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 September 3rd, 2016, 12:40 AM #1 Newbie   Joined: Dec 2014 From: Italy Posts: 5 Thanks: 0 Is a set with finite Lebesgue measure necessarily bounded? I consider $\mathbb{R}^n$ with the Lebesgue measure. Let be $\Omega$ subset of $\mathbb{R}^n$ with finite measure. Can I say that $\Omega$ is also bounded? Thank you!
 September 3rd, 2016, 06:56 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 No, because it could have infinite points which don't contribute to the measure. Thanks from Glo
 September 3rd, 2016, 01:46 PM #3 Global Moderator   Joined: May 2007 Posts: 6,766 Thanks: 698 The set could consist of intervals of form (n,n+1/n^2), so the total measure is finite, but the set is not bounded even when excluding subset of measure 0. Thanks from Glo and romsek
 September 3rd, 2016, 06:06 PM #4 Senior Member   Joined: Aug 2012 Posts: 2,324 Thanks: 715 Looking at zylo's and mathman's respective counterexamples (both good!), one is tempted to conjecture that any connected set of reals of finite measure must be bounded. Indeed, a connected set of reals is an interval, finite or infinite. An infinite interval (a ray or the entire real line) has infinite measure. Therefore a connected set of finite measure is a finite interval, which is bounded. Of course connectedness is not necessary, since there are disconnected sets of finite measure that are bounded. Thanks from Glo Last edited by Maschke; September 3rd, 2016 at 06:19 PM.
September 9th, 2016, 02:19 PM   #5
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 Originally Posted by Maschke Looking at zylo's and mathman's respective counterexamples (both good!), one is tempted to conjecture that any connected set of reals of finite measure must be bounded. Indeed, a connected set of reals is an interval, finite or infinite. An infinite interval (a ray or the entire real line) has infinite measure. Therefore a connected set of finite measure is a finite interval, which is bounded. Of course connectedness is not necessary, since there are disconnected sets of finite measure that are bounded.
Hi I'm new here so forgive me if the text formatting isn't quite right. I'm not sure connectedness is the right thing to consider one way or another. What you say is certainly true but this fails if $n > 1$. Lines in $\mathbb{R}^2$ have measure zero. Even more pathologically, connected subsets of $\mathbb{R}^n$ may not even be measurable.

September 9th, 2016, 04:06 PM   #6
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 Originally Posted by SDK Hi I'm new here so forgive me if the text formatting isn't quite right. I'm not sure connectedness is the right thing to consider one way or another. What you say is certainly true but this fails if $n > 1$. Lines in $\mathbb{R}^2$ have measure zero. Even more pathologically, connected subsets of $\mathbb{R}^n$ may not even be measurable.
Right you are. Wasn't aware of the example of a nonmeasurable connected set, do you happen to have a reference? I only know the standard example of the Vitali set in the reals.

 September 9th, 2016, 06:06 PM #7 Senior Member   Joined: Aug 2012 Posts: 2,324 Thanks: 715 ps ... found this: geometry - Does simply-connected imply measureable? - Mathematics Stack Exchange There are indeed nonmeasurable connected sets.
September 10th, 2016, 02:11 PM   #8
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 Originally Posted by Maschke ps ... found this: geometry - Does simply-connected imply measureable? - Mathematics Stack Exchange There are indeed nonmeasurable connected sets.
That is a great example. I had in mind a proof using indicator functions for an unmeasurable set in $\mathbb{R}$ and obtaining a contradiction to Fubini's theorem. This somehow seems nicer. Thanks for sharing.

 September 11th, 2016, 07:52 AM #9 Senior Member   Joined: Sep 2016 From: USA Posts: 621 Thanks: 394 Math Focus: Dynamical systems, analytic function theory, numerics Great Example. Much clearer than what I had in mind. Thanks for sharing it.

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