Approximating a σ\sigma-algebra by a generating algebra

Theorem. Let (X,B,μ) a finite measure space, where μ is a positive measure. Let AB an algebra generating B.

Then for all BB and ε>0, we can find AA such that μ(AΔB)=μ(AB)μ(AB)<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×BR+, 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 1p<, 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:={ABε>0,AA,μ(AΔA)ε}.
We have to prove that S is a σ-algebra, as it contains by definition A.

  • XS since XA.
  • If AS and ε>0, let AA such that μ(AΔA)ε. Then μ(AcΔAc)=μ(AΔA)ε and AcA.
  • First, we show that A is stable by finite unions. By induction, it is enough to do it for two elements. Let A1,A2S and ε>0. We can find A1,A2A such that μ(AjΔAj)ε/2. As
    (A1A2)Δ(A1A2)(A1ΔA1)(A2ΔA2),
    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

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