Let $\mu$ be a finite Borel measure in $\mathbb{R}^N$. Then for all $A\in\mathcal{B(\mathbb{R}}^N)$ we have
$$\mu(A)=\sup\{\mu(F)\;|\; F\subseteq A, F\;\text{closed}\}=\inf\{\mu(V)\;|\; V\supseteq A,\;V\;\text{open}\}\tag1$$
$$\forall \varepsilon >0\; \exists V\;\text{open}, F\;\text{closed}\;\text{such that}\; F\subseteq A \subseteq V\;\text{and} \;\mu(V\setminus F)<\varepsilon\tag2$$
I must prove that $(1)$ and $(2)$ are equivalent.
$(1)\implies (2)$ Fixed $\varepsilon >0$ exists $F\subseteq A$ closed and $V\supseteq A$ open such that $$\mu(A)-\varepsilon <\mu(F)\quad\text{and}\quad \mu(A)+\varepsilon >\mu(V).$$ Since $\mu(V)=\mu(V\setminus F)+\mu(F)$ and $\mu$ is a finite measure we have that $$\mu(V\setminus F)=\mu(V)-\mu(F)<\mu(A)+\varepsilon -\mu(A)+\varepsilon=2\varepsilon.$$
$(2)\implies (1)$ Since for all $\varepsilon > 0$ $$\mu(A)=\mu(F)+\mu(A\setminus F)<\mu(F)+\varepsilon$$ we obtain that $\mu(A)\le \mu(F)$ for all closed set $F\subseteq A$ and therefore $$\mu(A)\le \sup\{\mu(F)\;|\; F\subseteq A,\;F\;\text{closed}\}$$
I can't get the other inequality. Could you give me some hints?