Hints about two measure related tasks (almost everywhere problems). I got these two tasks as exercises on my measure theory lectures, and somehow I have no idea where to start. I tried a few approaches (and made some research for similar problems, but it didn't help me anyway). I'll greatly appreciate any hint, or solution to these :)
1) Let $f_n$ be sequence of measurable functions, such that $f_n \rightarrow f$ in terms of given measure, and $|f_n|<M$. Prove that $|f|\le M$ almost everywhere.
2) Let $f_n$ be sequence of measurable functions, such that $f_n \rightarrow f$ in terms of given measure, and sequence $f_n$ is nondecreasing. Prove that $f_n \rightarrow f$ almost everywhere.
With the first one, I tried to rewrite the $|f_n(x)-f(x)|$ property using triangle inequality, but this lead me to nowhere. Also I tried to prove it by contradiction, but it seems even harder then original task.
With the second one I don't even have idea where to start. 
 A: Regarding Question 1: As mentioned in a comment, we know that $f_n \to f$ in measure implies the existence of a subsequence $(f_{n_k})_{k \in \mathbb{N}}$ such that $f_{n_k} \to f$ almost everywhere. This easily proves the assertion.
However, there is also a more elementary proof based on the triangle inequality. Let $\epsilon>0$. Since
$$\{x; |f(x)| \geq M+\epsilon\} \subseteq \{x; |f(x)-f_n(x)| \geq \epsilon\} \cup \{x; |f_n(x)| \geq M\}$$
we have
$$\mu(|f| \geq M+\epsilon) \leq \mu(|f-f_n| \geq \epsilon) + \mu(|f_n| \geq M).$$
By assumption, the second term on the right-hand side equals $0$ and the first one converges to $0$ as $n \to \infty$. Hence, $\mu(|f| \geq M+\epsilon)=0$. As $\epsilon>0$ is arbitrary, this proves $\mu(|f|>M)=0$.

Regarding Question 2: I take it that you mean by "non-decreasing sequence" that for any $x$ we have $f_1(x) \leq f_2(x) \leq f_3(x) \leq \ldots$. Because of this monotonicity, the function
$$g(x) := \sup_{n \in \mathbb{N}} f_n(x)$$
is well-defined and $f_n \to g$ pointwise. Since pointwise convergence implies convergence in measure and limits in measure are unique, this gives $\mu(f \neq g)=0$; in particular, $f_n \to f=g$ almost everywhere. 
