What makes the elements of sigma algebra measurable (and measurable w.r.to which measure)? I'm familiar with the definition of sigma algebra defined over a set, which is a collection of subsets that is closed under countable unions and also under complements. While reading about probability theory, it was mentioned that by limiting ourselves to the sigma-algebra, we are avoiding some pathological behaviours caused by what are called as non-measurable sets. 


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*From the definition of sigma algebra, I don't see how it only consists of measurable sets. Is it an implication of the definition? If yes, how is it avoiding admitting non-measurable sets into sigma algebra? 

*When they say measurable/non-measurable, what is the measure they are talking about? Lebesgue, counting, probability? It seems there is an implicit measure every time someone says a set is measurable/non-measurable.

*Can anyone give an example of a set which is measurable w.r.to one measure but not w.r.to another measure?

*How important is it to really worry about sigma algebras and measurable/non-measurable sets?
Most of the continuous distributions are defined on real line and almost every subset of real line that I can think of is a part of sigma algebra and has a definite measure on it.  It feels like these concepts are mostly used in definitions to have a concrete theory of probability. I don't remember needing any of those concepts while working with probability/random variables. Are these concepts only useful in developing further theory of probability?

 A: *

*This is by the definition of measure (see @DEATH_CUBE_K 's answer)

*If not specified, it is usually the Lebesgue measure. If you are dealing with a random variable $X:(\Omega,\mathcal{F},P)\to\mathbb{R}$, the measure might be $P\circ X^{-1}$ given by its cumulative distribution function.

*Vitali sets $V$ are non-measurable w.r.t. Lebesgue measure,but are measurable w.r.t. cardinal measure (i.e. $\operatorname{card}(V)=\infty$)

*Non-measurable sets are constructed in a "non-constructive" way (using Axiom of Choice), so it is understandable that you never meet such sets in real-life applications. From a theoretically point of view, Ulam's theorem says that one cannot extend a continuous measure on real line such that all subsets of $\mathbb{R}$ are measurable, in particular there are subsets of $\mathbb{R}$ which are not measurable Lebesgue. Working witn non-measurable sets leads to pathological behavior, such as Banach–Tarski paradox. You may also read Non-measurable set, it gives more information.

A: *

*Let $(X, \Sigma, \mu)$ be a measure space, that is, $X$ is a non-empty set, and $\Sigma \subset \mathcal{P}(X)$ is a $\sigma$-algebra of subsets of $X$, and $\mu: \Sigma \to [0,\infty]$ is a measure. By definition, a set $E \subset X$ is measurable with respect to $\mu$ if and only if $E \in \Sigma$, i.e., we declare which sets are to be measurable by constructing $\Sigma$. Thus, a set is not measurable if it does not belong to $\Sigma$. 

*When we say a set is measurable or non-measurable, we always have the measure space $(X, \Sigma, \mu)$ in mind. Sometimes the measure space can be understood from the context. 

*If two measures $\mu, \nu$ are defined on the same $\sigma$-algebra, say $\Sigma$, then there exists no set which is $\mu$-measurable by not $\nu$-measurable. So you would need to be looking at totally different measure spaces which makes the question somewhat enlightening. 
