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Hi,
Hi,
I was wandering whether one can drop the sigma-finite assumption when trying to apply the dominated convergence theorem under convergence in measure conditions (as showed in this entry).
I was wondering whether one can drop the sigma-finite assumption when trying to apply the dominated convergence theorem under convergence in measure conditions (as showed in this entry).
The reason is that the support of the dominating function is sigma-finite, for:
The reason is that the support of the dominating function is sigma-finite, for:
:<math> Supp(g) = \cup_{n=1}^\infty g^{-1} ([ \frac{1}{n} , n]) </math>
:<math> Supp(g) = \cup_{n=1}^\infty g^{-1} ([ \frac{1}{n} , n]) </math>

Revision as of 22:33, 5 December 2011

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Something is not quite right here: let X=R and fn be the indicator function for the interval ; obviously the sequence fn converges to zero (pointwise/everywhere), but it does not satisfy the condition given as the definition of convergence in measure. [Take ; then all the sets have infinite measure and (therefore) do not tend to zero.]

This appears to contradict two assertions in the rest of the article: 1) the characterisation (in the sigma-finite case) of convergence in measure via a.e.-convergence; and 2) the characterisation of convergence in measure via pseudometrics. [In the example above, does tend to zero for every set of finite measure F.]

Perhaps one should say that a sequence fn converges "locally in measure" to f if, for every measurable set of finite measure F

.

Then the issues raised above would seem to be resolved by replacing "convergence in measure" by "local convergence in measure".

Boy Waffle (talk) 20:09, 5 January 2008 (UTC)[reply]

Welcome, Mr. Waffle! It seems you're right. I remember proving at one time that a.e./Lp-convergence implies convergence in measure on [0,1], but I don't remember anything else well. I imagine it generalizes to finite measures The first section, before the topology, is cited from statements throughout Royden, but perhaps not correctly. I don't have the book with me, unfortunately. As for the pseudometrics, I never personally verified that the topology describes the convergence, but the concepts do have the same name. —vivacissamamente (talk) 02:27, 7 January 2008 (UTC)[reply]
Thanks for the welcome, Vivacissamamente! For the moment, unfortunately, I'm nowhere near a decent library, but I checked Fremlin online, and at one point he writes
warning! the phrase "topology of convergence in measure" is also used for [...some other topology...] I have seen the phrase "local convergence in measure" used for [...the topology described here...]
So, I've rather substantially re-written the article to account for the fact that different authors seem to use the same phrase to mean different things. I have also added that the dominated convergence theorem holds w.r.t. local convergence of measure (at least in the sigma-finite case); this is an essentially trivial corollary of its characterization in terms of a.e. convergence. Finally, I have temporarily deleted the line about "Cauchyness in measure" (which is nonsensical as it stands) and replaced it with a general comment about Cauchyness in any topology generated by pseudo-metrics. It seems likely to me that there are also global and local versions of this notion. [I will eventually do some homework on that score...] —Preceding unsigned comment added by Boy Waffle (talkcontribs) 02:15, 9 January 2008 (UTC)[reply]

Hi, I was wondering whether one can drop the sigma-finite assumption when trying to apply the dominated convergence theorem under convergence in measure conditions (as showed in this entry). The reason is that the support of the dominating function is sigma-finite, for:

thanks — Preceding unsigned comment added by 132.65.16.64 (talk) 11:52, 5 December 2011 (UTC)[reply]