# a set which is not measurable in relation to an outer measure

given an outer measure $\eta: 2^X \to [0, \infty]$ we call a subset $E$ of $X$ $\eta$-measurable (measureable with respect to the outer measure), if for every subset $Q$ of $X$, the following holds:

$$\eta(Q) = \eta(Q \cap E) + \eta(Q \cap E^C)$$

so heres my question: do you know any examples of sets E that are not $\eta$-measurable? Thanks!

-

Whether or not $\eta$-nonmeasurable sets exist depends on $\eta$. For example, if $X$ is a finite set and $\eta(Q)$ is the number of elements in $Q$, then every set is $\eta$-measurable. For the Lebesgue measure, there are nonmeasurable sets such as Vitali set.

-
Ah, thank you! i started learning measure theory and your answer was helpful for me. – Stefan Jun 16 '12 at 17:59
@Stefan Since Vitali set is quite an abstraction, I'll add an example of the world's simplest non-measurable set. Let $X=\{1,2\}$ (or any set, really) and define $\eta(Q)=0$ if $Q$ is empty and $\eta(Q)=1$ otherwise. This is an outer measure. Which sets are $\eta$-measurable? Only $\varnothing$ and $X$. In particular, $\{1\}$ is not $\eta$-measurable. – user31373 Jun 16 '12 at 18:40

As Leonid mentioned, it depends entirely on what the measure $\eta$ is. As far as Lebesgue measure on $\mathbb{R}$ is concerned, sets like the Vitali set exist which are certainly not Lebesgue measurable, but are understood to exist. It is important to notice that the description of a Vitali set relies on the axiom of choice, and there is no constructive way to show what a non-measurable set looks like. More information can be found here.

The fact that Vitali sets use the axiom of choice is not arbitrary. As the Wikipedia page mentions, Solovay showed that there are versions of ZF where the axiom of choice doesn't hold and where each subset of $\mathbb{R}$ is indeed measurable with respect to Lebesgue measure. The Banach–Tarski paradox, which states that a solid sphere can be decomposed into subsets and then reassembled (using translations and rotations) to form two identical copies of the original ball, involves the axiom of choice and uses non-measurable sets in its decomposition.

-
There has been quite the discussion about non-measurable sets in this site as well. – Asaf Karagila Jun 16 '12 at 18:34

I like this more explicit example. (Explained in more detail in my book Measure, Topology, and Fractal Geometry, beginning of section 5.4.) On $X = \mathbb R$, let $H^{1/2}_1$ be the "method I" outer measure: $$H^{1/2}_1(E) = \inf \sum_i \mathrm{diam}(U_i)^{1/2}$$ where the inf is over all countable covers $E \subseteq \bigcup_i U_i$ by sets $U_i$ with diameter $\mathrm{diam}(U_i) \le 1$. But then consider sets $A=[0,1)$, $B=[-1,0)$ and $Q=[-1,1)$. Both $[0,1)$ and $[-1,0)$ have outer measure $1$, but $[-1,1)$ has outer-measure ${}\le \sqrt{2}$. So $A$ is not measurable since $1+1 > \sqrt{2}$.

-