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If $f$ is integrable on $[0,1]$, then $\int_{0}^{x}f$ is bounded variation.

I know that every bounded function is integrable over a set of finite measure. And I know that if $f$ is bounded variation, $f$ is Riemann integrable. Any help is appreciated.

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  • $\begingroup$ The indicator function of a non-measurable set, if they exist, is bounded but not measurable. $\endgroup$ – user553213 May 9 '18 at 0:51
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For any partition $x_0<x_1<\cdots<x_n$, you have $$ \sum_{j=0}^n \left|\int_0^{x_{j+1}}f-\int_0^{x_j}f\right| =\sum_{j=0}^n\left|\int_{x_j}^{x_{j+1}}f\right|\leq\sum_{j=0}^n \int_{x_j}^{x_{j+1}}|f|=\int_0^1|f|<\infty. $$

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