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I'm looking for a pair of sequences $f_n \to f$ in measure, $g_n \to g$ in measure, where $f_ng_n \not\!\to fg$ in measure.

I've tried a number of things with characteristic functions that move around with $n$, and nothing seems to pan out. I've also tried looking at non-Lebesgue measures, like the counting measure. At least, I realize that whatever measure one uses must not be finite, or else $f_ng_n \to fg$ is always true.

Any hints?

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@cardinal: Why the change to the title? The "not" sign looks out of place (to me). –  Zev Chonoles Dec 4 '11 at 3:23
    
I TeXed it the same as in the post, but it's not aligned right... –  JeremyKun Dec 4 '11 at 3:25
    
Curious; this is what I see in the question (out of place), but this is what I see on the main page (just fine). –  Zev Chonoles Dec 4 '11 at 3:27
    
@Zev: I was trying to fix the alignment, but it seems there is a MathJax rendering error that is preventing this from working. Sorry if the edit was disruptive. Interestingly, if you click edit, remove the \! and add it back, the real-time rendering looks correct, but then after refreshing the page it doesn't. Sigh. –  cardinal Dec 4 '11 at 3:29
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Also, I'm seeing the error you describe as well (remove and add the \! and the misalignment doesn't reappear). I assume this is a bug of some sort and could be reported here. –  Zev Chonoles Dec 4 '11 at 3:36

2 Answers 2

up vote 5 down vote accepted

The classic counterexample I'm aware of is with the Lebesgue measure on $\mathbb{R}$, $f_n=g_n=x+\frac{1}{n}$, and $f=g=x$. Then obviously we have $f_n\to f$ and $g_n\to g$ in measure, but $$f_ng_n=x^2+\tfrac{2}{n}x+\tfrac{1}{n^2}\not\to x^2=fg$$ in measure because for any $\epsilon>0$ and $n\in\mathbb{N}$. $$\{x\in \mathbb{R}\mid \tfrac{2}{n}x+\tfrac{1}{n^2}\geq\epsilon\}=[\tfrac{n^2\epsilon-1}{2n},\infty)$$ has infinite measure, so for any $\epsilon>0$, $$\lim_{n\to\infty}\mu(\{x\in\mathbb{R}\mid |f_ng_n(x)-fg(x)|\geq\epsilon\})=\lim_{n\to\infty}\infty=\infty$$ I'm sorry I don't know how to give a good hint for this one; I didn't see it on my own either, someone pointed it out to me. In retrospect the intuition is that we set things up so that the gap between the $f_n$'s and $f$, and $g_n$'s and $g$, when multiplied, becomes "amplified" (since $\frac{2}{n}x+\frac{1}{n^2}$ depends on $x$, whereas $\frac{1}{n}$ doesn't).

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It's probably worth noting that convergence in measure is the "wrong" concept for infinite measures. If you replace it by local convergence in measure (that is, for every set $E$ of finite measure $f_n|_E \to f|_E$ in measure) then the space $L^0$ of classes of measurable functions modulo null sets becomes a topological algebra. –  t.b. Dec 4 '11 at 5:37

You can also do it with counting measure on $\mathbb{N}$ and functions from $\mathbb{N}$ to $\mathbb{R}$. Let

$$\begin{align*} x_n&:\mathbb{N}\to\mathbb{R}:k\mapsto 2^{-n}\;,\\ x&:\mathbb{N}\to\mathbb{R}:k\mapsto 0\;,\text{ and}\\ y,y_n&:\mathbb{N}\to\mathbb{R}:k\mapsto 2^k\;. \end{align*}$$

For $m,n\in\mathbb{N}$ we have

$$\begin{align*} \mu\big(\{k\in\mathbb{N}:|x_n(k)-x(k)|\ge 2^{-m}\}\big)&=\mu\big(\{k\in\mathbb{N}:2^{-n}\ge 2^{-m}\}\big)\\ &=\begin{cases} \infty,&n\le m\\ 0,&n>m\;, \end{cases} \end{align*}$$

so $\langle x_n:n\in\mathbb{N}\rangle\to x$ in measure, and obviously $\langle y_n:n\in\mathbb{N}\rangle\to y$ in measure. But

$$x_ny_n:\mathbb{N}\to\mathbb{R}:k\mapsto 2^{-n}2^k=2^{k-n},$$

so for $m,n\in\mathbb{N}$ we have

$$\begin{align*} \mu\big(\{k\in\mathbb{N}:|(x_ny_n)(k)-(xy)(k)|\ge 2^{-m}\}\big)&=\mu\big(\{k\in\mathbb{N}:2^{k-n}\ge 2^{-m}\}\big)\\ &=\mu\big(\{k\in\mathbb{N}:k\ge n-m\}\big)\\ &=\infty\;, \end{align*}$$

and $\langle x_ny_n:n\in\mathbb{N}\rangle$ does not converge to $xy$ in measure.

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