$\sum_{n = 1}^\infty f_n =f.$ Prove that $f′_n =f′$ a.e. I am studying for a real analysis qualifying exam.  Was hoping that there was a very slick proof for this?  Thanks.
Let $f_1, f_2, . . . , f : [0, 1] → \mathbb{R}$ be non-decreasing right-continuous functions such that $\sum_{n = 1}^\infty
f_n =f.$ Prove that $\sum_{n = 0}^\infty$ $f′_n =f′$ a.e.

You can write $f_n(x)=\mu_n([0,x])$ for some nonnegative measure $\mu_n$. Write that as $d\mu_n=f'_n\,d\lambda+d\nu_n$, where $\lambda$ is Lebesgue measure and $\nu_n$ is singular. Do the same for $f$: $d\mu=f'\,d\lambda+d\nu$. Let $N$ be a (Borel) null set where $\nu_n([0,1]\setminus N)=0$ for all $n$, and likewise $\nu([0,1]\setminus N)=0$.
By assumption $\sum_n\mu_n([0,x])=\mu([0,x])$ for all $x$, and it follows that $\sum_n\mu_n(E)=\mu(E)$ for all Borel sets $E$. When $E\cap N=\emptyset$ then (using the montone convergence theorem in the first part)
$$ \int_E\sum_nf_n'\,d\lambda=\sum_n\int_E f_n'\,d\lambda=\sum_n\mu_n(E)=\mu(E)=\int_E f'\,d\lambda,$$
and we're done.
