A proof that if $f_n \rightarrow f$ in measure and $g_n \rightarrow g$ in measure, then $f_n\cdot g_n \rightarrow f\cdot g$. The problem is to prove that if $f_n \rightarrow f$ in measure and $g_n \rightarrow g$ in measure, then $f_n\cdot g_n \rightarrow f\cdot g$. Where these functions are defined on a set that has a finite measure. The portion I am confused about is when they say $$\{x: |f_n(x)| > M+1\} \subset \{x: |f(x)| > M \} \cup \{x: |f(x)-f_n(x)| 
\geq 1\},$$ this is on page (1) of the file. I feel like this should be simple, and it makes sense intuitively, however when trying to prove this rigorously I fail.
Link: https://www.math.arizona.edu/~friedlan/teach/523/sol.pdf
Any help or hints would be much appreciated.
Edit: Sorry I just realized is this, but is it because we are considering the case for when $|f(x)| > M$, where the definition of $M$ is given in the previous lines in the proof.
 A: This can definitely be a fiddly question but hopefully I can help you out. I do not know your mathematical level so please forgive me if anything here is patronising I shall talk through things throughly.
Let us come up with a way to show generally that:
$\{x: \text{Condition 1 holds} \} $ $ \subset \{x: \text{Condition 2 holds} \} \cup \{x: \text{Condition 3 holds} \} \ $
In general to show one set is a subset of another mathematicians tend to show if anything is in the first set then it has to belong to the other!
In our case that means that if ever Condition $1$ holds then at least one of Condition 2 or 3 must hold as union means or.
One one to do this would be via a contradiction. That is to assume Condition $1$ holds and that neither Condition $2$ or $3$ hold and then arrive at a contradiction. i.e showing it impossible for Condition $1$ to hold without Condition $2$ or $3$
In our case:
Condition $1$ is $ |f_n(x)| > M+1$

Condition $2$ is $|f(x)|>M$

Condition $3$ is $|f(x)-f_n(x)| \geq 1$
Let us proceed by assuming Condition $1$ holds and neither $2$ or $3$, hoping to find a contradiction.
Firstly we have that $|f_n(x)| = |f_n(x) -f(x) +f(x)|$
This adding then instantly subtraction $f(x)$ may seem odd but I promise it will help us out!
Via the triangle inequality, that is $|a-b| \leq |a| + |b|$ we have that:

$|f_n(x)| = |f_n(x) -f(x) +f(x)| \leq |f_n(x)-f(x)| + |f(x)| $
However as we assumed both Condition $2$ and $3$ are false we have that:

$|f(x)|\leq M$ and  $|f(x)-f_n(x)| < 1$
Hence:

$|f_n(x)| = |f_n(x) -f(x) +f(x)| \leq |f_n(x)-f(x)| + |f(x)| < M+1 $
i.e
$f_n(x) < M+1$
However this violates Condition $1$!
Hence if $ |f_n(x)| > M+1$ then either $|f(x)|>M$ or $|f(x)-f_n(x)| \geq 1$ (or both)
Hope this helps,
Oskar
Edit the tricky part of this question would be showing we have a proper subset and not $\subseteq$ that is there are some points where condition $2$ or $3$ hold without condition $1$ just on the assumption that they are measurable functions. Is this needed for you?
