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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.

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  • $\begingroup$ Look here: math.stackexchange.com/questions/228998/… $\endgroup$ – NCh Jan 18 at 16:28
  • $\begingroup$ This is the right approach. You can find the proof in the other question. The basic idea is to use continuity of the measure. $\endgroup$ – tomasz Jan 18 at 17:09