Let functions $f_n$ be measurable, $n \in N$, $f_n\rightarrow f$ almost everywhere. Prove that $\operatorname{arctg}f_n \rightarrow \operatorname{arctg}f$ almost everywhere.
Honestly speaking, we have had so many theorems on different types of convergence and relationships between them that I find it very hard to mix them in order to solve problems.
First of all, we are given that the measure of the set of points for which $f_n$ does not converge to $f$ is zero.
How should I think about the problem next? How do we show that $arctg$ preserves convergence?
I have my lecture notes, but when it comes to practice I often feel helpless and do not know how to approach the problem. Probably, you could recommend some good problem book with solved examples.
Thank you
Suppose that $g : E \subset \mathbb{R}^l \to \mathbb{R}$ is continuous, and let $(f_n)_n :\mathbb{R}^k \to E$ converge almost everywhere to some function $f : \mathbb{R}^k \to E$. Now, by continuity, if $x \in \mathbb{R}^k$ is such that $f_n(x) \to f(x)$, then $gf_n(x) \to gf(x)$. By the contrapositive, then, if $gf_n(x) \not \to gf(x)$ we have that $f_n(x) \not \to f(x)$. That is,
$$ \{x : gf_n(x) \to gf(x)\}^c \subset \{x : f_n(x) \to f(x)\}^c $$
and so
$$ \mu(\{x : gf_n(x) \to gf(x)\}^c) \leq \mu(\{x : f_n(x) \to f(x)\}^c) = 0. $$
Hence $gf_n \to gf$ a.e.