exists $A \in \mathcal{F}$ such that $\mu(B\triangle A) < \epsilon$ Let $\mu$ be a probability measure on $(S, \mathcal{S})$, where $\mathcal{S} = \sigma(\mathcal{F})$ for a field $\mathcal{F}$. How do I go about showing that for each $B \in \mathcal{S}$ and $\epsilon > 0$ there exists $A \in \mathcal{F}$ such that $\mu(B \triangle A) < \epsilon$? I am totally stuck, any help would be appreciated.
 A: Here is a hint. Define$$\mathcal{L} = \{B \in \sigma(\mathcal{F})\text{ }|\text{ for all }\epsilon > 0\text{ there exists }A \in \mathcal{F}\text{ such that }\mu(B \triangle A) < \epsilon\}.$$Show that the above class is a $\sigma$-field. Use the Dynkin $\pi-\lambda$ Theorem.

In the interests of completeness, here is a complete solution. Clearly $\emptyset \in \mathcal{L}$, where $\mathcal{L}$ is as defined above. Moreover, $\mathcal{L}$ is closed under complement since $B^c\Delta A^c = B \Delta A$. Now consider $B = \bigcup_n B_n$ where $B_n \in \mathcal{L}$ for all $n$. Given $\epsilon > 0$, take $N \in \mathbb{N}$ such that $\mu\left(B \setminus \bigcup_{n \le N} B_n\right) \le {{\epsilon}\over{2}}$. Then for $n \le N$, take $A_n$ such that $\mu(B_n \Delta A_n) < {{\epsilon}\over{2N}}$. Since $\bigcup_{n < N} B_n \Delta \bigcup_{n < N} A_n \subset \bigcup_{n \le N}(B_n \Delta A_n)$. We have then $\mu(B \Delta \bigcup_{n \le N} A_n) < \epsilon$. Therefore, $\mathcal{L}$ is a $\sigma$-field containing $\mathcal{L}$, which permits to conclude.
We remark that it is also possible to appeal to the notion of outer measure to solve the problem.
