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September 3rd, 2016, 01:40 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, 07:56 AM  #2 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
No, because it could have infinite points which don't contribute to the measure.

September 3rd, 2016, 02:46 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,683 Thanks: 658 
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.

September 3rd, 2016, 07:06 PM  #4 
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630 
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. Last edited by Maschke; September 3rd, 2016 at 07:19 PM. 
September 9th, 2016, 03:19 PM  #5  
Senior Member Joined: Sep 2016 From: USA Posts: 556 Thanks: 321 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
 
September 9th, 2016, 05:06 PM  #6  
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630  Quote:
 
September 9th, 2016, 07:06 PM  #7 
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630 
ps ... found this: geometry  Does simplyconnected imply measureable?  Mathematics Stack Exchange There are indeed nonmeasurable connected sets. 
September 10th, 2016, 03:11 PM  #8  
Senior Member Joined: Sep 2016 From: USA Posts: 556 Thanks: 321 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
 
September 11th, 2016, 08:52 AM  #9 
Senior Member Joined: Sep 2016 From: USA Posts: 556 Thanks: 321 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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bounded, finite, lebesgue, measure, necessarily, set 
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