I am trying to prove the following:

Suppose for every $x \in (0,1)$, the function $f$ is absolutely continuous on $[0,x]$, $f \in \text{BV}([x,1])$, and $f$ is continuous at $x=1$. Prove that $f$ is absolutely continuous on $[0,1]$.

I've been toying around with a few different approaches, none of which I am particularly confident about.

  • Somehow applying the Banach-Zarecki theorem: $f : I \rightarrow \mathbb R$ is absolutely continuous $\iff$ $f$ is continuous, of bounded variation, and has the Luzin-N property on $I$. Obviously $f$ is of bounded variation but the best I could get is that $f$ is continuous a.e.; I'm not sure if this is sufficient, I only encountered this result via the Wikipedia page for absolute continuity.
  • Doing a normal $\epsilon-\delta$ type of argument: however, I get stuck on trying to bound the total variation $V_x^1(f)$ by some $\epsilon$.
  • Another definition of absolute continuity would be showing that if $f'$ exists a.e. and is integrable on $[x,1]$ (which should be true because $f \in \text{BV}([x,1])$) then $f(y) = f(x) + \int_x^y f'(t) dt$ for all $y \in [x,1]$... I think this could be valid by just applying the fundamental theorem of calculus, but that feels almost too elementary.

Any suggestions on how to proceed would be appreciated.

  • 1
    $\begingroup$ If $x \lt 1$, choose $y$ such that $x \lt y \lt 1$. Then $f$ is absolutely continuous on $[0, y]$ by hypothesis, so a fortiori $f$ is continuous at $x \in [0, y]$. And you're given that $f$ is continuous at $x=1$. $\endgroup$ Commented Jul 12, 2023 at 22:14
  • $\begingroup$ @RobertShore, isn't absolute continuity stronger than continuity? $\endgroup$ Commented Jul 12, 2023 at 23:27
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    $\begingroup$ Yes, but in your first bullet point, that's how you get that $f$ is continuous everywhere, not just a.e. $\endgroup$ Commented Jul 12, 2023 at 23:32
  • $\begingroup$ Oh, that makes sense. Thank you! $\endgroup$ Commented Jul 13, 2023 at 0:12

1 Answer 1


For an increasing function $g$ you have that $$\int_a^bg’(x)\,dx\le g(b)-g(a)$$ Let $g(x)=-V^1_x(f)$. Since $$|f(x)-f(y)|\le g(x)-g(y)$$ if you consider a point $x$ where both $f$ and $g$ are differentiable, you get that $|f’(x)|\le g’(x)$. Hence, $$\int_{1/2}^1|f’(x)|\,dx\le \int_{1/2}^1|g’(x)|\,dx\le -g(1/2)$$ This shows that $f’$ is integrable near $1$. Now, if you apply the ftc in $[0,x]$ where $x<1$, you get $$ f(x)=f(y)+\int_y^x f’(t)\,dt$$ Now you want to pass to the limit as $x\to 1$. On the left hand side you use continuity of $f$ at $1$ and on the right hand side the Lebesgue dominated convergence theorem, which you can use because $f’$ is integrable.


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