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Let $(X,S,\mu)$ be a measure space s.t. $\mu(X)=1$.

Let $\mu^{*}$ be defined on $X$ by:

$$\forall E\subseteq X:\,\mu^{*}(E):=\text{inf}\left\{\sum_{i=1}^{\infty}\mu(A_{i})\,|\, A_{i}\in S,E\subseteq\cup A_{i}\right\}$$

I have a set $E$ s.t. $\mu^{*}(E)=1$, does this mean $\mu^{*}(E^{c})=0$?

I have tried to work with the definition, given $\epsilon>0$ there is $N\in\mathbb{N}$ and $\{A_{i}\}_{i=1}^{N}\subseteq S$ s.t. $\sum_{i=1}^{N}\mu(A_{i})\geq1-\epsilon$ and $E\subseteq\cup_{i=1}^{N}A_{i}$

I want to use the $A_{i}$'s to get some set $B$ s.t $E^{c}\subseteq B$ and $\mu(B)<\epsilon$, but I didn't manage to find such a set.

Can someone please help me understand if this claim is true, and if so how to prove it?

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1 Answer 1

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Answer: No, it doesn't. There exists a set $\mathbf{E \subset [0,1]}$ with $\mathbf{\mu^*(E) = 1}$ and $\mathbf{\mu^*(E^c) > 0}$.

Note that if for a set $E\subset [0,1]$ we have $\mu^*(E) = 1$ then by definition $$\mu^*(E^c) = 0 \text{ if and only if } E \text{ is measurable.}$$

There exists a non-measurable set $E\subset [0,1]$ with $\mu^*(E) = 1$ as explained in this math.SE post Vitali-type set with given outer measure. For this set $E$, we have $\mu^*(E) = 1$ and $\mu^*(E^c) > 0$. (In fact, we can construct $E\subset [0,1]$ s.t. $\mu^*(E) = \mu^*(E^c) = 1$.)

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  • $\begingroup$ How exactly is the procedure for the last statement, the $\mu^*(E^c)=1$ $\endgroup$
    – George
    Aug 12, 2021 at 18:21

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