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Suppose $E \subseteq [0,1]$ is a lebesgue measurable set with $m(E)=1$. Show that $E$ is dense in $[0,1]$.
I would appreciate any useful hints.

zhw.
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Aref
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2 Answers2

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Suppose there exists a non-empty open $\;A\subset[0,1]\;$ such that $\;E\cap A=\emptyset\;$, and let $\;a,b\in[0,1]\;,\;\;a<b$ be such that $\;(a,b)\subset A\;$ . Observe that $\;m(a,b)=b-a>0\;$, so then also $\;m(A)>0\;$ , and thus

$$1=m(E)<m(E)+m(A)=m(E\cup A)\le m[0,1]=1$$

which is a contradiction

DonAntonio
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Hint: $E$ is dense in $X$ iff $E \cap A \neq \emptyset$ for any open set $A$ of $X$.

Lionel Ricci
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    This is the definition of a dense subset of $X$ in the abstract topological setting. A useful hint (in my opinion) needs to tie the definition to the measurability property of $E$ posited as a subset of $[0,1]$. Please improve your overly terse answer by adding material to make such a connection. – hardmath Apr 10 '16 at 18:19
  • TS asked for any useful hint and not a full answer. Recalling the definition of what a dense set is would to me, at least, be a useful hint. – Lionel Ricci Apr 10 '16 at 21:27
  • FYI, I agree with @hardmath That said, what you're wanting to do would work better if you defined dense in a more concrete way, such as $E$ is not dense in $[0,1]$ if and only if there exists an interval $I$ such that $I \subseteq [0,1]$ and $I$ is of positive length and $E \cap I = \emptyset.$ – Dave L. Renfro Apr 11 '16 at 16:09
  • Because I though about phrasing it this way, I have to agree with you. However putting anything more into it would make it more of a solution than a hint in my opinion. – Lionel Ricci Apr 11 '16 at 16:17