I'm a little bit in truble with definition of measurable set. In one definition, I have :
Let $X$ a set $\mathcal F$ a $\sigma -$algebra on $X$. Then a set $A\subset X$ is measurable if $A\in \mathcal F$.
I guess that in "measurable set", we can measure them, no ? So in such a space, what could be a measure ?
For example, $\mathcal P(\mathbb R)$ (the power set of $\mathbb R$) is a a $\sigma -$algebra on $\mathbb R$. And with this $\sigma -$algebra, all sets are measurables. But in such space, what could be a measure ? I sa somewhere (If my memory is good), that there are no measure on $(\mathbb R,\mathcal P(\mathbb R))$ because for all function $\mu:\mathcal P(\mathbb R)\to \mathbb R^+$ s.t.
1) $\mu(\varnothing )=0$
2) $A\subset B\implies \mu(A)\leq \mu(B)$, $A,B\in \mathcal P(\mathbb R)$
3) $\mu\left(\bigcup_{n\in\mathbb N} A_i\right)\leq\sum_{n\in\mathbb N}\mu(A_i)$, $A_i\in \mathcal P(\mathbb R)$
then there are always sets $C,D\in \mathcal P(\mathbb R)$ s.t. $$\mu(A\cup B)<\mu(A)+\mu(B),$$
and thus there is no measure on $(\mathbb R, \mathcal P(\mathbb R))$.
- So how can we prove that a space $(X,\mathcal F)$ has a measure or not, and if there is no measure, can we talk about measurable space ?
Best Answer
Yes, a measure space has a measure function that is defined on a $\sigma$-algebra. So having a $\sigma$-algebra is a prerequisite for having a measure, and hence a set with just a $\sigma$-algebra is "measurable" (we could define a measure on it, but in some cases we cannot..) The $\sigma$-algebra is the "wish list": all sets we would like to be able to measure using a measure function. We already anticipate on our wish and call all sets in the $\sigma$-algebra "measurable").
It's analogous to having a set with a topology and calling its members "open". We have a set and a $\sigma$-algebra and call its members "measurable", or we have a set with a bornology and call its members "bounded", or convexity and "convex" etc.
On the powerset of a set $X$ we can always define a measure pick $p \in X$ and define: $\mu(A) = 1$ iff $p \in A$, $\mu(A) = 0$ if $p \notin A$. Pretty boring, but a valid measure.
On the power set of the reals we cannot define a measure that gives all intervals the Lebesgue measure (so $\mu([a,b]) = b-a$ for $a < b$) and also is translation-invariant (so that $\mu(A+x) = \mu(A)$ for all $A$ and all $x \in \mathbb{R}$). We always have measures we can define, but not always "nice" measures with extra good properties.