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Let $f:[a,b]\rightarrow \mathbb{R}$ a bounded variation function and define $g:[a,b]\rightarrow \mathbb{R}$ by $g(a)=0$ and $g(x)=\frac{1}{x-a}\int_{a}^{x}f(t)dt$. g is a bounded variation function?

I know that $ | \int_{a}^{b}f(t)dt|\leq M(b-a)$. M is bound of $f$.

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Yes.

There exist two nondecreasing functions $f_1,f_2$ such that $f=f_1-f_2$. Then we have $g(x)=\frac{1}{x-a}\int_a^xf_1(t)\,dt-\frac{1}{x-a}\int_a^xf_2(t)\,dt$. In order to show that $g$ has bounded variation, it is sufficient to show that $x\mapsto\frac{1}{x-a}\int_a^xf_1(t)\,dt$ and $x\mapsto\frac{1}{x-a}\int_a^xf_2(t)\,dt$ are nondecreasing.

Since $f_1$ is nondecreasing, it is clear that the mean of $f_1$ on $[a,x]$ is not greater than the mean of $f_1$ on $[a,y]$ for $x\le y$, that is $\frac{1}{x-a}\int_a^xf_1(t)\,dt\le\frac{1}{y-a}\int_a^yf_1(t)\,dt$. If you want a more formal proof: let $x,y$ be such that $x\le y$. By the change of variables $u=\frac{(y-a)}{(x-a)}(t-a)+a$, we have $$ \begin{align*} \frac{1}{x-a}\int_a^xf_1(t)\,dt&=\frac{1}{x-a}\int_a^yf_1\left(a+(u-a)\frac{(x-a)}{(y-a)}\right)\frac{(x-a)}{(y-a)}\,du\\ &=\frac{1}{y-a}\int_a^yf_1\left(a+(u-a)\frac{(x-a)}{(y-a)}\right)\,du \end{align*} $$

For all $u\in[a,y]$, $a+(u-a)\frac{(x-a)}{(y-a)}\le u$ so by monoticity of $f_1$, we have $f_1\left(a+(u-a)\frac{(x-a)}{(y-a)}\right)\le f_1(u)$, hence $$ \frac{1}{x-a}\int_a^xf_1(t)\,dt\le\frac{1}{y-a}\int_a^yf(u)\,du $$

Of course the same holds for $f_2$, hence $g$ has bounded variation.

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