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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 $L^1$.

one potato two potato
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    If $f$ is increasing then $f'$ exists a.e and $\int_a^{b}f'(t)dt \leq f(b)-f(a)$. – geetha290krm Mar 28 '23 at 07:23
  • @geetha290krm The statement is true when $f$ is continuous. – one potato two potato Mar 28 '23 at 08:54
  • It is true even if $f$ is not continuous. – geetha290krm Mar 28 '23 at 09:17
  • @geetha290krm Why? I only know the case when $f$ is continuous. – one potato two potato Mar 28 '23 at 09:21
  • Just type this inequality in approach0.xyz and search. – geetha290krm Mar 28 '23 at 09:24
  • @geetha290krm I didn't found on approach0 but on Royden that if $f$ is increasing bounded function on $[a,b]$ then $\int_a^b f'(t)dt\leq f(b)-f(a)$. I edit the post. Is that the argument you have in mind? I assumed $f'$ is bounded. – one potato two potato Mar 28 '23 at 10:15
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    onepotatotwopotato 1) I don't understand why you inserted @Luca.b 's answer at the end of your post, instead of simply accepting it. 2) While copying it, you added a redundant hypothesis ("If $f_1,f_2$ are bounded"). – Anne Bauval Mar 28 '23 at 11:37
  • @AnneBauval I accepted before but retracted because I didn't know that If $f$ is increasing then $f'\in L^1$ without assuming $f$ is continuous, which I know it's true if $f$ is bounded. I posted Luca.b's answer to check if my understanding is correct (to geetha290km). – one potato two potato Mar 28 '23 at 12:08
  • This comment clarifies your intent but I do not think this edit of your post was a good idea. Why not ask simply and precisely in comments, either to geetha or to Luca, about what you are still missing? And again: your boundedness assumption is redundant (every monotone function on $[a,b]$ is obviously bounded). Anyway, the present post is a duplicate. – Anne Bauval Mar 28 '23 at 12:17
  • @onepotatotwopotato Added in my answer a source for the proof of the theorem you didn't know – Luca.b Mar 28 '23 at 12:23

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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

Luca.b
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  • There's an example given by @robjohn that Heaviside function is of bounded variation but its derivative is dirac delta function which is not $L^1$. – one potato two potato Mar 28 '23 at 09:59
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    It's importante that the conclusion Is true only almost everywhere. The derivative of Heaviside function Is $0\in L^1$ a.e. (since it Is constant except in zero). – Luca.b Mar 28 '23 at 11:21