I am trying to prove this statement which appears in the real analysis text by Stein which he just passes as a remark.
If $F$ is absolutely continuous on $[a,b]$, then the total variation of $F$ is continuous and to be specific, absolutely continuous.
I believe that it suffices to show that the total variation of $F$ is absolutely continuous.
The total variation of $F$ on $[a,x]$ is defined to be $T_F(x)=\sup\sum_{j=1}^N|F(t_j)-F(t_{j-1})|$ where the $\sup$ is over all partitions of $[a,x].$ Does this mean that I have to prove that $\sum_{k=1}^N|T_F(b_k)-T(a_k)|<\epsilon$ if $\sum_{k=1}^N(b_k -a_k)<\delta?$
This looks a bit strange to me…can anyone clarify this matter? thanks!
Best Answer
Yes, basically that is what you have to prove. What makes the proof more manageable is the additive property of the total variation,
$$T_F(c) - T_F(b) = T_{F\lvert_{[b,c]}}(c) = \sup \left\lbrace \sum_{i=1}^n \lvert F(x_i) - F(x_{i-1})\rvert : b = i_0 < i_1 < \dotsb < i_n = c;\; n\in\mathbb{Z}^+\right\rbrace$$
for $b < c$.
The proof becomes still shorter if you know that a function is absolutely continuous iff it is an integral, and if
$$F(x) = F(a) + \int_a^x f(t)\,dt,$$
then
$$T_F(x) = \int_a^x \lvert f(t)\rvert\,dt.$$