# 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

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.

• When the subsequence $n_{k_j}$ which works for $f$ might not also work for $g$ immediately; maybe one needs a common refinement of one for $f$ and another for $g,$ or the like,however there is still the worry the intersection of the two sets of subscripts might be empty, or at least not contain arbitrarily large naturals. Jun 27 '16 at 22:32
• @coffeemath Yeah, I glossed over that to avoid too many subscripts. There is a subsequence of $n_k$ which works for $f$, and a subsequence of that subsequence which works for $g$. This latter subsequence is the subsequence of $n_k$ that I am using.
– user169852
Jun 27 '16 at 22:34
• @Usermat The answer I linked in my question above includes a counterexample for Lebesgue measure on $\mathbb R$.
– user169852
Nov 30 '20 at 7:27
• Does this answer your question? Convergence in measure - product Jun 10 at 18:47
• I have voted to reopen, but if it doesnt work you can try math.meta.stackexchange.com/questions/32975/… Jun 14 at 9:34