# Let $F$ be an algebra. Prove that $\forall A\in\sigma(F), \varepsilon>0, \exists B\in F\mid\mathbb P((A\setminus B)\cup(B\setminus A))< \varepsilon$. [duplicate]

I tried approaching the problem by defining $$S = \{A\in\sigma(F) : \forall\varepsilon >0 \exists B\in F\mid\mathbb P((A\setminus B)\cup(B\setminus A))<\varepsilon\}$$ and then showing that $$S$$ itself is a $$\sigma$$-algebra that contains $$F$$ and thus it must be $$\sigma(F)$$.

I was able to show that $$\Omega\in S$$ and that $$\forall A\in S, \overline{A}\in S$$. And that $$F\subseteq S$$. But I got stuck trying to show that $$S$$ is closed under countable union. I would love help in proving the last part of if someone has another approach, I would love to hear it.

• – NCh Jan 18 at 16:28
• This is the right approach. You can find the proof in the other question. The basic idea is to use continuity of the measure. – tomasz Jan 18 at 17:09