Can a non-measurable set be measurable? (Seriously)

Question

Perhaps I don’t speak English as well as I thought I did. In Folland, and other sources, I have encountered the following definitions: A “measurable space” consists of a set X together with a sigma-algebra M, which is a subset of the power set of X. And a “measurable set” is any element of that sigma-algebra.

In these definitions, there is nothing that says the measurable sets in question can be measured. Thus, for example, consider any set X that has a non-measurable set E which is a subset of its power set. Then the set X, together with the set E, its compliment, and the null set form a sigma-algebra, and thus a “measurable space.” Since E is an element of this sigma-algebra it is “measurable”, although by assumption it is non-measurable. A contradiction.

The definitions thus appear to be utter non-sense. But perhaps, I do not understand English.

Answer 1

I will venture to try to answer my own question:

For any given “measurable space,” and for any “measurable set” in that space, there always exists at least one perfectly good measure for which the set can be measured, namely m(.) = 0 for every set.

Answer 2

If $X$ is a set, and we fix a sigma-algebra $\mathcal{M}$ on $X$, then the pair $(X,\mathcal{M})$ is called a measurable space.

Further, if $\mu:\mathcal{M}\rightarrow \mathbb{R}_+$ satisfies a short list of properties (related to the notion of area) then $\mu$ is called a measure, and the triple $(X,\mathcal{M}, \mu)$ is called a measure space. So, with every measure space comes a measurable space. A set $E\subset X$ is called measurable if it belongs to $\mathcal{M}$ because, in the case of the measure space, $E$ would be in the domain of the measure $\mu$. The non-measurable sets are the rest of the subsets of $X$, which do not belong to $\mathcal{M}$.

My point is, to determine the non-measurable sets of $X$, we do not need a measure, only the measurable space $(X,\mathcal{M})$: they are the sets in the power set of $X$ not in $\mathcal{M}$.