Bounded Variation and the Riemann–Stieltjes Integral: Examples and Counterexamples

By [L1], $f$ is $1$-Lipschitz, and [L2] makes it absolutely continuous.

At zero, $(\mathrm{\mid }h\mathrm{\mid }-\mathrm{\mid }0\mathrm{\mid })\mathrm{/}h$ equals $1$ for $h>0$ and $-1$ for $h<0$. The two one-sided limits differ, so [L3] shows that $f^{\mathrm{\prime }}(0)$ does not exist. Therefore $f$ is not $C^1$ on $[-1,1]$.

Let $[u_j,v_j]$ be pairwise nonoverlapping subintervals. Fix $r>0$ and split at $r$ the one family interval, if any, that crosses $r$; monotonicity makes this split preserve its endpoint increment. The resulting pieces contained in $[0,r]$ contribute at most $\sqrt{r}$ in total, because their increments telescope after gaps are filled. On every remaining interval, whose left endpoint satisfies $u_j\ge r$, [given] $\sqrt{v_j}-\sqrt{u_j}=\frac{v_j-u_j}{\sqrt{v_j}+\sqrt{u_j}}\le \frac{v_j-u_j}{2\sqrt{r}}\mathrm{.}$

Given $\epsilon >0$, choose $r>0$ with $\sqrt{r}<\epsilon \mathrm{/}2$, then require ${\sum }_j(v_j-u_j)<\epsilon \sqrt{r}$. Steps 1.1 and 1.2 make the total endpoint increment less than $\epsilon$, proving absolute continuity by [L2].

If a Lipschitz constant $K$ existed, the pair $0$ and $1\mathrm{/}{n}^2$ would give $n^{-1}\le K{n}^{-2}$, hence $n\le K$ for every positive integer $n$, contradicting the Archimedean property. Thus [L3] fails.

Monotonicity makes every variation sum telescope after its absolute values are removed, so ${\mathrm{Var}}_{[0,1]}(c)=c(1)-c(0)=1$. Thus $c$ is BV.

For the finite disjoint family of the $2^n$ surviving intervals, the total length is $(2\mathrm{/}3{)}^n$ but the sum of endpoint increments is $2^n{2}^{-n}=1$. By [L3], the former is eventually smaller than every prescribed $\delta >0$, while the latter never falls below, say, $1\mathrm{/}2$. This contradicts the definition of absolute continuity.

By [L1] and the algebra of limits, $f$ is continuous for $x>0$. By [L2], $\mathrm{\mid }f(x)\mathrm{\mid }\le x\mathrm{/}2$, so $f(x)\to 0=f(0)$ as $x\downarrow 0$. Hence $f$ is continuous on $[0,1]$.

For every integer $n\ge 1$, $f(1\mathrm{/}n)=0$, while [given] $f\left(\frac{1}{n+1\mathrm{/}2}\right)=\frac{1}{2n+1}\mathrm{.}$ The ordered partition containing these alternating zeros and peaks for $1\le n\le N$ therefore has variation sum at least $2{\sum }_{n=1}^N(2n+1{)}^{-1}$, apart from at most one endpoint term.

Since $(2n+1{)}^{-1}\ge (3n{)}^{-1}$ and the harmonic partial sums are unbounded by [L3], the variation sums in step 1.2 are unbounded. Thus the continuous function $f$ is not BV.

In every tagged partition, all increments of $H_c$ vanish except the one on the interval across its jump; that increment is one. Its tag $\xi$ satisfies $\mathrm{\mid }\xi -c\mathrm{\mid }\le \parallel P\parallel$, so the complete Stieltjes sum is $f(\xi )$ and tends to $f(c)$ by [L2].

The same computation counts a jump at $b$. A jump at $a$ is represented by $H(a)=0$ and $H(x)=1$ for $x>a$ and likewise contributes $f(a)$. By contrast, merely assigning a constant endpoint value on the whole interval creates no increment and hence has integral zero.

The constant term has every increment equal to zero. By [L1], each $H_{c_j}$ contributes $f(c_j)$, and finite linearity [L2] gives the displayed sum.

