Let $A$ be a Lebesgue measurable subset of $\mathbb{R}$ with $m(A)>0$. Show that there is a bounded measurable $B$ subset $A$ with $m(B)>0$

Help. Need to get proof because we just learned it. How do we prove this.

  • $\begingroup$ The answers below use that $\mathbb{R}$ is $\sigma$-finite, i.e. it is a countable union of sets of finite measure. If you were looking for a word to go with the property that makes this work. $\endgroup$ – Francis Adams Aug 17 '13 at 23:34


Let $F_n=A\cap[n,n+1)$ for $n\in\mathbb{Z}$.

Then, you get a sequence of bounded measurable subset of $A$.

Can $m(F_n)=0$ for all $n$?

  • $\begingroup$ I do not think that it equals 0 for all n. How would I write this sequence to show that. Not sure how the sequence would look like. $\endgroup$ – 9959 Mar 18 '13 at 5:45
  • $\begingroup$ If they are all zero, use countable additivity to show that there is a contradiction with $m(A)>0$. $\endgroup$ – NECing Mar 18 '13 at 14:02

Here is a similar proof that uses continuity of measure.

Let $A_n = [-n,n] \cap A$. Since $\cup_n A_n = A$, we have $\lim_n m A_n = mA > 0$. Hence $m A_n >0$ for some $n$, and since $A_n \subset [-n,n]$, it is bounded.

  • $\begingroup$ Got this. How to show that B is a subset A bounded measure? $\endgroup$ – 9959 Mar 18 '13 at 6:47
  • $\begingroup$ Choose $B=A_n$, where $n$ is the index mentioned above. Since $B = [-n,n] \cap A$ it is both a subset of $A$ and bounded by $n$. $\endgroup$ – copper.hat Mar 18 '13 at 7:20

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