# $f$ is integrable implies $F(x)$ is bounded variation.

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.

• The indicator function of a non-measurable set, if they exist, is bounded but not measurable. – user553213 May 9 '18 at 0:51

For any partition $x_0<x_1<\cdots<x_n$, you have $$\sum_{j=0}^n \left|\int_0^{x_{j+1}}f-\int_0^{x_j}f\right| =\sum_{j=0}^n\left|\int_{x_j}^{x_{j+1}}f\right|\leq\sum_{j=0}^n \int_{x_j}^{x_{j+1}}|f|=\int_0^1|f|<\infty.$$