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Let $(\Omega,\mathcal{F},P)$ be a probability space and $\mathcal{A}$ be an algebra of subsets of $\Omega$ such that $\sigma(\mathcal{A})=\mathcal{F}$. Prove that for all $B\in \mathcal{F}$ and for all $\varepsilon >0$ there is $A\in \mathcal{A}$ such that $P(B\bigtriangleup A)<\varepsilon,$ where $B\bigtriangleup A:=(A\setminus B)\cup (B\setminus A)$ is the symmetric difference of $A,~B.$

I have tried proof by contraposition, by assuming the existence of a set $B \in \mathcal{F}$ and an $\varepsilon>0$ such that $P(B\bigtriangleup A)\geq \varepsilon,$ but this doesn't seem to get me somewhere. On the other hand, for a straight proof, the construction of the desired set $A$ seems foggy, since I don't know how to start.

Thanks a lot for the help!

Nikolaos Skout
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    http://math.stackexchange.com/questions/228998/approximating-a-sigma-algebra-by-a-generating-algebra –  Feb 01 '16 at 21:57

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Show first that the family $\mathcal C$ of all sets $B\in \mathcal{F}$ such that for each $\varepsilon >0$ there exists $A\in \mathcal{A}$ with $P(B\bigtriangleup A)<\varepsilon$ is a $\sigma$-algebra. Since $\mathcal C\subset \mathcal F$ and $\mathcal F$ is the $\sigma$-algebra generated by $\mathcal A$, you can conclude that $\mathcal C=\mathcal F$ thus giving the desired property.

John B
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    Thanks for the idea @Jonas! My issue is in the part we show that $\mathcal{C}$ is closed under countable unions: if $(B_n)\subseteq \mathcal{C}$, then for all $n$ and $\varepsilon>0$ there is $A_n\in \mathcal{A}$ such that $P(B_n\vartriangle A_n)<\varepsilon/2^n$ and therefore: $$P(B\vartriangle A)\leq P\bigg(\bigcup_{n=1}^{\infty}B_n\vartriangle A_n\bigg)\leq \sum_{n=1}^{\infty}P(B_n\vartriangle A_n)\leq \varepsilon$$ for $B:=\bigcup_{n=1}^{\infty}B_n\in \mathcal{F},~A:=\bigcup_{n=1}^{\infty}A_n$. Since $\mathcal{A}$ is an algebra (not a sigma algebra), why does $A$ belongs to $\mathcal{A}$? – Nikolaos Skout Feb 01 '16 at 20:51
  • If you assume that $\mu$ restricted to $\mathcal{A}$ is $\sigma$-finite then $A\subset \cup_n A_n$. – user2820579 Jul 20 '22 at 13:33
  • If not the trick is in the book Walsh, Knowing the Odds, AMS, p6. For each $B_n$ one chooses $A_\epsilon = \cup_{i=1}^n A_n$, i.e. $A_\epsilon \in \mathcal{A}$. – user2820579 Jul 20 '22 at 14:37