Absolute Continuity and the Sharp Fundamental Theorem of Calculus: Examples

The difference quotient at $0$ is $x\sin (1\mathrm{/}{x}^2)$, which tends to $0$; for $x>0$, $F^{\mathrm{\prime }}(x)=2x\sin (1\mathrm{/}{x}^2)-2\cos (1\mathrm{/}{x}^2)\mathrm{/}x$. Thus $F$ is differentiable everywhere.

Put $x_n=(\pi \mathrm{/}2+n\pi {)}^{-1\mathrm{/}2}$. Then $F(x_n)=(-1{)}^n{x}_n^2$, and the finite partitions containing $x_N<x_{N-1}<\cdots <x_0$ have variation at least ${\sum }_{n<N}(x_n^2+x_{n+1}^2)$. This diverges as $N\to \mathrm{\infty }$, so $F$ is not of bounded variation.

Every absolutely continuous function has bounded variation by $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation. Therefore $F$ is not absolutely continuous.

$F(x)={\int }_0^x(2\sqrt{t}{)}^{-1}dt$ and $G(x)={\int }_0^{x}1dt$. Both integrands are in $L^1[0,1]$, so The indefinite integral of an $L^1$ function is absolutely continuous makes $F$ and $G$ AC; their derivatives are respectively $(2\sqrt{x}{)}^{-1}$ and $1$ almost everywhere.

Apply Integration by parts for absolutely continuous functions: its endpoint term is $F(1)G(1)-F(0)G(0)=1$.

The two integrands are respectively $\sqrt{x}$ and $\sqrt{x}\mathrm{/}2$, whose integrals are $2\mathrm{/}3$ and $1\mathrm{/}3$.

The integrand ${\mathbf{1}}_{[0,1]\setminus E}$ lies in $L^1[0,1]$. Thus The indefinite integral of an $L^1$ function is absolutely continuous makes $g$ AC, and The indefinite integral of an $L^1$ function is differentiable almost everywhere gives $g^{\mathrm{\prime }}={\mathbf{1}}_{[0,1]\setminus E}$ a.e.; hence $g^{\mathrm{\prime }}=0$ on $E$ a.e. and $g(1)=1\mathrm{/}2$.

The hypotheses of Change of variables for an increasing absolutely continuous function hold for $g$ and $f$.

Its formula gives the displayed equality, while direct integration gives its common value $1\mathrm{/}4$.

$C$ is continuous, nondecreasing, and maps the ternary Cantor set $K$ onto $[0,1]$.

The set $K$ has Lebesgue measure zero, but $\lambda (C(K))=1$.

Thus $C$ fails $(N)$ as defined in Luzin's property $(N)$ on a compact interval; Banach--Zarecki confirms it cannot be AC.

$g^{\mathrm{\prime }}$ is bounded on $(0,1]$ and $g^{\mathrm{\prime }}(0)=0$, so $g$ is Lipschitz and AC; $h^{\mathrm{\prime }}=1\mathrm{/}(2\sqrt{y})$ lies in $L^1[0,1]$, so $h$ is AC.

$(h\circ g)(x)=\mathrm{\mid }x\sin (1\mathrm{/}x)\mathrm{\mid }$ for $x>0$. On alternating half-waves its total variation has a positive contribution comparable to $1\mathrm{/}n$; the harmonic sum diverges.

Thus the composite is not BV and hence not AC, proving the general failure announced in The composition of two absolutely continuous functions need not be absolutely continuous.

On each $[1\mathrm{/}(n+1),1\mathrm{/}n]$, $F$ is $C^1$, hence maps null sets to null sets. Together with the singleton $\{0\}$ this countable cover proves that $F$ has $(N)$ in the sense of Luzin's property $(N)$ on a compact interval.

$F^{\mathrm{\prime }}(x)=\sin (1\mathrm{/}x)-\cos (1\mathrm{/}x)\mathrm{/}x$ for $x>0$; alternating subintervals again give infinite variation. Thus $F$ is not BV.

The reverse implication in Banach--Zarecki characterisation of absolute continuity requires BV, so this $(N)$ function is not AC.