Prove that $\int_0^1f'(x)dx \leq f(1) - f(0)$.

Question

Let $f(x)$ be a non-decreasing function on $[0, 1].$ You may assume that $f$ is differentiable almost everywhere. Prove that

$\int_0^1f'(x)dx \leq f(1) - f(0)$.

I am having a hard time with this question. Obviously we know that $f$ is continuous. It looks a lot like absolute continuity.

Thanks for any help

Answer 1

More generally, it holds for $f$ a non-decreasing function on $[a,b]$ ($f$ is then differentiable a.e. on $[a,b]$):

For $x>b$, let $f(x)=f(b)$.

For each $n$, let $$f_n(x)={f(x+1/n)-f(x)\over 1/n}.$$

Note that $$\eqalign{ \int_a^b f_n(x)\,dx&=n\int_b^{b+1/n} f(x)-n\int_a^{a+1/n}f(x)\,dx\cr &\le n\int_b^{b+1/n} f(b )\,dx-n\int_a^{a+1/n}f(a)\,dx\cr &= f(b)-f(a). } $$

Now use this result, the fact that the sequence $(f_n)$ of nonnegative functions converges to $f'$ almost everywhere, and Fatou's Lemma.