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.

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.)

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 $y(\epsilon,t)<t$. 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 $[0,1]$.