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  "body": "\n## Statement\n\nLet $(X, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]] with $\\mu(X) \u003c \\infty$.\nLet $f_1, f_2, \\ldots$ and $f$ be [[measurable functions|YYNCJSR0PJ]] from $X$ to $\\R$ such that $f_n(x) \\to f(x)$ for [[almost every|RARTMA0ZEC]] $x \\in X$.\nThen for every $\\varepsilon \u003e 0$ there is a set $E \\in \\Sigma$ with $\\mu(E) \u003c \\varepsilon$ such that $f_n \\to f$ [[uniformly|08VZ57PT26]] on $X \\setminus E$.\n\n## Remarks\n\nFor $k, n \\ge 1$ let $E_{n,k} = \\bigcup_{m \\ge n} \\set{x : \\abs{f_m(x) - f(x)} \\ge 1/k}$.\nFor fixed $k$ these sets decrease in $n$ to a null set, so continuity from above, which needs $\\mu(X) \u003c \\infty$, gives $n_k$ with $\\mu(E_{n_k, k}) \u003c \\varepsilon / 2^k$.\nThe set $E = \\bigcup_k E_{n_k, k}$ works.\nExample: $f_n(x) = x^n$ on $[0, 1]$ with Lebesgue measure tends to $0$ almost everywhere, not uniformly; it does converge uniformly on $[0, 1 - \\delta]$ for each $\\delta \u003e 0$.\nThe finite measure hypothesis is necessary: on $\\R$ with Lebesgue measure, $f_n = \\mathbf{1}_{[n, \\infty)}$ tends to $0$ everywhere.\nEvery set $E$ of finite measure misses points of $[n, \\infty)$ for each $n$, so $\\sup_{X \\setminus E} f_n = 1$ and the convergence is never uniform there.\n\n## Explanations\n\n### L1 human\n\nA sequence of functions may converge at every point but at very different speeds, so that it is not uniformly convergent.\nEgorov's theorem says that on a space of finite measure this is a small problem.\nAfter removing a set as small as one likes, the convergence becomes uniform on what is left.\n\n### L2 human\n\nThis is the third of Littlewood's three principles: pointwise convergence is nearly uniform convergence.\nIt is used to prove convergence theorems for integrals, such as the bounded convergence theorem and the Vitali convergence theorem.\nThe convergence it gives, uniform outside sets of small measure, is called almost uniform convergence.\nAlmost uniform convergence implies [[convergence in measure|G5WT6QN8WR]], so on finite measure spaces almost everywhere convergence implies convergence in measure.\n",
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