Let $f_1, f_2,f_3,...$ be increasing function on $[a,b]$. If $\sum_k f_k$ converges point wise to f on [a,b], show that $\sum_k f'_k$ converges to $f'$ a.e on $[a,b]$.
This is not a statement question. I tried to solve this question but I doubt there is a missing information! I can not understand where the a.e. convergence of the series of derivatives comes from? should I use Egoroff's theorem to obtain the uniform convergence of the main series and then use the term by term derivation theorem? Thanks.
My Proof
Consider the partial sum $F_n (x)= \sum_{k=1}^n f_k (x)$. We are give that $F_n \to f$ pointwise as $n\to \infty$. Since $f_k$ is increasing for each $k$. Then $F_n$ is increasing for each $n$. Then $\{F_n\}$ a sequence of measurable (because increasing) functions that converges pointwise to $f$. By Egoroff's theorem, there is subset $A \subset [a,b]$ and $|A|=0$ such that $F_n \to f$ uniformly on $[a,b] \setminus A$. Then by term-by-term differentiation we have $F'_n \to f'$ pointwise on $[a,b] \setminus A$. Then $F'_n \to f'$ a.e. on $[a,b]$.
Notation
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Let $$r_n(x) = \sum_{k>n} f_k(x)$$ Let $$f(x) = \sum_{k=1}^n f_k(x) + r_n(x)$$ Hence $$f'(x) = \sum_{k=1}^n f'_k(x) + r'_n(x) $$
$$\int_a^b f'(x) - \sum_{k=1}^n f'_k(x) = \int_a^b r'_n(x) dx$$
$$f(b) - f(a) - \left(\sum_{k=1}^n (f_k(b) - f_k(a))\right) = \int_a^b r'_n(x) dx$$
$$\left(f(b) - \sum_{k=1}^n f_k(b)\right) - \left(f(a) - \sum_{k=1}^n f_k(a))\right) = \int_a^b r'_n(x) dx$$
By choose $n$ large enough such that $\left|f(b) - \sum_{k=1}^n f_k(b)\right| < \epsilon/2$ and $\left|f(a) - \sum_{k=1}^n f_k(a)\right| < \epsilon/2$ we have $$\int_a^b r_n'(x) dx \leq \epsilon$$
Let $$A_m = \cap_{n = 1}^{\infty} \cup_{k \geq n} \left\{x: r_k'(x) > \frac{1}{m} \right \}$$
for every $x \in [a,b] \setminus A_m$, we have for all $n \geq N$ for some $N$, $$r'_n(x) \leq \frac{1}{m} \implies |f'(x) - \sum_{k=1}^n f'_k(x)| \leq \frac{1}{m}$$
Hence its enough if we prove that $$\lim_m \mu(A_m) = 0$$
Hence $$\mu(A_m) = \mu\left(\cap_{n = 1}^{\infty} \cup_{k \geq n} \left\{x: r_k'(x) > \frac{1}{m} \right \}\right) \leq \lim_n \sum_{k \geq n} \mu(\left\{x: r_k'(x) > \frac{1}{m} \right \})$$
By choosing a subsequence $n_k$ we can make sure $\int_a^b r'_{n_k}(x) dx \leq \frac{\epsilon^2}{2^k}$
and hence if we have $$B_{1/m} = \cap_{n = 1}^{\infty} \cup_{k \geq n} \left\{x: r_{n_k}'(x) > \frac{1}{m} \right \}$$
Then, $$\mu(B_{\epsilon}) = \mu\left(\cap_{n = 1}^{\infty} \cup_{k \geq n} \left\{x: r_{n_k}'(x) > \epsilon \right \}\right) \leq \lim_n \sum_{k \geq n} \mu(\left\{x: r_{n_k}'(x) > \epsilon \right \})$$
Since $\int_a^b r'_{n_k}(x) dx \leq \frac{\epsilon^2}{2^k} \implies \mu(\left\{x: r_{n_k}'(x) > \epsilon \right \}) \leq \frac{\epsilon}{2^k}$
Hence $$\mu(B_{\epsilon}) \leq \lim_n \sum_{k \geq n} \mu(\left\{x: r_{n_k}'(x) > \epsilon \right \}) \leq \lim_n \sum_{k \geq n} \frac{\epsilon}{2^k} = 0$$.
Hence we have that $$r'_{n_k}(x) = f'(x) - \sum_{m=1}^{n_k} f_m'(x) \rightarrow 0$$.
$$|r'_{n_k}(x)| = |r'_{n}(x) + \sum_{m = n}^{n_k} f_m'(x)| \leq \epsilon$$ But all terms in the above are positive, Hence $$|r'_n(x)| \leq \epsilon$$.
Hence $$r_n'(x) = f'(x) - \sum_{m = 1}^n f_m'(x) \rightarrow 0$$ for all $x$ except for a set of measure $0$ since we have proved convergence only for all the points except points in $\cap_{m = 1}^{\infty} B_{1/m}$ which is a set of measure $0$.