# Show $\left| \int_a^b f d\alpha \right| \le \int_a^b|f|dV$ if $V$ is variation of $\alpha$ on $[a, b]$

Let $\alpha$ be of bounded variation on $[a, b]$ and assume that $V(x)$ be the total variation of $\alpha$ on $[a, x]$, $a<x\le b$ and $V(a) = 0$. Let $f$ be defined and bounded on $[a, b]$. If $f\in R(\alpha)$ on $[a, b]$, then prove that
$i) \hspace{2 mm} f \in R(v)$ on $[a, b] \hspace {2 cm} ii) \left| \int_a^b f d\alpha \right| \le \int_a^b|f|dV$

I need help with part $ii)$. I know $\left| \int_a^b f d\alpha \right| \le \int_a^b|f|d\alpha$. Is $\Delta V_k \ge \Delta\alpha_k$? Does it imply the inequality? How do I proceed?

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$\Delta V_k$ is the total variation of $\alpha$ on the $k$th interval. As such, it is bounded from below by the difference of $\alpha$ at the endpoints: $\Delta V_k\ge |\Delta \alpha_k|$. Note the absolute value signs: you'll need them at the next step.
To estimate the modulus of $\int f \,d\alpha$, we consider a sum such as $\sum M_k \Delta \alpha_k$ and apply the triangle inequality: $\left|\sum M_k \Delta \alpha_k\right|\le \sum |M_k| |\Delta \alpha_k|$. Put this together with the above, and the rest should be easy.