# Given $x \in (0,1)$, $f$ is absolutely continuous on $[0,x]$, bounded variation on $[x,1]$, and continuous at 1 implies absolute continuity on $[0,1]$

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.

• 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 $x=1$. Commented Jul 12, 2023 at 22:14
• @RobertShore, isn't absolute continuity stronger than continuity? Commented Jul 12, 2023 at 23:27
• Yes, but in your first bullet point, that's how you get that $f$ is continuous everywhere, not just a.e. Commented Jul 12, 2023 at 23:32
• Oh, that makes sense. Thank you! Commented Jul 13, 2023 at 0:12

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.