UM Analysis Qual September 2013 I am trying to study for quals and I encountered this question from UM September 2013.   
Let $\space f_1, f_2,..., f: [0,1] \rightarrow \mathbb{R}$ be non-decreasing right-continuous functions such that $\sum_{n=1}^{\infty} f_n =f$.  Prove that $\sum_{n=1}^{\infty} f'_n =f'$.  
I want to think of the right continuous functions as induced Lebesgue-Stieltjes measure and apply Lebesgue-Radon-Nikodym Theorem.  I am not sure how the usual function derivative will play a role with the Radon-Nikodym derivative.  
 A: Key steps, with some results being merely alluded to that may require more work to show rigorously:


*

*First, extend the functions so that $f_j(x)=f_j(0)$ for all $x<0$ and $f_j(x)=f_j(1)$ for all $x>1$. Let $\mu_{j}$ be the Borel measure for which $\mu_{j}((-\infty,x])=f_j(x)-f_j(0)$ for all $x$. It can be shown that such a Borel measure exists and is unique. From now on, assume wlog that $f_j(0)=0$.

*Consider the Lebesgue-Radon-Nikodym representation of $\mu_j$: $$\mu_j(E)=\lambda_j(E)+\int_{E}g_j\,\mathrm{d}m$$ for any Borel set $E$, where $m$ is the Lebesgue measure and $\lambda_j\perp m$ and $g_j\in L^1(m)$. (Note that $\mu_j$ is a finite nonnegative measure.)

*Let $\mu\equiv \sum_j\mu_j$, $\lambda\equiv\sum_j\lambda_j$ (these are legit measures), and $g\equiv\sum_j g_j$. Show that $\lambda\perp m$ and $g\in L^1(m)$. Hence, $$\mu(E)=\lambda(E)+\int_E g\,\mathrm{d} m$$ is the unique LRN representation of $\mu$.

*On the other hand, $f(x)=\sum_j f_j(x)=\sum_j\mu_j((-\infty,x])=\mu((-\infty,x])$. Since $\mu$ is a finite measure, it can be shown that $$\lim_{r\to0}\frac{\mu((x,x+r])}{m((x,x+r])}=\lim_{r\to0}\frac{f(x+r)-f(x)}{r}=g(x)$$ for $m$-a.e. $x$, and similarly for the limit from the left. Hence, $f$ is differentiable a.e. and $f'=g$ a.e.

*Repeating the same procedure for each $j$, $f_j$ is differentiable a.e. and $f_j'=g_j$ a.e.

*Conclusion: since the set of points where one or more of the results above does not hold (i.e., points outside any of the a.e. sets) is a countable union of null sets and hence a null set itself, $f'=g=\sum_{j}g_j=\sum_j f_j'$ a.e. (Note: the a.e. stipulation cannot be weakened if $f$ is merely right-continuous or is continuous but not differentiable everywhere.)
This proof is basically exercise 3.39 in Folland (1999), and every result needed for it can be found in chapter 3 and theorem 1.16 in this book. See also here.
