TITLE: One approach to showing the total variation of an absolutely continuous function f is the integral of |f'|
QUESTION [9 upvotes]: I'm trying to solve the following problem in preparation for an exam:
Suppose $f$ is absolutely continuous on [a,b]. Show that $$ V[f; a,b] = \int_{[a,b]} |f'| dx$$
There is a suggestion to define $F(x) = V[f; a,x]$ and show $F \pm f$ are absolutely continuous and nondecreasing, then obtain a relationship between $F'$ and $f'$. It's the last part that is perplexing -- I have succeeded (I think) in demonstrating $F \pm f$ AC and nondecreasing -- but I don't see how to derive a (useful) relationship between $F'$ and $f'$. Any suggestions? Am I missing something obvious?

REPLY [10 votes]: First note that, since $F$ is nondecreasing, $$\int_{[a,b]} F' \le F(b) - F(a) = F(b) = V[f;a,b].$$
Next observe that since $|f(x) - f(y)| \le |F(x) - F(y)|$ for any two points $x,y \in [a,b]$ you have $$|f'(x)| \le |F'(x)| = F'(x)$$ whenever both functions are differentiable. This implies $$\int_{[a,b]} |f'| \le V[f;a,b].$$
All that was required so far was for $f$ to have bounded variation. If in addition $f$ is absolutely continuous then for any partition $\{t_k\}$ of $[a,b]$ you have $$f(t_{k+1}) - f(t_k) = \int_{[t_k,t_{k+1}]} f'$$ so that $$\sum_k |f(t_{k+1}) - f(t_k)| \le \int_{[a,b]} |f'|.$$
Now take the supremum over all partitions.