# Prove that $F'(x) = \sum_{n=1}^\infty F_n'(x)$ almost everywhere.

Suppose $F_n$ is a sequence of increasing non-negative right continuous functions on $[0,1]$ such that $\sup_n F_n(1) < \infty$. Let $F = \sum_{n=1}^\infty F_n$ and suppose that $F(1) < \infty$. Prove that $$F'(x) = \sum_{n=1}^\infty F_n'(x)$$ for almost every $x$.

My Attempt: Let $m$ be Lebesgue measure on $[0,1]$ and let $\mu$ be the Lebesgue-Steiltjes measure induced by the increasing function $F$. Since $\mu$ and $m$ are both finite, we can apply the Lebesgue Decomposition Theorem to get the decomposition $\mu = \lambda + \rho$. Also, for each $n$, we get an induced Lebesgue-Steiltjes measure, $\mu_n$. Applying LDT for each $n$, we get $\mu_n = \lambda_n+\rho_n$, $m\perp\lambda$, and $\rho \ll m$. Since $F'$ exists a.e., we have \begin{align*} F'(x) &= \lim_{r\to 0} \frac{F(x+r) - F(x)}{(x+r)-x}\\ &= \lim_{r\to0}\frac{\sum_n[F_n(x+r) - F_n(x)]}{(x+r)-x}\\ &= \lim_{r\to0} \frac{\sum_n\mu_n((x,x+r])}{m((x,x+r])}\\ &= \lim_{r\to0} \frac{\sum_n\lambda_n((x,x+r])}{m((x,x+r])}+ \frac{\sum_n\rho_n((x,x+r])}{m((x,x+r])}\\ &=0 + \lim_{r\to0}\frac{1}{m((x,x+r])}\int_{(x,x+r]}\sum_nF_n'(y)dm \end{align*} We know that our Radon-Nikodym Derivatives are unique and thus, justifying the use of $F_n'(x)$ inside the integral. Since our measure space is finite and $(x,x+r] \subset [0,1]$, our functions $F_n'$ are locally integrable. Thus, we get that $$F'(x) = \sum_nF_n'(x).$$

My problem is that I don't quite see where I'm using the conditions $\sup_n F_n(1) < \infty$ and $F(1) < \infty$.