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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $\\lambda$ be [[Lebesgue measure|2Z2QVWZ262]] on $\\R^n$, and let $f \\colon \\R^n \\to \\R$ be locally [[integrable|X7GTF3D98F]].\nFor $x \\in \\R^n$ and $r \u003e 0$, let $B(x, r)$ be the [[open ball|0YE5868HZB]] of centre $x$ and radius $r$.\nA point $x \\in \\R^n$ is a *Lebesgue point* of $f$ if\n$$\\lim_{r \\to 0^+} \\frac{1}{\\lambda(B(x, r))} \\int_{B(x, r)} \\abs{f(y) - f(x)} \\, d\\lambda(y) = 0.$$\n\n## Remarks\n\nExample: if $f$ is continuous at $x$, then $x$ is a Lebesgue point.\nGiven $\\varepsilon \u003e 0$, choose $r_0$ with $\\abs{f(y) - f(x)} \u003c \\varepsilon$ on $B(x, r_0)$; then every average with $r \u003c r_0$ is at most $\\varepsilon$.\n\nNon-example: let $n = 1$ and $f = \\mathbf{1}_{[0, \\infty)}$.\nAt $x = 0$, the integrand equals $1$ on $(-r, 0)$ and $0$ on $[0, r)$, so the average is $1/2$ for every $r$, and $0$ is not a Lebesgue point.\n\nThe set of Lebesgue points depends on the chosen function, not only on its class modulo null sets.\nChanging $f$ at one point $x$ can destroy the Lebesgue point property at $x$ but at no other point.\n",
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