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  "body": "\n## Definition\n\nLet $n \\ge 3$ and let $D \\subset \\R^n$ be a bounded domain whose boundary points are all [[regular|B7948M365K]].\nFor $x \\in D$, let $\\omega_x$ be the [[harmonic measure|CFD7BPM51W]] of $D$ at $x$, a Borel probability measure on $\\partial D$.\nLet $\\mu$ be a finite positive [[Borel measure|DYR1QFNGNJ]] with compact support in $D$.\nThe *balayage* of $\\mu$ onto $\\partial D$ is the Borel measure $\\mu'$ on $\\partial D$ given by\n$$\\mu'(A) = \\int_D \\omega_x(A)\\, d\\mu(x) \\qquad \\text{for every Borel set } A \\subseteq \\partial D.$$\n\n## Remarks\n\nWrite $U^\\mu(y) = \\int \\abs{y - x}^{2-n}\\, d\\mu(x)$ for the [[Newtonian potential|DCMFQ01QGS]].\nThen $\\mu'(\\partial D) = \\mu(D)$ and $U^{\\mu'} = U^\\mu$ on $\\R^n \\setminus D$.\nIndeed, for $y \\notin \\overline{D}$ the function $x \\mapsto \\abs{y - x}^{2-n}$ is harmonic on $D$ and continuous on $\\overline{D}$, so it is the integral of its boundary values against $\\omega_x$.\nInside $D$ one has $U^{\\mu'} \\le U^\\mu$.\nExample: let $D$ be the ball of radius $R$ about $0$ and $\\mu = \\delta_0$.\nThen $\\omega_0$ is the normalised surface measure on the sphere, so $\\mu'$ is that measure.\nIts potential is $\\min(R^{2-n}, \\abs{y}^{2-n})$, which equals $\\abs{y}^{2-n} = U^{\\delta_0}(y)$ for $\\abs{y} \\ge R$.\nNon-example: $\\mu$ itself is not a balayage of $\\mu$ onto $\\partial D$, since it is not carried by $\\partial D$.\nMore generally, for any set $E \\subseteq \\R^n$ the balayage $\\mu^E$ is the measure whose potential is the regularised [[reduced function|909B1WYQ4Q]] $\\widehat{R}^E_{U^\\mu}$.\n",
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