Ordering the distinct jump points prevents double counting. The jump points are interior because [L1] places them strictly inside, and the restriction is not cosmetic: $H_a$ takes the value $1$ at every point of $[a,b]$, so a jump placed at $a$ makes every increment zero and contributes nothing, while the weighted sum would still count $wf(a)$.

For each positive integer $n$, $f(1\mathrm{/}n)=n$, so the range of $f$ is not bounded above.

Every Stieltjes sum equals $f(\xi )$ for a tag $\xi$ in the interval across $1\mathrm{/}2$; as the mesh tends to zero, $\xi \to 1\mathrm{/}2$. By [L2], these sums tend to $f(1\mathrm{/}2)=2$. The unbounded behavior near zero is multiplied only by zero increments of $\alpha$.

Monotonicity and [L1] make $c$ BV, so existence follows from [L2]. Every tagged sum for the constant integrand telescopes: [L1, L2] ${\sum }_{i}1(c(t_{i+1})-c(t_i))=c(1)-c(0)=1.$

Thus its common limit is one. This computation invokes neither a derivative of $c$ nor measure theory.

By [L2], both $f$ and $\alpha$ have total variation one. For every $h>0$ small enough, choose a partition containing $c-h$ and $c$. The only nonzero integrator increment occurs on $[c-h,c]$.

Tag that interval first at $c-h$ and then at $c$. The corresponding sums are respectively $f(c-h)=0$ and $f(c)=1$, although both partitions have mesh tending to zero after the other intervals are refined. Thus no common mesh limit exists, contradicting the necessary condition [L1].

Directly from the difference quotient, [given] $\frac{(x+h{)}^2-x^2}{h}=2x+h⟶2x,$ so ${\alpha }^{\mathrm{\prime }}(x)=2x$. Likewise, $(2x^3\mathrm{/}3{)}^{\mathrm{\prime }}=2x^2$ by expanding the difference quotient.

Therefore [L1] and [L2] give [L1, L2] ${\int }_0^{1}xd(x^2)={\int }_0^{1}2x^{2}dx={\left[\frac{2x^3}{3}\right]}_0^1=\frac{2}{3}\mathrm{.}$ ∎

The map $\varphi (t)=t^2$ is a strictly increasing continuous bijection of $[0,1]$ onto itself. Applying [L1] gives ${\int }_0^{1}xdx={\int }_0^1{t}^{2}d(t^2)$.

Direct difference quotients give $(t^2{)}^{\mathrm{\prime }}=2t$ and $(t^4\mathrm{/}2{)}^{\mathrm{\prime }}=2t^3$. Hence [L2], [L3], and the FTC give [L2, L3] ${\int }_0^{1}xdx=\frac{1}{2},{\int }_0^1{t}^{2}d(t^2)={\int }_0^{1}2t^{3}dt=\frac{1}{2}\mathrm{.}$ ∎

By [L1], $0<S<\mathrm{\infty }$ and ${\sum }_n{w}_n=1$, so the intervals tile $(0,1]$. On one tent, the linear slope estimate and [L2] give $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le 2\mathrm{\mid }x-y{\mathrm{\mid }}^{3\mathrm{/}4}$. If $x,y$ lie in different tents, let $z_x$ be the endpoint of $x$'s tent toward $y$ and $z_y$ the endpoint of $y$'s tent toward $x$. Both have value zero and $\mathrm{\mid }x-z_x\mathrm{\mid },\mathrm{\mid }y-z_y\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }$, so the two one-tent estimates and the triangle inequality give $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le 4\mathrm{\mid }x-y{\mathrm{\mid }}^{3\mathrm{/}4}$. Taking the endpoints of successively smaller tents gives the same estimate at zero. Thus $f$ is $3\mathrm{/}4$-Hölder.

A partition through the endpoints and peaks of the first $N$ tents has variation at least [given] $2{\sum }_{n=1}^N{w}_n^{3\mathrm{/}4}=2S^{-3\mathrm{/}4}{\sum }_{n=1}^N\frac{1}{n}\mathrm{.}$ This is unbounded by [L1], so $f$ is not BV.

Since $3\mathrm{/}4+3\mathrm{/}4>1$, [L3] nonetheless gives existence of ${\int }_0^{1}fdf$. This is genuinely outside the BV existence theorem.