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Let $g_k$ be a sequence of absolutely continuous functions on $[a,b]$. Suppose $\sum_{k=1}^ \infty g_k(x)$ is convergent for all $x \in [a,b]$ and define $f(x) = \sum_{k=1}^ \infty g_k(x)$. Also suppose that $\sum_{k=1}^\infty \int_a^b |g_k'(x)|dx < \infty$. From these hypotheses, I have shown that $f$ is absolutely continuous on $[a,b]$. I am trying to show now that $f'(x) = \sum_{k=1}^\infty g_k'(x)$ for a.e. $x \in [a,b]$ but I am not sure how to proceed.

I tried applying this theorem as every absolutely continuous function is a difference of increasing functions: https://en.wikipedia.org/wiki/Fubini%27s_theorem_on_differentiation. But I'm unsure about the convergence of the increasing functions so don't think that works. I'd appreciate any help on this problem!

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Recall that AC functions are indefinite integrals of their derivatives. It is shown in Rudin's RCA (Theorem 7.19) that any AC continuous function on $[a,b]$ is the difference of two AC increasing functions. So the proof reduces to the case when the functions are all increasing. In this case $\int_c^{d} \sum g_k'(t)dt= \sum\int_c^{d} g_k'(t)dt$ by Tonelli's Theorem. This gives $\int_c^{d} f'(t)dt=\int_c^{d} \sum g_k'(t) dt$ whenever $a\leq c <d \leq b$. This implies that $ f'(t)= \sum g_k'(t) $ almost evrywhere.

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  • $\begingroup$ I have a few questions about the proof. Why can you assume that $\sum g'_k(t)$ is nonnegative in your application of Tonelli's Theorem? Also, why does this imply that $\int_c^d f'(t) dt = \int_c^d \sum g_k'(t)dt$? I'm also unclear on where the increasing property of the functions is being used. $\endgroup$
    – Nick
    Dec 22, 2020 at 14:47
  • $\begingroup$ @Nina If $f$ is increasing then $f'(x) \geq 0$ whenever $f'(x)$ exists. $\int_c^{d} f'(t)dt=\int_c^{d} \sum g_k'(t)dt=\sum\int_c^{d} g_k'(t)dt$ (by Tonelli's Theorem) since the terms are non-negative. $\endgroup$ Dec 22, 2020 at 23:14
  • $\begingroup$ Thank you! I understand the increasing property and use of Tonelli's Theorem now. Are you applying the theorem I linked to obtain that first equality in your comment then? $\endgroup$
    – Nick
    Dec 23, 2020 at 15:55
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    $\begingroup$ @Nina $\sum\int_c^{d} g_k'(t)dt=\sum [g_k(d)-g_k(c)]=f(d)-f(c)=\int_c^{d} f'(t)dt$. $\endgroup$ Dec 23, 2020 at 23:15
  • $\begingroup$ I see, thank you very much for clarifying! $\endgroup$
    – Nick
    Dec 24, 2020 at 0:26

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