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  "body": "\n## Statement\n\nLet $a \u003c b$ be real numbers and let $f \\colon [a, b] \\to \\R$ be a function.\nThen $f$ has [[bounded variation|A8251BHCMP]] on $[a, b]$ if and only if there are nondecreasing functions $g, h \\colon [a, b] \\to \\R$ with $f = g - h$.\nMoreover, when $f$ has bounded variation, one can take $g(x) = V_a^x(f)$ and $h(x) = V_a^x(f) - f(x)$, where $V_a^x(f)$ is the [[total variation|8QA5DEWNYD]] of $f$ on $[a, x]$ and $V_a^a(f) = 0$.\n\n## Remarks\n\nA nondecreasing function has bounded variation, and a difference of two functions of bounded variation has bounded variation, which gives one direction.\nFor the other direction, $g(x) = V_a^x(f)$ is nondecreasing because variation is additive over adjacent intervals.\nFor $x \u003c y$ one has $f(y) - f(x) \\le \\abs{f(y) - f(x)} \\le V_x^y(f) = g(y) - g(x)$, so $h = g - f$ is nondecreasing too.\nFor example, $f(x) = \\sin x$ on $[0, 2\\pi]$ is the difference of $g(x) = V_0^x(\\sin)$ and $g - \\sin$.\nThe decomposition is not unique: one may add the same nondecreasing function to $g$ and to $h$.\n\n## Explanations\n\n### L1 human\n\nA function of bounded variation may go up and down, but it can always be split into a part that only goes up and a part that only goes down.\nThis reduces many questions about such functions to [[monotone functions|0WFF6AZPBP]], which are much easier to handle.\n\n### L2 human\n\nWith the decomposition, results about monotone functions pass to functions of bounded variation.\nBy [[Lebesgue's theorem on monotone functions|1Y4N1DDQNR]], a function of bounded variation is differentiable almost everywhere.\nIt also has one-sided limits at every point and at most countably many discontinuities.\nThe theorem is the one-variable model of the [[Jordan decomposition of a signed measure|WAGJVK2NWE]].\n",
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