Differentiating the sum of increasing nonnegative functions Let $f_n:[0,1] \to \mathbb R_{+}$ be a sequence of nondecreasing functions such that $f_n(0) = 0$ and $\sum_{n=1}^{\infty} f_n(1) \leq \infty$. Show that the sum can be differentiated term by term almost everywhere, i.e. $f' = \sum^{\infty}_{n=1} f'$.

It appears that the set up of the problem is hinting Dominated Convergence Theorem. So I want to view the summation as the integration against the counting measure. In order for the theorem to apply, I need a bound on the derivative of $f_n$. Is it possible to derive a bound from the information given?
Since $f_n$ is non-decreasing and $f_n(0)=0$, we have $f_n \ge 0$. Moreover $\sum_{n=1}^\infty f_n(1) &lt; \infty$ (the series should convergence?) implies that the function $g(x):= \sum_{n=1}^\infty f_n(x)$ converges on $[0,1]$ and is also non-decreasing. Any monotonic function is (by a Theorem of Lebesgue) almost everywhere differentiable.

Now, let $g_m(x):= \sum_{n=1}^m f_n(x)$. Since $f_n$ is non-decreasing, we have $f_n' \ge 0$ almost everywhere. This implies that $$g_m' \leq g_{m+1}' \leq g'$$
almost everywhere. Thus $g'_m(x) = \sum_{n=1}^m f_n'(x)$ converges almost everywhere. In order to get $g_m' \rightarrow g'$ it is enough to show this for a subsequence.

For this, choose $m_k$ such that $g(1) - g_{m_k}(1) \le 2^{-k}$. Because $H_k(x) := g(x) - g_{m_k}(x)$ is non-negative, non-decreasing with $H_k \le 2^{-k}$, we can apply the same argument as above for $H_k$. Thus $L:=\sum_{k=1}^\infty H_k$ converges almost everywhere and is differentiable with $\sum_{j=1}^k H_j' \le L'$ for any $k \ge 1$. Hence $\sum_{j=1}^k H_j'$ converges almost everywhere. In particular $H_j' \rightarrow 0$ almost everywhere.

This statement was proven by Fubini and sometimes stated as "Fubini's Theorem on termwise differentiabilty of series with monoic coefficients".
I have myself a different solution to the same problem.
Since each $F_n$ is increasing, $F$ is increasing (on $[0, 1]$). Furthermore, this means $F'$ exists on $(0, 1)$. Since we're only asked for a derivative formula to hold a.e. I'm going to discard the endpoints $0$ and $1$ now to avoid technicalities with one-sided derivatives/limits.
Now, fix $x \in (0, 1)$. By our given formula for $F$, we have the following formula for the symmetric difference quotient where $h &gt; 0$ is sufficiently small so that $x + h, x - h \in (0, 1)$. Such $h$ exists since $(0, 1)$ is an open set. Then
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int_{\mathbb{N}}\frac{F_n\left(x + \frac{h}{2}\right) - F_n\left(x - \frac{h}{2}\right)}{h}d\mu(n),$$
where $d\mu(n)$ denotes the counting measure on $\mathbb{N}$. Now, by applying the Fundamental Theorem of Calculus on the right-hand side,
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int_{\mathbb{N}}\int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{F_n'(y)}{h}dyd\mu(n)$$
Since the functions $F_n$ are all increasing, the integrand on the right-hand side is nonnegative, and measurable with respect to the product measure (which I'll denote $d(y \times n)$). Then by the Fubini Theorem,
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{1}{h}\int_{\mathbb{N}}F_n'(y)dyd\mu(n)$$
The limit on the left-hand side as $h \rightarrow 0$ exists since $F$ is differentiable at $x$, so
$$F'(x) = \lim_{h \rightarrow 0}\int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{1}{h}\int_{\mathbb{N}}F_n'(y)dyd\mu(n)$$
The limit on the right-hand side actually exists as well. The inner integral is independent of $h$, so we only concern with the outer integral. The length of the interval is $h$, and we are dividing by $h$, so this is a mean-value integral in one dimension. Thus, as $h \rightarrow 0$, the integrand approaches the value at the center of the interval. Thus
$$F'(x) = \int_{\mathbb{N}}F_n'(x)d\mu(n),$$
completing the proof. $\square$