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?


1 Answer 1


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. $\mu^*(E) = \mu^*(E^c) = 1$.)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.