If $f_n,g_n$ converge in measure, then $f_n g_n$ converges in measure if $\mu(X)<\infty$, but not always if $u(X)=\infty$ More precicely, the problem is in Folland's Real Analysis 2nd edition, 1999, problem 38b in chapter 2(integration)
Let $f_n$ be a sequence of measurable, complex valued functions on a measure space $(X, M, \mu)$
The problem states, if $f_n\rightarrow f$ in measure, and $g_n\rightarrow g$ in measure, then so does $f_n g_n\rightarrow fg$ if $\mu(X)<\infty$, but not always if $\mu(X)=\infty$
Ok, so I have attempted the problem, I have some work but I am struggling to finish it. Hopefully someone can help. This is my work so far:
My strategy is to work with $\mu(X)<\infty$ first, and to show that, for a fixed n and some $\delta> 0$ (preferabbly an expression of $\epsilon$ in some way) $B_{n,\delta}=\{x\;|\;|f(x)g(x)-f_n(x)g_n(x)|>\epsilon\}$ is contained in the set $F_{n,\epsilon} \cup G_{n,\epsilon}$ where $F_{n,\epsilon}=\{x\;|\;|f(x)-f_n(x)|>\epsilon\}$ and $G_{n,\epsilon}=\{x\;|\;|g(x)-g_n(x)|>\epsilon\}$, where $\epsilon > 0$ and we have that
$$\mu(F_{n,\epsilon} \cup G_{n,\epsilon})\leq 2\epsilon$$
for all $n\geq N$ for some N
by the fact that $f_n,g_n$ converge in measure. If we can do that, then the statement follows from montonictiy of $\mu$.
So far so good. We can do this now. If $x\in B_{n_\delta}$ (we can go back and change $\delta$ to suit our needs later) then
$$\delta < |fg(x)-f_ng_n(x)|\leq (|g|+|f_n|)|f-f_n|$$
Now here is where I am stuck. I have not yet used the fact that $\mu(X)<\infty$, the only idea i had so far was, that possible this fact implied that (|g|+|f_n|) < C for some constant, which would be nice, but I'm not sure if this is true, I suspect it's not true but I'm not 100% sure. I was thinking also that we could use integration at this point, but I'm just not sure how to apply the integral in this setting to my advantage. All this seems to say to me is that
$$ \int_X 1 <\infty $$
I don't see how this would imply anything useful about our functions.
Can someone please help? I would prefer a solution which builds off of my previous work/ideas, although this is not required and it is preferable to start over if if my ideas lead nowhere.
 A: Let $f_{n,k}$ be a truncation of $f_n$. Specifically, $f_{n,k}(x) = f_n(x)$ is $|f_n(x)| \le k$ otherwise $f_{n,k} = k$ (depending on the sign of $f_n(x)$.
Now I claim that $f_{n,k} \to f_n$ if and only if $\mu(X) < \infty$.
First, if $\mu(X) \not< \infty$. Consider $X = \mathbb{C}$ and $\mu$ be the Lebesgue measure. Then let $f_n(x) = f(x) = x$. Then for any $\epsilon > 0$ we have that for all $k$, the set $\{ x : |f_{n,k}(x) - f(x)| > \epsilon\}$ has infinite measure.
On the other hand, suppose $\mu(X) < \infty$. Then define $A_{n,k} = \{ x : f_n(x) = f_{n,k}(x) \}$. Now we see that $A_{n,k} \subset A_{n,k+1}$ and that their union is equal to $X$. Thus, we have that $\mu(A_{n,k}) \to \mu(X)$. Then since $\mu(X) < \infty$, we have that $\mu(X-A_{n,k}) = \mu(X) - \mu(A_{n,k}) \to 0$ (don't have $\infty - \infty$ here).
Thus, $f_{n,k} \to f_n$ in measure.
Now for your question, note that $f_{n,k} \to f_k$ (the truncation of $f$) in measure and $g_{n,k} \to g_k$ (the truncation of $g$). Now there are bounded function and we can apply your argument to show that $f_{n,k}g_{n,k} \to f_k g_k$ (the product of the truncations) in measure.
Now $f_kg_k \to fg$ via the previous arguments as well (since these are bounded as well).
A: Hint
Let $\varepsilon >0$ and $\eta>0$. Since $\mu(X)<\infty $, there is $K>0$ s.t. $\mu\{|f|>K\}<\eta $ and $\mu\{|g|>K\}<\eta$.
You have that
\begin{align*}
\mu\{|f_ng_n-fg|>\varepsilon \}&\leq \mu\{|g||f_n-f|>\varepsilon/2 \}+\mu\{|f_n||g_n-g|>\varepsilon/2 \}\\
&\leq \underbrace{\mu\{|g||f_n-f|>\varepsilon /2\}} _ {I_1^n}+\underbrace{\mu\{|f||g_n-g|>\varepsilon /4\}}_{I_2^n}+\underbrace{\mu\{|f_n-f||g_n-g|>\varepsilon /4\}} _ {I_3^n}
\end{align*}
Moreover
\begin{align*}
I_1^n&\leq \mu\{|g||f_n-f|>\varepsilon/2, |g|\leq K\}+\mu\{|g||f_n-f|>\varepsilon/2, |g|> K\}\\
&\leq \mu\left\{|f_n-f|>\frac{\varepsilon }{2K}\right\}+\mu\{|g|>K\}\\
&\underset{n\to \infty }{\longrightarrow}\eta.
\end{align*}
Therefore $I_1^n\underset{n\to \infty }{\longrightarrow }0.$ The same applies to $I_3$.
For $I_3$, use the identity $ab\leq \frac{1}{2}(a^2+b^2)$ to get
$$I_3\leq \mu\{|f_n-f|>\sqrt \varepsilon \}+\mu\{|g_n-g|>\sqrt \varepsilon \},$$
and the claim follows.
A: If $\mu(X) < \infty$, then on the space $M(X)$ of measurable complex-valued functions (equality $M(X)$ is defined by $f = g$ iff $f = g$ a.e.), convergence in measure is convergence in the topology on $M(X)$ given by the metric
$$d(f, g) = \int_{X}\rho(|f - g|)\,d\mu,$$
where $\rho(x) = \frac{x}{1 + x}$. So if $f_n \to f$, $g_n \to g$ a.e., then by dominated convergence ($\rho \leq 1 \in L^1(X, \mu)$), $d(f_ng_n, fg) \to 0$. Hence $f_ng_n \to fg$ in measure.
If $f_n \to f$, $g_n \to g$ only in measure, then we can use the "subsequence principle" and the fact that convergence in measure implies a subsequence converges a.e. and the previous result to prove that $f_ng_n \to fg$ in measure.
