I am trying to directly prove from the definition of absolute continuity, that if a function $f:[a,b]\rightarrow\mathbb{R}$ is absolutely continuous, then it is of bounded variation, i.e $Vf$ is finite.
I have tried taking a uniform partition of [a,b] and played around with the definition. But it just does not seem intuitive to me. What is the main idea behind the proof?
image below is as described in the comment to the answer.
https://i.stack.imgur.com/TzljL.png
Best Answer
So then since $f$ is absolutely continous on $[a,b]$, for $\epsilon=1 >0$ there exists $\delta > 0$ such that for all finite disjoint open intervals on $[a,b]$, $\{(x_1,y_1),...,(x_n,y_n)\}$ with sum of their length less than $\delta$, we have $\sum_{i=1}^n |f(y_i)-f(x_i)|<\epsilon$.
let $P*=\{a_0=a,...,a_n=b\}$ be a partition of $[a,b]$ with tthe propertiy that $a_i-a_{i-1}=\frac{\delta}{2}$ for all $i\in \{1,2...,n-1\}$, and $a_n-a_{n-1} \leq \frac{\delta}{2}$.
Then $n= \lfloor \frac{2(b-a)}{\delta} \rfloor +1$. Now pick any partition $P$ and let $P'=P\cup P^*$ and $P'=\{z_0=a,...,z_n=b\}$. Let $P_i'$ be the set of points in $P_i$ contained in $[a_{i-1},a_i]$ for $i\in \{1,...,n\}$.
So variation $V(P,f)=\sum_{i=1}^s|f(c_i)-f(c_{i-1})|\leq V(P',f)=\sum_{i=1}^n\sum|f(z_{i_k})-f(z_{i_{k-1}})|\leq n$. Here $\{c_0,...,c_s\}$ is the partition $P$.