Example of absolutely continuous function $f$ with $sqrt{f}$ not absolutely continuous

No Cantor-Lebesgue type of example will work, at least not on a closed interval $[a, b]$. This is because if $f(x)$ is absolutely continuous on $[a, b]$, then so is $(f(x))^p$ for any $p > 0$. The argument goes as follows: $f(x)$ is absolutely continuous on $[a, b]$ iff $f'(x)$ exists and is Lebesgue integrable almost everywhere on $[a, b]$, and $$f(x) = f(a) + int_a^x f'(t) dt$$ for all $x in [a, b]$.

Suppose $f(x)$ is a nonnegative, absolutely continuous function on $[a, b]$. $f(x)$ is absolutely continuous only if it is absolutely continuous on every finite subcollection of closed intervals of $[a, b]$. Since $f(x) equiv 0$ on any subinterval of $[a, b]$ implies $sqrt{f(x)}$ is also identically zero on that subinterval, hence automatically absolutely continuous there, WLOG we can assume the support of $f(x)$ is $[a, b]$. This implies that the set $Z_f := { x in [a, b]: f(x) = 0 }$ has measure zero in $[a, b]$.

Since $f(x)$ is differentiable almost everywhere on $[a, b]$, $sqrt{f(x)}$ also has a Lebesgue integrable derivative almost everywhere on $[a, b]$, as $$frac{d}{dx}sqrt{f(x)} = frac{f'(x)}{2sqrt{f(x)}}$$

wherever $f'(x)$ exists and $f(x) neq 0$, and $Z_f$ has measure zero. So to show $sqrt{f(x)}$ is absolutely continuous, we just need to show that, if $g(x) = frac{f'(x)}{2sqrt{f(x)}}$ wherever this expression is defined and $g(x) = 0$ elsewhere, then $$sqrt{f(x)} = sqrt{f(a)} + int_a^x g(t) dt$$ for any $x in [a, b]$, since the set where $frac{f'(x)}{2sqrt{f(x)}}$ is undefined has measure zero. Since $[a, b] setminus Z_f$ is open and has full measure, it follows that we may write $(a, b) setminus Z_f$ as a finite or countable union of open intervals ${ I_j }_{j in S} = { (a_j, b_j) }_{j in S}$ such that:

• $sum_{j in S} (b_j - a_j) = 1$
• $f(a_j) = f(b_j) = 0$ for all $j in S$, except possibly if $a_j = a$ or $b_j = b$.

Here $S = {1, ..., n }$ for some $n$, or $S = Bbb{N}$.

Clearly $g$ is Lebesgue integrable on each of the $I_j$, and $sqrt{f(x)} - sqrt{f(y)} = int_y^x g(t) dt$ whenever $a_j < y < x < b_j$ for some $j in S$. If we let $y rightarrow a_j$, by continuity of $sqrt{f}$, we can evaluate the improper integral $int_{a_j}^x g(t) dt$ as

$$int_{a_j}^x g(t) dt = lim_{y rightarrow a_j} int_y^x g(t) dt = lim_{y rightarrow a_j} sqrt{f(x)} - sqrt{f(y)} = sqrt{f(x)} - sqrt{f(a_j)}.$$

A similar limiting operation letting $x rightarrow b_j$ shows that $int_{I_j} g(t) dt = int_{a_j}^{b_j} g(t) dt = 0$ whenever $f(a_j) = f(b_j) = 0$. If $U$ is the $I_j$ with left endpoint $a$, we also find that $int_U g(t) dt = -sqrt{f(a)}$. So either $x in U$, in which case $int_a^x g(t) dt = sqrt{f(x)} - sqrt{f(a)}$, or else $x in operatorname{cl}(I_j)$ for some $I_j = (a_j, b_j) neq U$, in which case

begin{align*} int_a^x g(t) dt &= int_a^{a_j} g(t) dt + int_{a_j}^x g(t) dt \ &= int_U g(t) dt + sum_{I_k neq U: I_k subseteq [a, a_j]} int_{I_k} g(t) dt + sqrt{f(x)} - sqrt{f(a_j)} \ &= -sqrt{f(a)} + 0 + sqrt{f(x)} \ &= sqrt{f(x)} - sqrt{f(a)}, end{align*}

which shows that $sqrt{f}$ is absolutely continuous. A similar proof shows that if $f(x)$ is absolutely continuous on $[a, b]$, then so is $(f(x))^p$ for any $p > 0$. That also implies that no positive power of the Cantor function can be absolutely continuous on an interval, since otherwise the Cantor function itself would be.

I believe $f(x) = x^2 (cos frac1x)^4$ is an example on the interval $(0,1)$. While a proof is certainly needed, the key observation is that sum of the infinitely many local maxima of $f$ converges (indeed $f'$ is uniformly bounded), but the sum of the infinitely many local maxima of $sqrt f$ does not converge.