# $\sigma$-algebra and measurable set.

Today I am reading David Williams's Probability with Martingales. In chapter one, He introduce the notion of Measurable space:

A pair $(S,\Sigma)$,where $S$ is a set and $\Sigma$ is a $\sigma$-algebra on $S$, is called a measurable space. An element of $\Sigma$ is called a $\Sigma$-measurable set of $S$.

I know, all the measurable sets form a $\sigma$-algebra. But for an arbitrary $\sigma$-algebra. for example , all the subsets of $S$ form a $\sigma$-algebra.But in this $\sigma$-algebra $2^S$, there exist a non-measurable set. Why author called elements in $\Sigma$ a $\Sigma$-measurable set of $S$? Does this make sense?

• It's a definition. There are many senses in which a set may be measurable, each corresponding to a different measure. – Alex G. May 28 '14 at 14:46

I think you're confusing measurability and measures. Measurability is determined by the $\sigma$ algebra. Usually, when introducing a probability space $(\Omega,\mathcal{F},P)$ or simply a measure space $(S,\Sigma)$, we agree that the "measurable sets" are the ones in $2^X$ we've chosen to be in $\Sigma$. That is, $\Sigma \subsetneq 2^X$ i.e. we do not take all possible subsets of $X$.

What you're thinking of when you say "there exists a non-measurable set in $2^X$" is the specific example of Lebesgue measure. There are sets in $2^X$ which, if I include in $\Sigma$, the measure cannot be translation invariant -- but this is an additional restriction you are imposing. À priori, there is nothing "non-measurable" about these sets: if I took $(\mathbb{R}, 2^{\mathbb{R}})$, then any $f : \mathbb{R} \rightarrow \mathbb{R}$ would be measurable.

"Non-measurable" is always taken in the context of a specific measure. There exist measures on $S$ for which every $E \in 2^S$ is measurable.

"$\Sigma$-measurable" is just a convenient way of saying "measurable for any measure on the pair $(S,\Sigma)$".

• I might misunderstand but I think it is misleading to say "non-measurable" is taken in the context of a specific measure. What do you mean by "measurable for any measure"? Measures are defined on measure spaces $(X, \mathcal{F})$, so you've already specified which sets should be measurable: the ones in $\mathcal{F}$. One should therefore say, "There exist measures on $(S,2^S)$"; the "for which ... " part is redundant. Question: for any set $X$, does there exist a $\sigma$-finite measure $\mu$ on $(X,2^X$)? – snar May 28 '14 at 15:11

"Non-measurable" completely depends on the $\sigma$-algebra. Specifically, given a measurable space $(S, \Sigma)$, a set $N \subseteq S$ is non-measurable iff $S \notin \Sigma$.

The confusion arises because a given set $S$ has multiple possible sigma algebras $\Sigma$. For example, take $S = \mathbb{R}$. We could take $\Sigma_1 =$ Lebesgue sigma algebra. But we could also take $\Sigma_2 =$ the power set of $\mathbb{R}$ (i.e. all subsets). Let $A$ be any subset of $\mathbb{R}$; a third sigma algebra is $\Sigma_3 = \{\emptyset, A, A^c, \mathbb{R} \}$. This last example is the smallest sigma algebra for which the set $A$ is measurable.