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Suppose that $(f_n)$ is a sequence of monotone non-decreasing functions on $[a,b]$ such that $f(x) = \sum_{n=1} ^\infty f_n (x)$ is finite for each $x\in [a,b]$. By Lebesgue's theorem on monotone functions, each $f_n$ is differentiable almost everywhere, and it is clear that $f$ is monotone as well, so $f$ is also differentiable a.e.

Must it be the case that $\sum_{n=1} ^\infty f_n' = f'$ almost everywhere? I know that for each $N$ and for each $h \in \mathbb R$, $ \frac{1}{h}\sum_{n=1} ^N [f_n(x+h) - f(x)] \leq \frac{1}{h}\sum_{n=1} ^\infty [f_n(x+h) - f(x)] $ and taking the limit as $h \rightarrow 0$ gives $\sum_{n=1} ^N f_n '\leq f'$, and so $\sum_{n=1} ^\infty f_n ' \leq f'$. Is the reverse inequality also true?

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Yes, equality holds almost everywhere. For the sake of convenience assume that $f_n \geq 0$ for all $n$, otherwise pointwise assume absolute convergence of the series so that we can replace $f_n$ by $f_n - f_n(a) \geq 0$.

Put $$F_N = \sum_{n=1}^{N} \; f_n$$ so that $F_N \to f$ everywhere and monotonically.

Choose an increasing sequence $\{N_k\}_{k=1}^{\infty}$ such that $0 \leq f(b) - F_{N_k}(b) \leq 2^{-k}$. Then we have $$ \sum_{k=1}^{\infty} \left(f(b) - F_{N_k}(b)\right) \leq 1. $$ Now put $$ g(x) = \sum_{k=1}^{\infty} \left(f(x) - F_{N_k}(x)\right) = \sum_{k=1}^{\infty} \sum_{n=N_k+1}^{\infty} f_{n}(x). $$ Observe that the inner sums $\sum_{n=N_k+1}^{\infty} f_{n}(x)$ are monotonically increasing functions of $x$, so $0 \leq g(x) \leq g(b) \leq 1$ for all $x \in [a,b]$, so $g$ is everywhere defined and monotonically increasing, too. Thus, $g$ is differentiable almost everywhere and your argument shows furthermore that $$ 0 \leq \sum_{k=1}^{\infty} \left(f'(x) - F_{N_k}^\prime(x)\right) \leq g'(x) $$ almost everywhere. Since the summands of a convergent series must tend to zero, this means that at these points $F_{N_{k}}^\prime (x) \to f'(x)$ as $k \to \infty$, as desired.

Note: I learned this argument from Fremlin, Measure Theory, Volume 2, Exercise 222Y a), page 62.

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