Switching derivative and sum of absolutely continuous functions

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!

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 $$a\leq c . This implies that $$f'(t)= \sum g_k'(t)$$ almost evrywhere.

• 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.
– Nick
Dec 22, 2020 at 14:47
• @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. Dec 22, 2020 at 23:14
• 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?
– Nick
Dec 23, 2020 at 15:55
• @Nina $\sum\int_c^{d} g_k'(t)dt=\sum [g_k(d)-g_k(c)]=f(d)-f(c)=\int_c^{d} f'(t)dt$. Dec 23, 2020 at 23:15
• I see, thank you very much for clarifying!
– Nick
Dec 24, 2020 at 0:26