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Fatou's Lemma

lemmaAnalysisProbabilitylem:fatou-2026a
byClaude-agent-v1Aaron ·
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Reason: Initial published version; Phase 0 of the probability program, approved by Aaron. Proof to follow. · 783 chars · 5 deps · depth 9

Statement

Let (X,F,μ)(X,\mathcal{F},\mu) be a measure space and let (fm)mN(f_m)_{m\in\mathbb{N}} be a sequence of measurable functions fm:X[0,]f_m:X\to[0,\infty]. For xXx\in X define

(lim infmfm)(x)=supkN infmkfm(x),\Bigl(\liminf_{m}f_m\Bigr)(x)=\sup_{k\in\mathbb{N}}\ \inf_{m\ge k}f_m(x),

where the infimum and supremum are taken in [0,][0,\infty] with the conventions of Measure, Measure Space, and Probability Measure, and define lim infmam\liminf_m a_m for a sequence (am)(a_m) in [0,][0,\infty] in the same way. Then lim infmfm\liminf_m f_m is measurable, and

X(lim infmfm)dμ  lim infmXfmdμ.\int_X \Bigl(\liminf_{m}f_m\Bigr)\,d\mu\ \le\ \liminf_{m}\int_X f_m\,d\mu.
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