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I have proven the following:

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $\mathcal{A}$ an algebra such that $\sigma(\mathcal{A})=\mathcal{F}$. Then for every $\varepsilon>0$ and $B\in\mathcal{F}$ we got that there is a set $A\in\mathcal{A}$ such that $\mathbb{P}(A\Delta B)\leq \varepsilon$. (here $\Delta$ is the symmetric difference)

I want to show:

Let $\mathbb{P}$ be a probability measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, where $\mathcal{B}(\mathbb{R}^{n})$ is the Borel algebra of subsets of $\mathbb{R}^{n}$. Using the previous fact, show that for any $\varepsilon>0$ and $B\in\mathcal{B}(\mathbb{R}^{n})$, that there is a compact set $A\in\mathcal{B}(\mathbb{R}^{n})$ such that $A\subseteq B$ and $\mathbb{P}(B\setminus A) \leq \varepsilon$.

Any suggestions?

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Hints: consider $\{A\in \mathcal B(\mathbb R^{n}): \text {for every} \epsilon >0 \, \text {there exists a closed set}\, C \subset A \, \text {and an open set} \, U \text {with} \,A\subset U \, \text {with} \, P(U\Delta C) <\epsilon\}$

Verify that this is a sigma algebra which contains all closed sets. [You will need the fact that any closed is an intersection of a decresing sequence of open sets containing it]. Conclude that or any Borel set $A$ and any $\epsilon >0$ there exists a closed set $C\subset A$ such that $P(A\Delta C) <\epsilon\}$. Now use the fact that $C=\cup_n (C\cap \{x:\|x\|\leq n\}$ and $C\cap \{x:\|x\|\leq n\}$ is compact for each $n$.

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