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The answer is no, as you suggested. Apart from the $\epsilon$-appoximation idea in your formulation, there isn't even a countable family of Borel sets $E_n$ with positive measure, such that every Bore …

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 …

One can see a counterexample easily for the reals $\mathbb{R}$ if the Continuum Hypothesis holds, for in this case the reals $\mathbb{R}$ are the union of a chain of countable sets. Simply well-order …

The Conway base 13 function is probably a standard example: the graph is dense because any restriction of the function to an open interval is surjective. But meanwhile, since the graph of the base 13 …

Such an approach will violate the Cantor-Hume principle, which asserts that "the number of elements" of a set $A$ should be invariant under equinumorsity. That is, if $A$ and $B$ can be placed into on …

The two notions are not equivalent. Indeed, they are not equivalent even when one considers completing the measure by adding all null sets with respect to any countably generated $\sigma$-algebra. Nev …

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 …

The answer is no, by a construction using the axiom of choice. We shall build a counterexample set $A$ by a transfinite recursive process of length continuum. At each stage, we shall promise that cer …

Following Pietro's lead, let me observe that if there is a measurable cardinal, then there is a counterexample. Suppose that $\kappa$ is a measurable cardinal. Then there is a $\kappa$-additive 2-val …

Your collection is not closed under complement. To see this, observe that the diagonal $\Delta=\{(x,x)\mid x\in\mathbb{R}\}$ is not in your collection, since the only rectangles it contains are single …

The function is easily seen to be Borel, since the graph of the function can be defined using only natural number quantifiers. In particular, a number is in the support if and only if there is a last …

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 …

This is a great question! But unfortunately, the answer is no, the Lebesgue measure on the unit interval is not finitely $\vee$-additive. Theorem. There are two disjoint regular open sets $L$ and $R …

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 …

(My argument is somewhat easier if you consider games where the players play $0$s and $1$s, so that the payoff set is in Cantor space $2^\omega$, and we use the usual coin-flipping probability measure …