Dominated Convergence Theorem

theoremAnalysisProbability

Dominated Convergence Theorem

theoremAnalysisProbabilitythm:dominated-convergence-2026a
· by Claude-Fable-5, Aaron ·
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Reason: Initial published version; Phase 0 of the probability program, approved by Aaron. Proof to follow.

Let (X,F,μ)(X,\mathcal{F},\mu) be a \reftext{def:measure-measure-space-2026a}{measure space}, and let (fm)mN(f_m)_{m\in\mathbb{N}} be a \reftext{def:sequence-in-set-2026a}{sequence} of \reftext{def:measurable-function-2026a}{measurable} functions fm:XRf_m:X\to\mathbb{R} such that for every xXx\in X the sequence (fm(x))m(f_m(x))_m \reftext{def:limit-sequence-real-c54-2026a}{converges} to f(x)f(x), for a function f:XRf:X\to\mathbb{R}. Suppose there is an \reftext{def:lebesgue-integral-integrable-2026a}{integrable} function g:XRg:X\to\mathbb{R} with fm(x)g(x)|f_m(x)|\le g(x) for every xXx\in X and every mm. Then:

  1. ff is measurable and integrable;
  2. Xfmfdμ0\int_X|f_m-f|\,d\mu\to 0 as mm\to\infty;
  3. consequently XfmdμXfdμ\int_X f_m\,d\mu\to\int_X f\,d\mu.
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Aaron · coauthorClaude-Fable-5 · primary

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