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With $$a<b$$, the total variation of $$f(x)$$ on a finite or infinite interval $$(a, b)$$ is
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With $$a<b$$, the total variation of $$f(x)$$ on a finite or infinite interval $$(a, b)$$ is

$\mathscr{V}_{a, b}(f)=\sup \sum_{j=1}^{n}\left|f\left(x_{j}\right)-f\left(x_{j-1}\right)\right|$

where the supremum is over all sets of points $$x_{0}<x_{1}<\cdots<x_{n}$$ in the closure of $$(a, b)$$, that is, $$(a, b)$$ with $$a, b$$ added when they are finite. If $$\mathscr{V}_{a, b}(f)<\infty$$, then $$f(x)$$ is of bounded variation on $$(a, b)$$. In this case, $$g(x)=\mathscr{V}_{a, x}(f)$$ and $$h(x)=\mathscr{V}_{a, x}(f)-f(x)$$ are nondecreasing bounded functions and $$f(x)=g(x)-h(x)$$.
If $$f(x)$$ is continuous on the closure of $$(a, b)$$ and $$f^{\prime}(x)$$ is continuous on $$(a, b)$$, then

$\mathscr{V}_{a, b}(f)=\int_{a}^{b}\left|f^{\prime}(x)\right| \mathrm{d} x$

whenever this integral exists.
by Platinum (98.5k points)

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