the measure of a set to be smaller than it ``actually'' is. What am I missing?