10
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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. $\endgroup$
    – coffeemath
    Jun 27 '16 at 22:32
  • $\begingroup$ @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. $\endgroup$
    – user169852
    Jun 27 '16 at 22:34
  • 1
    $\begingroup$ @Usermat The answer I linked in my question above includes a counterexample for Lebesgue measure on $\mathbb R$. $\endgroup$
    – user169852
    Nov 30 '20 at 7:27
  • 2
    $\begingroup$ Does this answer your question? Convergence in measure - product $\endgroup$
    – user0
    Jun 10 at 18:47
  • 1
    $\begingroup$ I have voted to reopen, but if it doesnt work you can try math.meta.stackexchange.com/questions/32975/… $\endgroup$ Jun 14 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy