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Could one give an example of a strictly increasing, continuous but not absolutely continuous, function on $[0,1]$ into $[0,1]$ or on $[0,1)$ into $R$ or any of the related combinations of 1-d domain and range?

Martin Sleziak
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Hans
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    the Cantor function is not strictly increasing. – mathemagician Dec 17 '13 at 18:10
  • The Cantor function is the most classical example. See

    http://math.stackexchange.com/questions/125345/why-is-the-cantor-function-not-absolutely-continuous

     – Yiorgos S. Smyrlis Dec 17 '13 at 18:17
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    When the interval is not compact, any continuous function which is not uniformly continuous will do (since absolute continuity implies uniform continuity). –  Dec 17 '13 at 18:47
  • @user43687 not in my book, there a function is absolutely continuous if and only if it's the integral of a locally integrable function. – Daniel Fischer Dec 17 '13 at 19:12
  • hmm...I'm not exactly sure what this has to do with my claim. In one dimension, the claim that absolute continuity implies uniform continuity follows immediately from the definition using partitions (just use a partition with one interval). –  Dec 17 '13 at 19:44
  • I believe the theorem that you are sighting only holds for functions defined on $compact$ intervals. You are talking about the fundamental theorem, no? –  Dec 17 '13 at 19:48
  • Wouldn't $x \mapsto 1/(1-x)$ on $[0,1)$ work for this? – Frank Conry Dec 17 '13 at 18:19
  • @user43687 $\frac1x$ on $(0,1)$ is not uniformly continuous, but it is the integral of the locally integrable function $-\frac{1}{x^2}$. Its restriction to every compact subinterval of its domain is AC, and that is one possible definition of an AC function on a non-compact interval. (And to notify somebody you're addressing in a comment, use @whoever, I just happened by here coincidentally.) – Daniel Fischer Dec 17 '13 at 20:47
  • @DanielFischer So you are saying one definition of AC function on an open interval is that its restriction to compact inervals is AC? This seems to be in conflict with the definition in terms of variation. –  Dec 17 '13 at 21:04
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    I say that's one possible definition, @user43687. Whether it's a good definition depends on what you want to do with it. Sometimes, the really important property is being an integral of an $L^1_{\text{loc}}$ function, then it's convenient to just say AC. Other times, you're interested in a function mapping sets of small measure to sets of small measure, then it's not a good definition. – Daniel Fischer Dec 17 '13 at 21:11

3 Answers3

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Just add the identity function, $\text{id}(x) = x$, to the Cantor function, $\text{c}$. The sum of continuous functions are continuous, and the sum of an increasing function with a strictly increasing one is strictly increasing.

As in the proof that $\text{c}$ is not absolutely continuous choose $\epsilon < 1$. For every $\delta > 0$ there is a finite pairwise disjoint sequence of intervals $(x_k,y_k)$ covering the zero-measure Cantor set with

$$ \sum_{k} |y_{k} - x_{k}| < \delta $$

And since the $\text{c}$ only changes on the Cantor set

$$\sum_{k} |\text{c}(y_{k}) - \text{c}(x_{k})| = 1$$

But

$$\begin{align} (\text{id}(y_{k}) + c(y_{k})) - (\text{id}(x_{k}) + c(x_{k})) &= (\text{id}(y_{k}) - \text{id}(x_{k})) + (c(y_{k}) - c(x_{k})) \\ &\ge c(y_{k}) - c(x_{k}) \end{align}$$

So a fortiori

$$\sum_{k} |(\text{id}(y_{k}) + c(y_{k})) - (\text{id}(x_{k}) + c(x_{k}))| \ge 1$$

Community
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A. Webb
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Let $f:[0,1)\to \mathbb{R}$, $f(x)=\tan(\pi x/2)$. This function is continuous and strictly increasing but not absolutely continuos.

Just to show that this function is in fact not absolutely continuous. Take $\epsilon=1$, and suppose there is a $\delta>0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_{k},y_{k})$ of $[0,1)$ satisfies $\sum_{k}|x_k - y_k| < \delta$ then we have $\sum_{k} |f(y_k)-f(x_k)| < 1$.

Since $\lim_{x \to 1-}f(x) = +\infty$ and $f$ is continuous, let $x_0 \in [1-\delta,1)$ so we can get $y_0 \in [1-\delta,1)$ such that $f(y_0)-f(x_0)>1$.

Using only the interval $(x_0,y_0)$ to test the definition of absolute continuity we have then that $|y_0 - x_0|<\delta$ but $|f(y_0) - f(x_0)|>1$. Therefore, $f$ is not absolutely continuous.

Hans
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Raul C. de Assis
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Counterexample number $8.30$ of "Counterexamples in Analysis" by Gelbaum and Olmsted (which can be found here) provides a continuous, strictly increasing function on $[0,1]$ which is singular. Since it is not constant, it can't also be absolutely continuous.

David Mitra
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yoknapatawpha
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  • Could you please explain in more detail your last sentence "Since it is not constant, it can't also be absolutely continuous."? – Hans Dec 17 '13 at 19:12
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    @Hansen See this. – David Mitra Dec 17 '13 at 19:19
  • @Hansen On compact intervals being absolutely continuous means the function has a derivative almost everywhere and that fundamental theorem of Lebesgue calculus applies. Singular functions have a zero derivative almost everywhere. – A. Webb Dec 17 '13 at 19:26
  • @DavidMitra, A.Webb and yoknapatawpha, thank you all. – Hans Dec 17 '13 at 19:49