How do I see that $E - \bigcup_{n \in \mathbb{N}} E_n$ is of measure zero? For $n \in \mathbb{N}$ let $f_n$ be a nondecreasing function on $[a, b]$. Assume that both $\sum_{n \in \mathbb{N}} f_n(a)$ and $\sum_{n \in \mathbb{N}}f_n(b)$ converge and let $f = \sum_{n \in \mathbb{N}} f_n$ on $[a, b]$. Let $A$ be the measure-zero set in $[a, b]$ consisting of all points $x$ such that either $f'(x)$ does not exist or $f_n'(x)$ does not exist for some $n \in \mathbb{N}$. For $n \in \mathbb{N}$ let $E_n$ be the set of points of $[a, b] - A$ where $f_n'$ is nonzero and let $E$ be the set of points of $[a, b] - A$ where $f'$ is nonzero.

Question. How do I see that $E - \bigcup_{n \in \mathbb{N}} E_n$ is of measure zero?

My thoughts on the problem so far are as follows. Assume the contrary. To get a contradiction, apply Vitali's covering technique to points $x$ of $E - \bigcup_{n \in \mathbb{N}} E_n$ where $f'(x) > \alpha$ for some appropriate $\alpha > 0$ in order to construct disjoint open intervals $(x, x + r)$ with$${{f(x + r) - f(x)}\over r} > \alpha$$and$${{f_n(x + r) - f_n(x)}\over r} < \beta$$for $n \le N$ for some appropriate $\beta > 0$ and some appropriate $N \in \mathbb{N}$.
However, I need some help with carrying this out and filling the details. Is it possible someone out there can help me fill in the details?
 A: Suppose the exterior measure $m_*(E - \bigcup_{n \in \mathbb{N}} E_n)$ of $E - \bigcup_{n \in \mathbb{N}} E_n$ is $\delta > 0$. For $m \in \mathbb{N}$ let $S_m$ be the set of points $x$ in $E - \bigcup_{n \in \mathbb{N}} E_n$ with $x \neq b$ such that $f'(x) > {1\over m}$. There exists some $m \in \mathbb{N}$ such that $m_*(S_m) \ge {\delta\over2}$. Choose $N \in \mathbb{N}$ such that$$f(b) - f(a) - \sum_{n = 1}^N (f_n(b) - f_n(a)) = \sum_{n = N + 1}^\infty (f_n(b) - f_n(a)) < {\delta\over{8m}}.$$For $x \in S_m$ there exists $r_x > 0$ with $x + r_x < b$ such that$${{f(x + r_x) - f(x)}\over{r_x}} > {1\over m}$$and$${{f_n(x + r_x) - f_n(x)}\over{r_x}} < {\delta\over{8Nm(b - a)}}$$for $1 \le n \le N$. By Vitali's covering technique we can find a finite number of points $x_1, \ldots, x_k$ of $S_m$ such that $(x_k, x_j + r_{x_j})$ are disjoint for $1 \le j \le k$ and the total measure $\sum_{j = 1}^k r_j$ of $\bigcup_{j = 1}^k (x_j, x_j + r_{x_j})$ is $\ge {\delta\over3}$. By the choice of $r_x$ for a given $x$, we know that$${{f(x_j + r_{x_j}) - f(x_j)}\over{r_{x_j}}} > {1\over m} \text{ and } {{f_n(x_j + r_{x_j}) - f_n(x_j)}\over{r_{x_j}}} < {\delta\over{8Nm(b - a)}}$$for $1 \le j \le k$. Thus,\begin{align*} {\delta\over{3m}} & \le \sum_{j = 1}^k (f(x_j + r_{x_j}) - f(x_j)) \\ & \le \sum_{n = N + 1}^\infty (f_n(b) - f_n(a)) + \sum_{n = 1}^N \sum_{j = 1}^k (f_n(x_j + r_{x_j}) - f_n(x_j)) \\ & \le {\delta\over{8m}} + N{\delta\over{8Nm(b - a)}} \sum_{j = 1}^k r_{x_j} \le {\delta\over{4m}},\end{align*}which is a contradiction.
