A problem about convergence in measure of product I am just confused about the proof for convergence in measure of product.
In this proof, he claimed that "When $μ$ is finite, fix $ε>0$. There is $A>0$ such that $μ\{|f|>A\}+μ\{|g|>A\}<ε$". Does it mean $μ\{|f|=\infty\}+μ\{|g|=\infty\}=0$? If it's true, then how to prove it when $μ\{|f|=\infty\}+μ\{|g|=\infty\}\neq0$? Do we just ignore this situation?
Also, I find another proof which doesn't require $μ\{|f|=\infty\}+μ\{|g|=\infty\}=0$ (I guess). So, which is better?
 A: In the same proof, the comments explain why:
Observe that for a given $x$, if $|f(x)|>n$, then $|f(x)|>n-1$ and so $...\subseteq\{x:|f(x)|>n\}\subseteq\{x:|f(x)|>n-1\}\subseteq...\subseteq\{x:|f(x)|>0\}$. We have for ourselves a descending sequence of sets and the "top" set, being a subset of a set $X$ of finite measure, is of finite measure. By continuity of measure, $$\lim_{n\rightarrow\infty}\mu\{x:|f(x)|>n\}=\mu\left(\bigcap_{n=1}^\infty\{x:|f(x)|>n\}\right)$$.
Second, note that $$\bigcap_{n=1}^\infty\{x:|f(x)|>n\}=\{x:|f(x)|=\infty\}$$
Finally, since $f$ (and $g$ are) is assumed to be real-valued, they are finite and so, the set on the right is empty and hence has measure zero. Thus, $\lim_{n\rightarrow\infty}\mu\{x:|f(x)|>n\}=0$. By definition of limit, this means that for every $\epsilon >0$, there exists $N$ such that, for all $n \geq N $, we have $\mu\{x:|f(x)|>n\}<\epsilon$ but since we have a decreasing sequence of sets, we might as well just say $\mu\{x:|f(x)|>N\}<\epsilon$. Now let $N=A$.
