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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.

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    $\begingroup$ You're correct that any set can belong to a sigma-algebra. When we refer to a non-measurable set, we have already fixed the sigma-algebra beforehand. $\endgroup$ – confused_wallet Mar 5 at 5:12
  • $\begingroup$ When you are dealing with multiple sigma fields, it is better to say with respect to which one. For example, $\mathscr{F}$-measurable as opposed to "Borel set" (that is, by definition, a Borel measurable set). $\endgroup$ – Will M. Mar 5 at 5:23
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    $\begingroup$ A set cannot be non-mesurable, it is non-measurable with respect to some measure. But this is often omitted. $\endgroup$ – Noel Lundström Mar 5 at 5:29
  • $\begingroup$ @Noel, yes thanks - a good point which is easy to forget. However, since the definitions above also do not specify the measure, we can assume any measure we please. $\endgroup$ – MPitts Mar 5 at 14:33
  • $\begingroup$ @confused. Ok. But unless you are saying we can fix the sigma-algebra beforehand, only in a way that avoids the non-measurable set, the problem I’ve described remains. $\endgroup$ – MPitts Mar 5 at 14:42
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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.

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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 $\mathcal{M}$.

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  • $\begingroup$ You may be right, but I don’t think “non-measurable set” are normally defined in that way. But even if that definition is correct, it does not mean “non-measurable sets” are “not measurable.” It only means they are not in M. Seems like a mis-use of the English language to me. $\endgroup$ – MPitts Mar 5 at 17:37
  • $\begingroup$ Well, in the measure space $(X, \mathcal{M}, \mu)$, "non-measurable" very literally means "not measurable". They are the sets not in the domain of the measure. Is there a certain example you're thinking of? $\endgroup$ – confused_wallet Mar 5 at 17:47
  • $\begingroup$ I don’t want to waste a lot of your time since this is all really a question of semantics rather than math. But to answer your question, start with X and a measure m, and take ANY set E that is not measurable by m. Then form the sigma algebra M by taking E, its compliment, X, and the null set. Then by definition (X,M) is a “measurable space” and by definition E is a “measurable set.” Yet, by hypothesis, E is not measurable, at least by m. $\endgroup$ – MPitts Mar 5 at 22:15
  • $\begingroup$ Right, if you change the sigma-algebra then it can change whether a set is measurable or not. As an analogy, it's kind of like how the function $f(x)=x^2$ can be surjective or not surjective depending on what you take as its domain and codomain. $\endgroup$ – confused_wallet Mar 5 at 22:28
  • $\begingroup$ I don’t see where I “changed” the sigma-algebra; but in any case you end with a “measurable set” that cannot be measured. $\endgroup$ – MPitts Mar 15 at 20:50

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