A countable pure-step integrator evaluates a continuous integrand as the absolutely convergent weighted sum of its values at the jumps

Fix $x\in [a,b]$. Each term $c_{n}I(x-s_n)$ lies in $[0,c_n]$, so the partial sums of ${\sum }_n{c}_{n}I(x-s_n)$ are nondecreasing and bounded above by ${\sum }_n{c}_n$. By [L1] the series converges and $\alpha (x)$ is defined, with $0\le \alpha (x)\le {\sum }_n{c}_n$.

Fix $s\in (a,b)$ and put $\sigma (x):=I(x-s)$ on $[a,b]$. Then $\sigma$ is nondecreasing, hence of bounded variation by [L2]. Let $\epsilon >0$ and take $\delta >0$ from [L9] for $f$ at $s$, so that $\mathrm{\mid }f(x)-f(s)\mathrm{\mid }<\epsilon$ whenever $\mathrm{\mid }x-s\mathrm{\mid }<\delta$. Let $(P,\xi )$ be a tagged partition of mesh below $\delta$. The increment $\sigma (t_{i+1})-\sigma (t_i)$ is $1$ when $t_i\le s<t_{i+1}$ and $0$ otherwise, and because $a\le s<b$ exactly one index $i$ satisfies $t_i\le s<t_{i+1}$. Hence $S(f,\sigma ;P,\xi )=f({\xi }_i)$ for that index, and $t_i\le {\xi }_i\le t_{i+1}$ with $t_i\le s<t_{i+1}$ give $\mathrm{\mid }{\xi }_i-s\mathrm{\mid }\le t_{i+1}-t_i<\delta$, so $\mathrm{\mid }S(f,\sigma ;P,\xi )-f(s)\mathrm{\mid }<\epsilon$. By [L7], ${\int }_a^{b}fd\sigma =f(s)$.

If $x\le y$ then $I(x-s_n)\le I(y-s_n)$ for every $n$, because $I$ is nondecreasing and $x-s_n\le y-s_n$. Multiplying by $c_n\ge 0$ and summing, every partial sum for $x$ is at most the corresponding partial sum for $y$, so the limits satisfy $\alpha (x)\le \alpha (y)$ by [L1]. Thus $\alpha$ is nondecreasing, and $\alpha =\alpha -0$ exhibits it as a difference of two nondecreasing functions, so [L2] gives $\alpha$ bounded variation.

For $N\in \mathbb{N}$ set ${\alpha }_N(x):={\sum }_{n<N}{c}_{n}I(x-s_n)$, a finite sum. Each summand is a nonnegative multiple of a function of the form treated in step 1.2, so applying [L4] finitely many times, with the integral of each summand supplied by step 1.2, gives ${\int }_a^{b}fd{\alpha }_N={\sum }_{n<N}{c}_{n}f(s_n)$.

By [L6] there is $M\ge 0$ with $\mathrm{\mid }f(x)\mathrm{\mid }\le M$ on $[a,b]$. Since $0\le \mathrm{\mid }{c}_{n}f(s_n)\mathrm{\mid }\le M{c}_n$ and $\sum c_n$ converges, [L8] makes ${\sum }_n{c}_{n}f(s_n)$ absolutely convergent, hence convergent. By step 2.1 and [L3] the integral ${\int }_a^{b}fd\alpha$ exists.

Set ${\beta }_N:=\alpha -{\alpha }_N$. For each $x$, ${\beta }_N(x)={\sum }_{n\ge N}{c}_{n}I(x-s_n)$, the tail of the series in step 1.1; the argument of steps 1.1 and 2.1 applies verbatim to it, so ${\beta }_N$ is nondecreasing with bounded variation. Since $s_n\in (a,b)$ we have $I(a-s_n)=0$ and $I(b-s_n)=1$, so ${\beta }_N(a)=0$ and ${\beta }_N(b)={\sum }_{n\ge N}{c}_n$. A nondecreasing function has every partition sum equal to ${\beta }_N(b)-{\beta }_N(a)$, because each increment is nonnegative and the sum telescopes, so [L2] gives ${\mathrm{Var}}_{[a,b]}({\beta }_N)={\sum }_{n\ge N}{c}_n$.

Both ${\alpha }_N$ and ${\beta }_N$ are of bounded variation, so [L3] makes ${\int }_a^{b}fd{\alpha }_N$ and ${\int }_a^{b}fd{\beta }_N$ exist, and $\alpha ={\alpha }_N+{\beta }_N$ with [L4] gives ${\int }_a^{b}fd\alpha ={\int }_a^{b}fd{\alpha }_N+{\int }_a^{b}fd{\beta }_N$. Using step 2.2 and then [L5] with the bound $M$ of step 3.1, $\mid {\int }_a^{b}fd\alpha -{\sum }_{n<N}{c}_{n}f(s_n)\mid =\mid {\int }_a^{b}fd{\beta }_N\mid \le M{\sum }_{n\ge N}{c}_n\mathrm{.}$

Convergence of $\sum c_n$ makes its tails ${\sum }_{n\ge N}{c}_n$ tend to $0$ as $N$ increases, so given $\epsilon >0$ the right side of step 4.1 is below $\epsilon$ for all large $N$. Hence the partial sums ${\sum }_{n<N}{c}_{n}f(s_n)$ converge to ${\int }_a^{b}fd\alpha$, and by step 3.1 that series converges absolutely. By [L1] and [L9] its sum is ${\int }_a^{b}fd\alpha$, which is the claimed identity.