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  "body": "\n## Definition\n\nLet $(X, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]], and let $f_1, f_2, \\ldots$ and $f$ be [[measurable functions|YYNCJSR0PJ]] from $X$ to $\\R$.\nThe sequence $(f_n)$ *converges in measure* to $f$ if for every $\\varepsilon \u003e 0$\n$$\\lim_{n \\to \\infty} \\mu\\big(\\set{x \\in X : \\abs{f_n(x) - f(x)} \\ge \\varepsilon}\\big) = 0.$$\n\n## Remarks\n\nExample: on $[0, 1]$ with Lebesgue measure, write $n = 2^k + j$ with $0 \\le j \u003c 2^k$ and let $f_n = \\mathbf{1}_{[j 2^{-k}, (j+1) 2^{-k}]}$.\nFor $0 \u003c \\varepsilon \\le 1$, the set where $\\abs{f_n} \\ge \\varepsilon$ has measure $2^{-k}$, which tends to $0$, so $f_n \\to 0$ in measure.\nBut each $x \\in [0, 1]$ has $f_n(x) = 1$ for infinitely many $n$, so $f_n(x)$ converges for no $x$.\nNon-example: on $\\R$ with Lebesgue measure, $f_n = \\mathbf{1}_{[n, n+1]}$ tends to $0$ at every point, but the set where $\\abs{f_n} \\ge 1/2$ has measure $1$ for all $n$.\nA local variant asks the limit to be $0$ only for the restriction to each set of finite measure.\nConvergence in $L^p$ for some $1 \\le p \u003c \\infty$ implies convergence in measure, by Chebyshev's inequality.\n",
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