If $f'\notin L^1([a,b])$ then $f$ is not a bounded variation on $[a,b]$

Question

The question is basically from here.

Suppose $f'\in L^1([a,b])$ if and only if $a>b$. If $0<a\leq b$, then $f'$ is not absolutely integrable on $[0,1]$, which implies that $f$ cannot be BV on $[0,1]$.

I know that if $f'\notin L^1([a,b])$ then $f$ cannot be absolutely continuous. But being bounded variation is more general than that. Does the above statement true?

Edit: If $f$ is of bounded variation on $[a,b]$ then $f'$ is absolutely integrable on $[a,b]$: First by Jordan decomposition, $f = f_1-f_2$ for some increasing functions $f_1,f_2$ on $[a,b]$. If $f_1,f_2$ are bounded, $f_1',f_2'\in L^1([a,b])$ so that $f' = f_1'-f_2'\in L^1([a,b])$.

@robjohn gave an example: If $f$ is the heaviside function then it's of bounded variation but $f'$ is a dirac delta function which is not $L^1$.

Answer 1

Yes it's true. The proof is the union of this two facts:

• If $f\in BV[a,b]$ then, by Jordan theorem, $f=f_1-f_2$, with $f_1$ and $f_2$ increasing functions on $[a,b]$.
• If $g:[a,b]\longrightarrow \mathbb{R}$ is a monotone function then it's a.e. (wrt Lebesgue measure) differentiable and $g'\in L^1[a,b]$

Edit: For the proof of second fact(which is true under this hypothesis, without assuming continuity) see R.L. Wheeden and A. Zygmund, Measure and Integral, chapter 7.4