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This is problem 25.15 Fubini's 'other' theorem from Rene Schilling's Measures, Integrals, and Martingales. Let $(f_n)_n$ be a sequence of monotone increasing functions $f_n: [a,b] \to \mathbb{R}$. If the series $s(x):= \sum_{n=1}^\infty f_n(x)$ converges, then $s'(x)$ exists a.e. and is given by $s'(x)=\sum_{n=1}^\infty f_n'(x)$ a.e.

Below is the solution to this problem. I can follow all steps of the proof except the last, where it says the first part of the proof applied to this series implies that we can differentiate this series term by term and $\sum_k (s'(x)-s_{n_k}'(x)) $ converges. I can't figure out which part of the proof justifies term by term differentiation here. I would greatly appreciate any help.

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Set

$$\varphi_k(x):= s(x)-s_{n_k}(x).$$

By the inequality from the first part of the proof,

\begin{align*} \frac{\varphi_{k}(x+h)-\varphi_k(x)}{h} &= \frac{s(x+h)-s(x)}{h} - \frac{s_{n_k}(x+h)-s_{n_k}(x)}{h} \geq 0 \end{align*}

for any $h>0$. This proves that $x \mapsto \varphi_k(x)$ is non-decreasing in $x$. Now consider

$$S_n(x) := \sum_{k=1}^n \varphi_k(x) = \sum_{k=1}^n (s(x)-s_{n_k}(x)).$$

By construction, $S(x)=\sum_{k \in \mathbb{N}} \varphi_k(x)$ converges. It follows from the first part of the proof that $S_n$ and $S$ are a.e. differentiable and

$$S_n'(x) \leq S'(x) \quad \text{a.e.}$$

for all $n \in \mathbb{N}$. In particular,

$$\sup_{n \in \mathbb{N}} S_n'(x) = \sup_{n \in \mathbb{N}} \sum_{k=1}^n (s'(x)-s_{n_k}'(x)) \leq S'(x)\quad \text{a.e.},$$

i.e.

$$\sum_{k \in \mathbb{N}} (s'(x)-s_{n_k}'(x)) < \infty\quad \text{a.e.}$$

(Recall that $s'(x)-s_{n_k}'(x) \geq 0$.)

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