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.
Best Answer
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.