TITLE: Convergence in measure - product
QUESTION [11 upvotes]: I'm trying to prove the following statements in Folland's book.
Let $(X,\mathcal{M},\mu)$ be a measure space. If $f_n\to f$ in measure and $g_n\to g$ in measure, then $f_n+g_n\to f+g$ in measure and $f_ng_n\to fg$ in measure (in this last case, assume $\mu(X)<\infty$).
The first one is already done, but the second one is giving me trouble. 
My attempt: We can take subsequences $f_{n_k}$ and $g_{n_k}$ such that $f_{n_k}\to f$ a.e. and $g_{n_k}\to g$ a.e. In this case, we know that $f_{n_k}g_{n_k}\to fg$ a.e. Note that $\mu(X)<\infty$, so we can use Egoroff's theorem to conclude that $f_{n_k}g_{n_k}\to fg$ almost uniformly. Finally, almost uniform convergence implies convergence in measure, therefore $f_{n_k}g_{n_k}\to fg$ in measure. 
The problem is that I just have convergence in measure for some subsequence of $f_ng_n$ and I need convergence in measure for the whole sequence. Also I don't know if using Egoroff's theorem is unnecessary, if this is the case, I would appreciate to see some more elementary prove.
Detail: We say that $f_n\to f$ almost uniformly if for every $\varepsilon > 0$ there is some $E\subset X$ measurable, such that $\mu(E) <\varepsilon$ and $f_n\to f$ uniformly on $E^c$.
Thank you!

REPLY [2 votes]: This is the alternative method mentioned by yi li's answer on this webpage.
Define functions $s_n,d_n,s,d$ on $X$ by\begin{align*}
s_n(x) & = (f_n(x) + g_n(x))/4,\\
d_n(x) & = (f_n(x) - g_n(x))/4,\\
s(x) & = (f(x) + g(x))/4,\\
d(x) & = (f(x) - g(x))/4\end{align*}
all of which are finite a.e. on $X$ and measurable. Moreover, $\{s_n\}\to s$ and $\{d_n\}\to d$ in measure. Then\begin{align*}
&\lim_{n\to\infty}\mu\{x\mid |f_n(x)g_n(x) - f(x)g(x)| >\eta\}\\
&\quad =\lim_{n\to\infty}\mu\{x\mid |s_n^2(x) - d_n^2(x) - s^2(x) + d^2(x)| >\eta\}\\
&\quad\le\lim_{n\to\infty}\mu\{x\mid |s_n^2(x) - s^2(x)| >\eta/2\}\\
&\quad+\lim_{n\to\infty}\mu\{x\mid |d_n^2(x) - d^2(x)| >\eta/2\}\\
&\quad = 0 + 0.\end{align*}
Thus, $\{f_n g_n\}\to fg$ in measure.