Theorem. Let (X,B,μ) a finite measure space, where μ is a positive measure. Let A⊂B an algebra generating B.
Then for all B∈B and ε>0, we can find A∈A such that μ(AΔB)=μ(A∪B)−μ(A∩B)<ε.
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 σ-algebra is.
- The map ρ:B×B→R+, ρ(A,A′)=μ(AΔA′) gives a pseudo-metric on B. This makes a link between generating for an algebra and dense for the pseudo-metric.
- We say that a σ-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 Lp(μ) is separable for 1≤p<∞, 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 σ-algebra.
Answer
Proof: Let S:={A∈B∣∀ε>0,∃A′∈A,μ(AΔA′)≤ε}.
We have to prove that S is a σ-algebra, as it contains by definition A.
- X∈S since X∈A.
- If A∈S and ε>0, let A′∈A such that μ(AΔA′)≤ε. Then μ(AcΔA′c)=μ(AΔA′)≤ε and A′c∈A.
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First, we show that A is stable by finite unions. By induction, it is enough to do it for two elements. Let A1,A2∈S and ε>0. We can find A′1,A′2∈A such that μ(AjΔA′j)≤ε/2. As
(A1∪A2)Δ(A′1∪A′2)⊂(A1ΔA′1)∪(A2ΔA′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.
Attribution
Source : Link , Question Author : Davide Giraudo , Answer Author : Davide Giraudo
