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  "body": "\n## Definition\n\nLet $(X, \\Sigma)$ be a measurable space and let $\\nu$ be a [[signed measure|VG2P7R91Z7]] or a [[complex measure|NY6TYHRGMS]] on it.\nThe *total variation* of $\\nu$ is the map $\\abs{\\nu} \\colon \\Sigma \\to [0, \\infty]$ given by\n$$\\abs{\\nu}(E) = \\sup \\set{ \\sum_{i=1}^{n} \\abs{\\nu(E_i)} : n \\ge 1, \\ E_1, \\ldots, E_n \\in \\Sigma \\text{ pairwise disjoint}, \\ E = E_1 \\cup \\cdots \\cup E_n },$$\nwhere the [[supremum|0BTMQ04556]] is over all finite partitions of $E$ into measurable sets.\nThe *total variation norm* of $\\nu$ is $\\norm{\\nu} = \\abs{\\nu}(X)$.\n\n## Remarks\n\nThe map $\\abs{\\nu}$ is a measure, and $\\abs{\\nu(E)} \\le \\abs{\\nu}(E)$ for all $E \\in \\Sigma$.\nFor a complex measure, $\\abs{\\nu}$ is a finite measure.\n\nExample: let $\\nu = \\delta_0 - \\delta_1$ on the subsets of $\\R$, where $\\delta_a$ is the Dirac measure at $a$.\nFor $E = \\set{0, 1}$, one has $\\nu(E) = 0$, but the partition $\\set{0}, \\set{1}$ gives $\\abs{1} + \\abs{-1} = 2$.\nNo partition gives more, so $\\abs{\\nu}(E) = 2$ and $\\norm{\\nu} = 2$.\n\nNon-example: the map $E \\mapsto \\abs{\\nu(E)}$ is not the total variation in general.\nFor the same $\\nu$ it gives $0$ on $\\set{0, 1}$, and it is not even additive: $\\abs{\\nu(\\set{0})} + \\abs{\\nu(\\set{1})} = 2 \\ne 0$.\n\nFor a signed measure with Jordan decomposition $\\nu = \\nu^+ - \\nu^-$, one has $\\abs{\\nu} = \\nu^+ + \\nu^-$.\nIf $\\nu(E) = \\int_E f \\, d\\mu$ with $f \\in L^1(\\mu)$, then $\\abs{\\nu}(E) = \\int_E \\abs{f} \\, d\\mu$.\n",
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