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Just came over this question in my mind: if I know a set has an infinite outer measure, does it mean it also has to have an infinite inner measure.

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  • $\begingroup$ This is related. The question isn't quite the same, but the answers provide the essential arguments. $\endgroup$
    – Xander Henderson
    Sep 17, 2018 at 18:07
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    $\begingroup$ A subset of the reals can have zero inner measure and infinite outer measure. In fact, a set with zero inner measure can be such that for each interval $I,$ the outer measure of its intersection with $I$ has outer measure equal to the length of $I.$ Seemingly even stronger (but I believe this will follow automatically from the interval property), a set with zero inner measure can "fill up the real line" so much that, for each set $E$ of positive (including infinite) measure, its intersection with $E$ will have outer measure equal to the measure of $E.$ $\endgroup$ Sep 17, 2018 at 18:39
  • $\begingroup$ Regarding the claims I just made, see Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure?. See also this 5 February 2006 sci.math post. $\endgroup$ Sep 17, 2018 at 18:41

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There exists $A\subset \Bbb R$ such that for any uncountable closed $C\subset \Bbb R$ we have $C\cap A\ne \emptyset \ne C \cap (\Bbb R\setminus A)$.

Any closed subset of $A$ is countable so $A$ has inner measure $0.$

Any closed subset of $\Bbb R\setminus A$ is also countable. So if $U$ is open and $U\supset A$ then $\Bbb R\setminus U$ is a closed subset of $\Bbb R\setminus A,$ so $\Bbb R\setminus U$ is countable, so $U$ has infinite measure.

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  • $\begingroup$ could you please explain why such a A exists? $\endgroup$
    – onRiv
    Aug 17, 2021 at 4:43
  • $\begingroup$ @Risto . Look up Bernstein Set on this site or on wikipedia. $\endgroup$ Aug 18, 2021 at 9:59
  • $\begingroup$ got it. thanks a lot! in fact i asked a duplicated question: math.stackexchange.com/q/4225931 $\endgroup$
    – onRiv
    Aug 18, 2021 at 13:24

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