# Approximate Borel set by compact set

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?

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 $$n$$.