Approximation of probability measure on the real line with borel sigma algebra. I came across this problem and could help where I'm stuck (details to follow)

Prove that if $\mathbf{P}$ is a probability measure on $(\mathbf{R}, \mathcal{B})$ then for any
  borel set $A$ and for any $\epsilon > 0$, there is an open set $G \subseteq \mathbf{R}$ such that $G \supseteq A$ and $\mathbf{P}(A) + \epsilon >
\mathbf{P}(G)$.

Notation: $\mathcal{B}$ is the borel $\sigma$-algebra on $\mathbf{R}$. 
What I've done/where I'm stuck:


*

*I see that this problem is equivalent to showing that 
$$
\mathbf{P}(A) = \inf\{\mathbf{P}(G) : G \supseteq A, G \text{ is open}  \}.  
$$

*One approach I thought could be fruitful is to show that 
$$ \mathcal{C} = \{ 
A \in \mathcal{B} : 
\mathbf{P}(A) = \inf\{\mathbf{P}(G) \mid G \supseteq A, G \text{ is open}  \} \}$$
is a $\sigma$-algebra and $\mathcal{B} \subseteq \mathcal{C}$. 

*It is clear that if $A$ is open that $A \in \mathcal{C}$ since by monotonicity, for any set we have $\mathbf{P}(A) \leq\inf\{\mathbf{P}(G) \mid G \supseteq A, G \text{ is open}  \} $. Additionally taking $G = A$ implies $\mathbf{P}(A) \geq \inf  \{\mathbf{P}(G) \mid G \supseteq A, G \text{ is open}  \}$. 

*The point above means that all it remains to show is that $\mathcal{C}$ is indeed a $\sigma$-algebra since $\mathcal{B}$ is the smallest $\sigma$-algebra containing all the open sets. 

*$\mathbf{R}$ is open, hence $\mathbf{R} \in \mathcal{C}$. 

*Where I'm stuck! : I'm trying to show that $\mathcal{C}$ is closed under complementation. I'm
having trouble with this, and could use help. 

 A: As suggested by KaviRamaMurthy, let $\mathcal{C}$ be the collection of Borel sets that can be approximated from above by open sets and approximated below by closed sets.
Precisely, we mean that $A$ is in $\mathcal{C}$ if and only if for each $\epsilon > 0$,


*

*${\bf P}(A) > {\bf P}(G) - \epsilon$ for some open $G \supseteq A$ and

*${\bf P}(A) < {\bf P}(F) + \epsilon$ for some closed $F \subseteq A$.


Now, establish the following facts:


*

*$\mathcal{C}$ is an algebra (easy).

*$\mathcal{C}$ contains a collection that generates the Borel sets (easy).

*$\mathcal{C}$ is a monotone class (proof below).


From the points above, we can conclude that $\mathcal{C}$ is the collection of Borel sets, as desired.
Proof that $\mathcal{C}$ is a monotone class: Let $A_1 \subseteq A_2 \subseteq \cdots$ be an increasing sequence in $\mathcal{C}$. Let $\epsilon > 0$. For each $j$, we can find an open set $G_j \supseteq A_j$ such that ${\bf P}(A_j) > {\bf P}(G_j) - \epsilon$. Therefore, $${\bf P}(\cup_j A_j) = \lim_j {\bf P}(A_j) \geq \lim_j {\bf P}(G_j) - \epsilon = {\bf P}(\cup_j G_j).$$ Note, in particular, that $\cup_j G_j$ is open and contains $\cup_j A_j$. Therefore, $\cup_j A_j$ is in $\mathcal{C}$. Similarly, we can show that if $B_1 \supseteq B_2 \supseteq \cdots$ is a decreasing sequence in $\mathcal{C}$, then $\cap_j B_j$ is in $\mathcal{C}$. $\square$
