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  "body": "\n## Definition\n\nLet $(X, \\Sigma)$ and $(Y, \\mathcal{T})$ be [[measurable spaces|N4KDT83V3S]], let $\\mu$ be a [[measure|PR11BZQQPG]] on $(X, \\Sigma)$, and let $f \\colon X \\to Y$ be a [[measurable function|YYNCJSR0PJ]].\nThe *pushforward* of $\\mu$ along $f$ is the map $f_* \\mu \\colon \\mathcal{T} \\to [0, \\infty]$ given by\n$$(f_* \\mu)(B) = \\mu\\big(f^{-1}(B)\\big) \\quad \\text{for } B \\in \\mathcal{T}.$$\n\n## Remarks\n\nThe set $f^{-1}(B)$ lies in $\\Sigma$ because $f$ is measurable, so the formula makes sense.\nThe map $f_* \\mu$ is a measure: $f^{-1}(\\varnothing) = \\varnothing$, and the preimage of a disjoint union of sets $B_n$ is the disjoint union of the sets $f^{-1}(B_n)$.\nOther notations are $\\mu \\circ f^{-1}$ and $f_\\# \\mu$.\nExample: let $\\mu$ be the measure on $\\R$ with $\\mu(A) = $ number of points of $A \\cap \\set{-1, 1}$, and $f(x) = x^2$.\nThen $f_* \\mu$ gives mass $2$ to the point $1$, since $f^{-1}(\\set{1}) = \\set{-1, 1}$.\nNon-example: if $f$ is not measurable, some $f^{-1}(B)$ is outside $\\Sigma$, and the formula does not define $f_* \\mu$ on all of $\\mathcal{T}$.\nFor a measurable $g \\colon Y \\to [0, \\infty]$ the change of variables formula holds:\n$$\\int_Y g \\, d(f_* \\mu) = \\int_X (g \\circ f) \\, d\\mu.$$\n",
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