# Showing $V_a^b\alpha = \sup \left\{\int_a^bfd\alpha:\|f\|_\infty\leq 1\right\}$

Let $\alpha:[a,b]\to\mathbb{R}$ be of bounded variation and right-continuous. Given $\varepsilon>0$ and a partition $P$ of $[a,b]$, construct $f\in C[a,b]$ with $\|f\|_\infty \leq 1$ such that $\int_a^bfd\alpha\geq V(\alpha,P)-\varepsilon$. Conclude that $V_a^b\alpha = \sup \left\{\int_a^bfd\alpha:\|f\|_\infty\leq 1\right\}$.

I am not sure how to show this but it seems to me that we want to construct $f$ that is some kind of characteristic function. Any hints would be appreciated.

-

Let $x_i$, $i=0,\dots n$ be the partition $P$. Let $h_j(t)=sgn(\alpha(x_{j+1})-\alpha(x_j))1_{(x_i,x_{i+1})}$ and $h=\sum h_j$. $h$ is $\alpha$ integrable since it's simple and $$\int_a^b hd\alpha= \sum_1^n |\alpha(x_{i+1})-\alpha(x_i)|$$ now we just want to approximate this function. For each $i$ and $\delta>0$ consider the intervals $[x_i+\delta,x_{i+1}-\delta]=J_{i,\delta}$ and consider $f_{i,\delta}=h_i|_{J_{i,\delta}}$. Take $f_{\delta}:\cup J_{i,\delta} \to \mathbb{R}$ as $f_{\delta}=\sum f_{i,\delta}$ and extend linearly (call the extension the same). It's easy to see that the $f_{\delta}$ are continuous, as $\delta \to 0, f_{\delta}(t) \to h(t)$ except possibly at the $x_i$ which is of $\alpha$-measure 0 , and $|f_{\delta}|\leq 1$, so you can apply the dominated convergence theorem to the integrals and conclude (formally you have to apply this last theorem to the integrals associated with the nonnegative measures given by the Jordan decomposition of $\alpha$).
Just to clarify, your $f$ will be a $f_{\delta}$ for $\delta(\varepsilon)$ sufficiently small.