Differentiation under an infinite sum for increasing function Let $f_n : \mathbb{R} \rightarrow [0, +\infty)$ be a sequence of increasing functions, and suppose that
$f(x) = \sum_{n\ge 1} f_n(x) < \infty$ for every $x \in \mathbb{R}$
Prove that
$f'(x) = \sum_{n\ge 1} f_n'(x)$ for almost every $x \in \mathbb{R}$
 A: With a little help from a friend, this one was solved: 
For the reversed inequality, let 
$$ f(x) = S_n(x) + r_n(x)$$, 
Where $S_n(x) = \sum_{k=1}^n f_k (x)$ and $r_n(x) = \sum_{k > n} f_k(x)$. Then we've that 
$$ f'(x) = S_n'(x) + r_n'(x) = \sum_{k=1}^n f_k'(x) + r_n'(x)$$
Now we claim that, for every $[a,b]$ bounded interval, we have that $ r_n' \rightarrow 0 $ in $L^1([a,b])$. Indeed, 
$$ \int_a^b r_n'(t) dt \le \sum_{k>n} \int_a^b f_k'(t) dt \le \sum_{k>n} [f_k(b)-f_k(a)]  $$  
And the last sum clearly goes to zero with $n$. Here we have used, in the first inequality, $f' \ge \sum_k f_k'$ and Fubini's Theorem. Thus, for almost every $x \in [a,b]$ we will have that there exist a subsequence $r_{n_k}' (x)\rightarrow 0$. But as $r_n' = r_{n_k}'-f_{n_k}' - \cdots - f_{n -1}' $, where the last terms are all nonpositive, we conclude that for these $x$ we also have that $r_n'(x) \rightarrow 0$. 
From this, we conclude that, for almost every $x \in [a,b]$, 
$$f'(x) \le \sum_{k=1}^{\infty} f_k'(x) + r_n'(x) \rightarrow \sum_{k=1}^{\infty} f_k'(x)$$
And this proves the claim for these $x$. 
To finish, pick the intervals as $[a,b] = [-N,N]$, and as the intersection of countably many full-measure sets is still a full-measure set, we finish the proof. 
