Question

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

Answer 1

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.