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Theorem. Let $(X,\mathcal B,\mu)$ a finite measure space, where $\mu$ is a positive measure. Let $\mathcal A\subset \mathcal B$ an algebra generating $\cal B$.

Then for all $B\in\cal B$ and $\varepsilon>0$, we can find $A\in\cal A$ such that $$\mu(A\Delta B)=\mu(A\cup B)-\mu(A\cap B)<\varepsilon.$$

I don't think there is a proof in this site.

It's a useful result for several reasons:

  • We know what the algebra generated by a collection of sets is, but not what the generated $\sigma$-algebra is.
  • The map $\rho\colon \cal B\times\cal B\to \Bbb R_+$, $\rho(A,A')=\mu(A\Delta A')$ gives a pseudo-metric on $\cal B$. This makes a link between generating for an algebra and dense for the pseudo-metric.
  • We say that a $\sigma$-algebra is separable if it's generated by a countable class of sets. In this case, the algebra generated by this class is countable. An with the mentioned result, we can show that $L^p(\mu)$ is separable for $1\leq p<\infty$, which makes a link between the two notions.
  • In ergodic theory, we have to test mixing conditions only an a generating algebra, not on all the $\sigma$-algebra.
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up vote 18 down vote accepted

Proof: Let $$\mathcal S:=\left\{A\in \mathcal{B}\mid \forall\varepsilon>0,\exists A'\in\mathcal A,\mu(A\Delta A')\leq \varepsilon\right\}.$$ We have to prove that $\cal S$ is a $\sigma$-algebra, as it contains by definition $\cal A$.

  • $X\in\cal S$ since $X\in\cal A$.
  • If $A\in\cal S$ and $\varepsilon>0$, let $A'\in\cal A$ such that $\mu(A\Delta A')\leq \varepsilon$. Then $\mu(A^c\Delta A'^c)=\mu(A\Delta A')\leq \varepsilon$ and $A'^c\in\cal A$.
  • First, we show that $\cal A$ is stable by finite unions. By induction, it is enough to do it for two elements. Let $A_1,A_2\in\cal S$ and $\varepsilon>0$. We can find $A'_1,A'_2\in\cal A$ such that $\mu(A_j\Delta A'_j)\leq \varepsilon/2$. As $$(A_1\cup A_2)\Delta (A'_1\cup A'_2)\subset (A_1\Delta A'_1)\cup (A_2\Delta A'_2),$$ and $A'_1\cup A'_2\in\cal A$, $A_1\cup A_2\in \cal A$.

    Now, let $\{A_k\}\subset\cal S$ pairwise disjoint and $\varepsilon>0$. For each $k$, let $A'_k\in\cal A$ such that $\mu(A_k\Delta A'_k)\leq \varepsilon 2^{-k}$.
    Let $N$ be such that $\mu\left(\bigcup_{j\geq N+1}A_j\right)\leq \varepsilon/2$ (such a choice is possible since $\bigcup_{j\geqslant 1}A_j$ has a finite measure and $\mu\left(\bigcup_{j\geq n+1}A_j\right)\leq \sum_{j\geq n+1}\mu\left(A_j\right)$ and this can be made arbitrarily small). Let $A':=\bigcup_{j=1}^NA'_j\in\cal A$. As $$\left(\bigcup_{k\geq 1}A_k\right)\Delta A'\subset \bigcup_{j=1}^N(A_j\Delta A'_j)\cup\bigcup_{k\geq N+1}A_k,$$ and we conclude by sub-additivity.

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If you put $\mathcal{S}' = \{A \in \mathcal{A}|\,\mu(A) = \infty\} \cup \mathcal{S}$, then $\mathcal{S}'$ is a $\sigma$-algebra even when the measure is not finite. Then you get that either $\mu(A) = \infty$ or $A$ can be "approximated" by some set in $\mathcal{A}$. Right? – André Caldas Sep 19 '13 at 16:08
Sorry. I think I missed the "complement" part $A^c \in \mathcal{S}'$, so my comment above is probably not right. – André Caldas Sep 19 '13 at 16:18
Did you find any textbook reference for this result? – Ilya Dec 2 '13 at 9:53
I guess the result is contained in Bogachev's book (as a theorem or an exercise). – Davide Giraudo Dec 2 '13 at 10:36
FWIW, for future references it seems to be Exercise 4.7.63 of the first volume. – Ilya Jul 11 '14 at 11:46

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