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This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some ti …

The bold statement is not true in the generality in which you state it. Nevertheless, something very like it is true, if one adopts the perspective and philosophy of large cardinal set theory and rest …

Since the existence of non-measurable sets is often seen as undesirable, we naturally want to have as many measurable sets as possible. With Lebesgue measure on the reals, for example, if we were to s …

Evidently, there are measure zero sets with a non measurable sum. The article begins as follows: Krzysztof Ciesielski, Hajrudin Fejzi´c, Chris Freiling, Measure zero sets with …

If you assume the countable axiom of choice, then most sets of reals are not Borel. Under AC, what you get is that there are continuum many Borel sets, that is, $2^{\aleph_0}$ many, but $2^{2^{\aleph_ …

The answer is yes. Every function on the reals is the sum of two injective functions, and this can be done in a highly effective manner, constructing the two functions $g,h$ from $f$ without any need …

This question arises from an issue in my post on Ashutosh's excellent question on Restrictions of the null/meager ideal. Question 1. Does every set of reals contain a measure-zero subset of the same …

No, the existence of a non-Lebesgue measurable set does not imply the axiom of choice. If ZF is consistent, then set-theorists can construct models of ZF having a non-Lebesgue measurable set, but stil …

The answer to the question is that it is independent of ZFC, if one is speaking of outer measure. The right context for the question and its answer is the very active research area known as Cardinal …

The assertion that $2^{\aleph_0}=2^{\aleph_1}$ is known as Luzin's hypothesis, and was presented by Luzin as an alternative to Cantor's continuum hypothesis. This is now known to be independent of ZFC …

The existence of such a measure is equiconsistent to the existence of a measurable cardinal, one of the large cardinal notions, and if ZFC is consistent, cannot be proved in ZFC. (See the notion of re …

There are several issues. On the one hand, any set can be made countable by forcing, and this process will certainly affect the measure of the set, if it did not have measure zero in the ground mode …

You are asking whether every set with inner measure $0$ has measure $0$ with respect to the completion measure. The Lebesgue measure, for example, does not have this property, since the usual Vitali s …

Let $D_0,D_1,\ldots$ enumerate a sequence of disjoint intervals in the unit interval with $\bigcup_n D_n$ open dense and having measure less than $1$. For example, place a very tiny interval around ea …

Let me mount the kind of cardinality argument to which you allude. You had asked for a proof that the $\sigma$-algebra of Lebesgue measurable sets is not countably generated. But in fact, a much stron …