# Example of continuous but not absolutely continuous strictly increasing function

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?

• the Cantor function is not strictly increasing. Dec 17, 2013 at 18:10
• The Cantor function is the most classical example. See math.stackexchange.com/questions/125345/… Dec 17, 2013 at 18:17
• Wouldn't $x \mapsto 1/(1-x)$ on $[0,1)$ work for this? Dec 17, 2013 at 18:19
• When the interval is not compact, any continuous function which is not uniformly continuous will do (since absolute continuity implies uniform continuity).
– user113529
Dec 17, 2013 at 18:47
• 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. Dec 17, 2013 at 21:11

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

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

• This maps to $[0,2]$ but scale by $1/2$ and choose $\epsilon < 1/2$ as well if $[0,1]$ is desired. Dec 17, 2013 at 19:13
• Excellent, thank you, A. Webb.
– Hans
Dec 17, 2013 at 19:48

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.

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.

• 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, 2013 at 19:12
• @Hansen See this. Dec 17, 2013 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. Dec 17, 2013 at 19:26
• @DavidMitra, A.Webb and yoknapatawpha, thank you all.
– Hans
Dec 17, 2013 at 19:49