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I just wanted to know the problem encountered in development of measure theory that gave rise to the condition of measurability. I mean, why exactly can't a measure be defined on any arbitrary set?

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  • $\begingroup$ You might want to consider looking at Vitali's theorem. $\endgroup$ – user305860 Dec 19 '16 at 9:17
  • $\begingroup$ Then you probably get sets for wich you cannot define a useful value for it's mass… e.g. the Vitali set: en.wikipedia.org/wiki/Vitali_set $\endgroup$ – Gono Dec 19 '16 at 9:19
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The Vitali set has already been provided as a proper example, but I'll try to explain it a bit further. I assume you know the construction of Vitali set, but I'll remind it: Consider equivalence relation on $\mathbb{R}$ given by a following formula: $$x\sim y \iff x-y\in \mathbb{Q}.$$ By axiom of choice, I can pick a single element from each equivalence class in such way, that it is in $[0,1]$. Set of all such elements is my Vitali set $V$.

Assume $V$ is measurable. Since Lebesgue measure is shift invariant, we can consider all sets of form $q+V:=\{q+v|\; v\in V\}$ which are measurable as well. If $V$ has Lebesgue measure $0$, then $$\bigcup_{q\in \mathbb{Q}} (q+V) = \mathbb{R}$$ has Lebesgue measure $0$ as well, as a countable union of null sets. If we consider now that $V$ has a positive measure $\epsilon>0$, then clearly $$[-1,2]\supset \bigcup_{q\in \mathbb{Q}\cap [0,1]} (V+q)$$ has infinite Lebesgue measure, a contradiction.

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Measurability (in the sense of Carathéodory) guarantees that the measure is additive. You can apply the outer measure to arbitrary sets, but you can't expect the resulting "measure" to be additive.

More precisely, suppose $A$ is a non-measurable set (with respect to a measure $\mu$). Then we can find a set $B$ such that $\mu^*$ restricted to the algebra of sets generated by $A$ and $B$ is not additive. This is by definition: by Carathéodory, we have $B$ such that $$\mu^*(A^c\cap B)+\mu^*(A\cap B)\neq \mu^*(B).$$ But $A\cap B$ an $A^c\cap B$ are disjoint, so this falsifies additivity.

This may not be entirely convincing: after all, who's to say that we have not chosen a particularly nasty $B$. However, it turns out, we may take it to be a nice set: by definition of the outer measure, we have a measurable set $C\supseteq B$ such that $\mu(C)=\mu^*(B)$. Note that since $\mu^*$ is subadditive, we have that $$\mu^*(A^c\cap B)+\mu^*(A\cap B)>\mu^*(B).$$ But since $C\supseteq B$ and $\mu^*$ is monotone, we have that in fact $$\mu^*(A^c\cap C)+\mu^*(A\cap C)\geq \mu^*(A^c\cap B)+\mu^*(A\cap B)>\mu^*(B)=\mu(C).$$

This shows the non-measurable sets not only do not play nice with arbitrary, possibly non-measurable sets, they don't even work together with the nice, measurable sets, not even the ones on which the original measure was defined.

In short, if you want to have a $\sigma$-algebra (or even an algebra!) of sets $\Sigma$ which contains the algebra $\mathcal M$ of measurable sets, and such that $\mu^*\mathord{\upharpoonright}_\Sigma$ is (finitely!) additive, then the only choice is $\Sigma=\mathcal M$. In the case of the Lebesgue measure, it means that if you take the Borel sets, and toss in any non-measurable set, the outer Lebesgue measure will not be additive on the resulting algebra of sets.

The point of the Vitali construction is to show that the Carathéodory condition is not vacuous, i.e. that there are, in fact, non-measurable sets (assuming axiom of choice).

(Note that it does not mean that you cannot possibly extend $\mu$ to a bigger $\sigma$-algebra, only that any such proper extension will disagree with $\mu^*$, and thus will not be “canonical”. Indeed, there are models of set theory (with choice) in which we have a measure on the whole $\mathcal P({\bf R})$, but such a measure is by necessity not regular.)

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