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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?

user144600
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  • COuld you remember what's the definition of $\text{Var}(F)|_a^b$? – Tryss May 01 '15 at 06:32
  • sup{$\sum_{i=1}^{i=k}|F(x_{i})-F(x_{i-1})|$},k can be any positive integer and $a=x_{0}<x_{1}<x_{2}.....<x_{k}=b$,which is any partition of [a,b] – user144600 May 01 '15 at 06:35
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    maybe this is helpful:http://math.stackexchange.com/questions/150579/question-about-riemann-integral-and-total-variation – user144600 May 02 '15 at 20:33

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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 $d\alpha$

Tryss
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  • But, f is not necessarily continuous in my case,also, since $\alpha$ is not the normal x,so maybe we coudn't figure out whether the set of the discontinuties of the function f should be measure 0 – user144600 May 02 '15 at 02:33
  • The set of discontinuity of $f$ can be of any measure. No, my idea is that you prove the propriety on a dense subset of $L^1$, then you extend the result to the whole space by density – Tryss May 02 '15 at 11:16