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I know that for every Lebesgue measurable set $A$ in $\mathbb{R}$, there exists a Borel set $B$ such that $A\setminus B$ is null.

May I ask is there an analogous statement for $\sigma$-algebra $\mathcal{E}$ and the $\pi$-system that generates $\mathcal{E}$? eg.

Let $(E, \mathcal{E}, \mu)$ be a measure space. If $M$ is a $\pi$-system that generates $\mathcal{E}$, then for every set $A$ in $\mathcal{E}$, there exists a countable union of sets $B_i$ from $M$ such that $\mu(A\setminus \bigcup_{i=1}^{\infty}B_i)=0$.

I always find it hard to do questions that extends properties of a $\pi$-system to the $\sigma$-algebra it generates, this gave the question above. Many thanks in advance!

HIH
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  • Related https://math.stackexchange.com/questions/228998/approximating-a-sigma-algebra-by-a-generating-algebra. The proof in this link also holds if one substitutes the algebra $\mathcal{A}$ with the set of finite unions of elements of the $\pi$-system. This should be enough to answer your question positively. – No-one Sep 13 '23 at 20:36

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