$Var(F)|_{a}^{b}=\int_{a}^{b}|f|d\alpha.$

Question

Let $[a,b]$ be an interval in $\mathbb R$,and $\alpha :[a,b]\to \mathbb R$ be monotone increasing.

Let $f:[a,b]\to \mathbb R$ be integrable, bounded and with respect to $\alpha$. Define $F:[a,b]\to \mathbb R$ by setting $$F(x)=\int_{a}^{x}fd\alpha$$ for all $x\in[a,b]$. Prove that $Var(F)|_{a}^{b}=\int_{a}^{b}|f|d\alpha.$

Certainly, I think $Var(F)|_{a}^{b}\le \int_{a}^{b}|f|d\alpha.$ But I have no idea how to prove another way. Also, I tried to use the definition to prove it, but I am stuck, can someone give me any help?

Answer 1

For the Lebesgue measure (instead of $d\alpha$), you can first work with continuous functions. Then

$$\text{Var}(F)|_a^b \geq \sum_{i=0}^{+\infty} \left| \int_{x_i}^{x_{i+1}} f(x) dx \right| $$

And if you take a partition of [a,b] such that $f$ is of constant sign on $[x_i, x_{i+1}]$ (possible because $f$ is continuous), you get

$$\text{Var}(F)|_a^b \geq \sum_{i=0}^{+\infty} \int_{x_i}^{x_{i+1}} \left| f(x) \right| dx = \int_a^b |f(x)| dx$$

THen by density of the continuous functions in $L^1$, you get the result.

I believe you can adapt this idea for a proof with $d\alpha$