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 $c_nI(x-s_n)$ lies in $[0,c_n]$, so the partial sums of $\sum_n c_nI(x-s_n)$ 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 $\varepsilon>0$ and take $\delta>0$ from [L9] for $f$ at $s$, so that $|f(x)-f(s)|<\varepsilon$ whenever $|x-s|<\delta$. 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 $|\xi_i-s|\le t_{i+1}-t_i<\delta$, so $|S(f,\sigma;P,\xi)-f(s)|<\varepsilon$. By [L7], $\int_a^b f\,d\sigma=f(s)$.

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 $c_n\ge0$ 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 $\alpha_N(x):=\sum_{n<N}c_nI(x-s_n)$, 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 $\int_a^b f\,d\alpha_N=\sum_{n<N}c_nf(s_n)$.

By [L6] there is $M\ge0$ with $|f(x)|\le M$ on $[a,b]$. Since $0\le|c_nf(s_n)|\le Mc_n$ and $\sum c_n$ converges, [L8] makes $\sum_n c_nf(s_n)$ absolutely convergent, hence convergent. By step 2.1 and [L3] the integral $\int_a^b f\,d\alpha$ exists.

Set $\beta_N:=\alpha-\alpha_N$. For each $x$, $\beta_N(x)=\sum_{n\ge N}c_nI(x-s_n)$, 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 $\operatorname{Var}_{[a,b]}(\beta_N)=\sum_{n\ge N}c_n$.

Both $\alpha_N$ and $\beta_N$ are of bounded variation, so [L3] makes $\int_a^b f\,d\alpha_N$ and $\int_a^b f\,d\beta_N$ exist, and $\alpha=\alpha_N+\beta_N$ with [L4] gives $\int_a^b f\,d\alpha=\int_a^b f\,d\alpha_N+\int_a^b f\,d\beta_N$. Using step 2.2 and then [L5] with the bound $M$ of step 3.1, $\Bigl|\int_a^b f\,d\alpha-\sum_{n<N}c_nf(s_n)\Bigr|=\Bigl|\int_a^b f\,d\beta_N\Bigr|\le M\sum_{n\ge N}c_n.$

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