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If $A\subset\mathbb{R}$ and $\mu(A)>0$ then exits $B\subset{A}$ bounded such that $\mu(B)>0$. I know the proof with $B$ a nonmensurable set, but can not B be simply subset of A without needing to be a nonmeasurable set?...and if it is true how do you prove it?

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  • $\begingroup$ You can't define $\mu(B)$ if $B$ is non-measurable (existence of $\mu(B)$ as an extended real number is precisely what it means to be measurable). It's a bit unclear what the question is. $\endgroup$ – User8128 Mar 4 '18 at 23:00
  • $\begingroup$ By definition a set A is measurable if $\mu(C)=\mu(C\cap{A})+\mu(C\cap{A^{c}})$ $\forall{C}\subset\mathbb{R}$, of course that there are sets that are nonmensurable because of the breach of equality $\endgroup$ – The Student Mar 4 '18 at 23:13
  • $\begingroup$ I understand that there are non-measurable sets. What is your question here? You first seem to want $B \subset A$ with $\mu(B)>0$, then you pivot from there to bring up non-measurable sets. I'm just trying to clarify what the question is. $\endgroup$ – User8128 Mar 4 '18 at 23:17
  • $\begingroup$ Well, as the question says I know the proof considering B a nonmensurable set, but can not B be simply subset of A without needing to be a nonmeasurable set?...and if it is true how do you prove it? $\endgroup$ – The Student Mar 4 '18 at 23:23
  • $\begingroup$ What do you mean by "I know the proof with B a nonmensurable set." Are you saying you can find a non-measurable set $B$ such that $\mu(B) > 0$? This is impossible since we don't define $\mu(B)$ for non-measurable sets. What do you mean by "can not B be simply subset of A without needing to be a nonmeasurable set?" Nowhere did it specify that $B$ needed to be non-measurable; it seems like you began talking about non-measurable sets for no particular reason when they aren't relevant to the problem. $\endgroup$ – User8128 Mar 4 '18 at 23:28
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$A$ is the union of the sets $A \cap (-n,n)$, $n=1,2,...$. If each set on the right has outer measure 0 then A would have outer measure 0 because outer measures are sub-additive. Hence there exists n such that $B \equiv A \cap (-n,n)$ has positive measure.

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  • $\begingroup$ Thanks for your answer! $\endgroup$ – The Student Mar 5 '18 at 20:23
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If I am interpreting the question correctly, you are asking the following: for arbitrary measurable $A \subset \mathbb R$ with $\mu(A) > 0$, can we find a measurable $B \subset A$ with $\mu(B) > 0$?

The answer is yes. Trivially, we could take $B = A$.

If instead, you want $0 < \mu(B) < \mu(A)$, this will depend on the measure. If the measure has atoms, then no, we can't necessarily this. If the measure is atomless (like the Lebesgue measure), then we can accomplish this. Define $$F(x) = \mu(A \cap (-\infty, x]), \,\,\,\, x \in \mathbb R.$$ This function is continuous with $$\lim_{x \to -\infty} F(x) = 0, \,\,\,\,\,\,\, \lim_{x\to \infty} F(x) = \mu(A)$$ so by the intermediate value theorem, for any $c \in (0,\mu(A))$, we can find $x$ such that $\mu(A \cap (-\infty, x]) = F(x) = c$.

Then put $B = A \cap (-\infty, x]$.

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  • $\begingroup$ Thanks, but the subset B need to be bounded according the question $\endgroup$ – The Student Mar 5 '18 at 0:18
  • $\begingroup$ Replace $(-\infty,x]$ with $[-x,x]$. The proof remains the same. $\endgroup$ – User8128 Mar 5 '18 at 5:48

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