Let $m^{*}(A)$ denote the outer-measure of a set $A\subset\mathbb{R}$ (w.r.t Lebesgue Measure). Now define $m_1(A)=\inf\{m^{*}(G)\mid G \ \text{is open and} \ A\subseteq G\}$ and $m_2(A)=\sup\{m^{*}(F)\mid F \ \text{is closed and} \ F\subseteq A\}$ .
Claim : $m_1(A)=m^{*}(A)$
Proof : By monotonicity of outer measure, we have $m^{*}(G)\ge m^{*}(A)$ $\forall G\supset A$. If $m^*(A)=\infty$, $m_1(A)=m^*(A)$. So, assume that $m^*(A)<\infty$. Then there exists open set $G$ s.t $G\supset A$ and $m^*(G)<m^*(A)+\epsilon$. Therefore, $m_1(A)\le m^*(A)+\epsilon$. So, we conclude that $m_1(A)=m^*(A)$.
Now, for the 2nd quantity $m_2$, we can say $m_2(A)\le m^*(A)$. Can we derive equality in this case?
No, generally $m_1(A)\ne m_2(A)$. $m_2(A)$ is known as inner measure of $A$. $m_1(A)= m_2(A)$ iff $A$ is Lebesgue measurable.