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I have this problem, but I can not solve, I hope someone can help me! thank you

Suppose $f$ is continuos and $\phi$ is of bounded variation . Prove that $$\psi= \int_a^x fd\phi $$ is a function of bounded variation on [a,b]

I don´t know how i could use Jordan descomposition, but here is my attempt but i am stuck. Let $$\Gamma $$ be a partition of [a,b] then $$S_\Gamma= \sum_{i=1}^{n} | \phi (x_i)-\phi(x_(i-1) $$ $$=\sum_{i=1}^{n} |\int_a^{x_i} fd\phi-\int_a^{x_{i-1}} fd\phi|$$ let $\Gamma_1=[a=x_0,x_1,...,x_{m-1},x_m=x_{i-1}]$ be a partition of $[a,x_{i-1}] $and $\Gamma_2= \Gamma_1 \bigcup {x_i}=[a=x_0,x_1,...,x_{m-1},x_m=x_{i-1}, x_{m+1}=x_i]$ so $\Gamma_2 $is a partition of $[a,x_{i}]$, then$$ \int_a^{x_i} fd\phi - \int_a^{x_{i-1}} fd\phi=f(\eta_{m+1})[\phi(x_i)-\phi(x_{i-1}]$$ then $$ S_\Gamma=f(\eta_{m+1})\sum_{i=1}^{n}[\phi(x_i)-\phi(x_{i-1}]$$, but $\phi$ is a bounded variation function so $$S_\Gamma \leq f(\eta_{m+1})*M$$ where am I wrong?

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up vote 1 down vote accepted

Hint: Consider the Jordan decomposition.

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