$\textbf{Question}$: Let $f$ be absolutely continuous on the interval $[\epsilon, 1]$ for $0<\epsilon<1$. Does the continuity of $f$ at 0 imply that $f$ is absolutely continuous on $[0,1]$? What if f is also of bounded variation on $[0,1]$?


My thoughts are that $f$ is NOT absolutely continuous on $[0,1]$.

The definition from my textbook states that a function $F$ defined on $[a,b]$ is absolutely continuous if for any $\epsilon>0$ there exists $\delta >0$ so that $\sum_{k=1}^{N}|F(b_{k})-F(a_{k})|<\epsilon$ whenever $\sum_{k=1}^{N}(b_{k}-a_{k})<\delta$ and the intervals $(a_{k},b_{k})$ are disjoint.

So, I think that the point $0$ is not included in this definition - $f$ is differentiable a.e. which does not necessarily include the boundary at the point 0, even if the point itself exists.

Is this correct? My guess is then that the bounded variation assumption will make $F$ absolutely continuous on $[0,1]$ but I am not sure why.

  • $\begingroup$ Hint: fundamental theorem of calculus $\endgroup$
    – user251257
    Dec 13, 2016 at 1:01
  • $\begingroup$ I did think about that. But I couldn't figure out the precise use, i.e. why does it help? $\endgroup$
    – PBJ
    Dec 13, 2016 at 1:03
  • $\begingroup$ You surely know some function that is Riemann improper integrable but not Lebesgue integrable. $\endgroup$
    – user251257
    Dec 13, 2016 at 1:05
  • $\begingroup$ Your example should have rough oscillation near $x = 0$ in order to make the variation diverge. So you may choose functions like $f(x) = x \sin(x^{-2016})$. $\endgroup$ Dec 13, 2016 at 1:08
  • $\begingroup$ Ok, that makes sense. The function $\frac{1}{x}$ on $[\epsilon,1]$ will work too, right? It is not absolutely continuous on $[0,1].$ So now for the bounded variation part...? I have seen from other sources that "Any function with bounded derivative on an interval is absolutely continuous there" but does that follow right from the definition of MVT? $\endgroup$
    – PBJ
    Dec 13, 2016 at 1:49

2 Answers 2


Let $h = \mathbb 1_{(0,1]}$. For $x\in \mathbb R$ defines $$ g(x) = \sum_{n=1}^\infty 2^n h(x \cdot 2^n - 1) \frac{(-1)^{n+1}}{n}$$ and $$ f(x) = \int_0^x g(t) dt. $$

  1. Then, $g$ is improper Riemann integrable and thus $f$ is continuous.

  2. Further, on every compact interval without $0$, $g$ is bounded and thus $f$ is absolutely continuous (in fact Lipschitz).

  3. But, on any neighborhood of $0$, $g$ is not Lebesgue integrable, thus $f$ is not absolutely continuous.

For the part that $f$ is additionally of bounded variation:

Notice that $f$ is AC if and only if $f$ is continuous, is BV, and maps measure zero set to measure zero set. Let $N\subset [0,1]$ have measure zero. Then, $N \cap [1/n, 1]$ has measure zero too and $$ |f(N)| = \lim_{n\to \infty} |f(N \cap [1/n, 1])| = 0. $$ That is, $f$ is AC on $[0,1]$.


Here's another counter example:

Let $f(x)=x^2\sin(1/x^2)$ on $(0,1]$ and $f(0)=0$. You can check that this function is continuous on $[0,1]$. On every $[\epsilon, 1]$, the derivative of $f$ is bounded so $f$ is Lipschitz on such intervals and therefore is also absolutely continuous on $[\epsilon,1]$. But $f$ is not of bounded variation (let me know if you would like to see details) on $[0,1]$ and thus is not absolutely continuous on this domain.
As the other poster wrote, if $f$ is bv on $[0,1]$, we can use the condition that it is absolutely continuous on $[\epsilon,1]$ to show that is is Luzin (sends measure $0$ sets to measure $0$ sets) on the whole interval, thus proving that it is AC.


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