# Absolute continuity inside the interval extends to the endpoint of the interval under some constraints

The below problem appeared on the UW-Madison Analysis qualifying exam in January 2020. The proof that $$f$$ is absolutely continuous (AC) on the whole interval is still not posted, although the requested counterexample has been given.

I don't seem to find version of this problem in the site, but I am sure this is pretty standard type of question.

The problem goes likes this: Let $$f$$ be of bounded variation on $$[0,1]$$ and AC on $$[\varepsilon,1]$$ for all $$\varepsilon >0$$. $$f$$ is also continuous at $$0$$. Prove $$f$$ is absolutely continuous on whole interval $$[0,1]$$.

Moreover I am looking for a counterexample in the case when the bounded variation of $$f$$ is dropped.

Here is what I think,

Using continuity, I can find $$\delta>0$$ for given $$\epsilon >0$$ which bounds the sum in the definition of AC upto $$\delta$$. Then in the interval $$[\delta,1]$$, I can use given hypothesis. But I am not totally comfortable writing this rigorously.

For the counter example I can use $$f(0)=0$$ and $$f(x)= x\sin (1/x)$$ for $$x$$ not equal to $$0$$. I would love to see the rigorous proof and rigorous proof of counterexample. Thank you in advance.

• I think the best way to show this is by showing that the total variance in $[0, \epsilon]$ is small for small epsilon. Jan 3 at 6:29

This is for the counterexample part. If $f$ is absolutely continuous on $[0,1]$, it is of bounded variation. So being of bounded variation is a necessary condition for the conclusion. Any function which is not of bounded variation but satisfies the other hypotheses will provide a counterexample.

As you indicated, the function $f(0)=0$ and $f(x)=x\sin(1/x)$ for $x\ne 0$ is not of bounded variation on $[0,1]$. This can be seen by evaluating $f$ at the points where $\sin$ is $1$ or $-1$, namely $x={1\over\pi(2k+1/2)}$ and $x={1\over\pi(2k+3/2)}$. It follows that the total variation of $f$ on $[0,1]$ is at least equal to a constant times the sum of a harmonic series, which diverges. So $f$ is not of bounded variation.

EDIT: The function $f$ above is also continuous at $0$, and is absolutely continuous on each interval $[\varepsilon,1]$ for $\varepsilon>0$. This last follows from the fact that on each such interval $f$ has a bounded derivative. (Apply the mean value theorem to each subinterval in the definition of absolute continuity.)

• You have to be careful when you say any function of bounded variation. You also want the function be absolutely continuous on $[\varepsilon,1]$. Dec 25, 2012 at 3:57
• Yes, I meant not of bounded variation. The condition on $[\varepsilon,1]$ is very relevant to the discussion at hand. We're looking for a counterexample to a particular theorem, we've only dropped the assumption that $f$ is of bounded variation. In particular we still want $f$ to be continuous on $[0,1]$ and absolutely continuous on $[\varepsilon,1]$, but absolutely continuity to fail on $[0,1]$. The function you give, indeed satisfies all of these. But any function not of bounded variation does not. Dec 25, 2012 at 4:25
• What @JacobSchlather is saying is absolutely correct. We do want the function which is continuous on $[0,1]$ and absolutely continuous on $[\epsilon,1]$. Dec 25, 2012 at 5:17
• I really want to see how one can prove the function mentioned above is absolutely continuous on $[\epsilon,1]$ Dec 25, 2012 at 5:19
• Thanks @JacobSchlather, I have edited the answer to make that clear. Dec 25, 2012 at 22:46

I would like to first show that $$TV([0,x])$$ is small for small $$x$$. Suppose $$TV([0,x])>C > 0$$ for all $$x> 0$$. Then $$C' = \inf_{x>0}{TV([0,x])} \geq C.$$ There is a partition of $$[0,x]$$ that yields a variation higher than or equal to $$C'$$. Let that partition be $$I_x=(x_i,y_i)_{i=1}^n$$, i.e. $$\sum_i|f(x_i)-f(y_i)|\geq C'.$$

Let $$\epsilon > 0$$. Because $$f$$ is continuous at $$0$$, there is a $$\delta(\epsilon) > 0$$ such that $$|f(0)-f(x)| < \epsilon$$ for all $$|x| < \delta(\epsilon)$$. Because $$f$$ is absolutely continuous on $$[t, 1],$$ there is $$\gamma(\epsilon,t)>0$$ such that $$\gamma(\epsilon, t)=\sup \gamma$$ where any finite disjoint set of intervals in $$[t,1]$$ with total length less than $$\gamma$$ has variation less than $$\epsilon$$. Let $$g(\epsilon,t)=\min(\gamma, \delta).$$ Let $$x_1=0, y_1=y(\epsilon,t)$$ in $$I_{g(\epsilon,t)}$$. $$\sum_{n=2}^N |f(x_n)-f(y_n)| \geq C'-\epsilon>C/2.$$ So it must be the case that $$\gamma(C/2, y(\epsilon,t)) < g(\epsilon,t)-y(\epsilon,t)<\gamma(\epsilon,t)$$. Choose $$\epsilon$$ such that $$y(\epsilon,t). Because $$\gamma$$ is non-increasing in the second variable, $$\gamma(C/2,t)\leq \gamma(C/2,y(\epsilon,t)) < \gamma(\epsilon,t).$$ $$\gamma$$ is nondecreasing in the first variable. So $$C/2 < \epsilon$$ which is false for small $$\epsilon$$. We have shown that $$TV([0,\epsilon])$$ goes to zero as $$\epsilon$$ gets smaller.

Once we have that fact about the total variation of $$[0,\epsilon]$$ the problem becomes straightforward. Choose $$t>0$$ such that $$TV([0,t]) < \epsilon/2.$$ Let $$\delta = \gamma(\epsilon/2,t)$$. Finally if $$\delta > \sum |x_i - y_i|$$, $$\sum_{x_i,y_i\in [t,1]} |f(x_i)-f(y_i)| + \sum_{x_i,y_i\in [0,t]} |f(x_i)-f(y_i)| < \epsilon/2 + \epsilon/2 = \epsilon.$$ So $$f$$ is AC on $$[0,1]$$.