If $f_n \to f$ and $g_n \to g$ in measure and $\mu$ is finite, then $f_n g_n \to fg$ in measure
Sebastian Wright 
This is Problem 3.1.5 in Cohn's Measure Theory, 2nd edition.
Let $\mu$ be a measure on $(X, \mathcal A)$, and let $f, f_1,f_2, \ldots$ and $g,g_1,g_2,\ldots$ be real-valued $\mathcal A$-measureable functions on $X$.
(a) Show that if $\mu$ is finite, if $f_n \to f$ in measure, and $g_n \to g$ in measure, then $f_n g_n \to fg$ in measure.
With some effort, I was able to work out a direct proof, similar to the one outlined in this answer.
After re-reading the section, I came up with an alternative argument similar to the one used by Cohn to prove another theorem. It results in a much quicker proof, and almost seems like a magic trick. I want to make sure that my argument is correct.
I use the following results, proved by Cohn in this section, for real-valued measurable functions:
Proposition 3.1.3: If $f_n \to f$ in measure, then there is a subsequence such that $f_{n_k} \to f$ almost everywhere.
Proposition 3.1.2: If $\mu$ is finite and $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.
Here is my proof:
Let $\{n_k\}$ be any subsequence of $\mathbb N$. Then $f_{n_k} \to f$ and $g_{n_k} \to g$ in measure, so by Proposition 3.1.3 there is a subsequence $\{n_{k_j}\}\subset \{n_k\}$ such that $f_{n_{k_j}} \to f$ and $g_{n_{k_j}} \to g$ almost everywhere. (To see this, take a subsequence of $\{n_k\}$ that works for $f$. Then that subsequence has a subsequence which works for both $f$ and $g$.) Therefore $f_{n_{k_j}} g_{n_{k_j}} \to fg$ almost everywhere.
As $\mu$ is finite, Proposition 3.1.2 implies that $f_{n_{k_j}} g_{n_{k_j}} \to fg$ in measure.
Now here is the magic part:
Suppose that $f_n g_n$ does not converge to $fg$ in measure. Then there is some $\epsilon > 0$ such that $\mu\{|f_n g_n - fg| > \epsilon\}$ does not converge to zero. Thus there is some $\delta > 0$ such that $\mu\{|f_n g_n - fg| > \epsilon\} > \delta$ for infinitely many $n$. In other words, there is a subsequence $\{n_k\} \subset \mathbb N$ such that $\mu\{|f_{n_k} g_{n_k} - fg| > \epsilon\} > \delta$ for all $n_k$. Clearly this subsequence cannot have any further subsequence $\{n_{k_j}\}$ for which $f_{n_{k_j}} g_{n_{k_j}} \to fg$ in measure, but this contradicts what we showed above. Consequently, $f_n g_n$ must converge to $fg$ in measure.
Is this argument legitimate? It almost seems too easy.
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