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Let I be the unit interval with the Borel $\sigma$-algebra. There is no $\sigma$-algebra on the set of measurable functions from I to I such that the evaluation functional $e:I^I\times I\to I$ given b …

Non-standard analysis has been quite successful in settling existence questions in probability theory. Hyperfinite Loeb spaces allow for several constructions that cannot be done on standard probabili …

Some reasons can be found here. Borel measurable functions are much nicer to deal with. Every continuous function is Borel measurable, but the inverse of a Lebesgue measurable set may not be Lebesgue …

A measure space $(\mathbb{P},\Omega,\mathcal{A})$ is atomless if for all $A\in\mathcal{A}$ with $\mathbb{P}(A)>0$ there exists $B\subset A, B\in\mathcal{A}$ such that $0<\mathbb{P}(B)<\mathbb{P}(A)$. …

There is an impossibility theorem: If you let $\mathcal{L}$ be the the space of Borel-measurable functions $f:[0,1]\to[0,1]$, and $e:\mathcal{L}\times [0,1]\to[0,1]$ the evaluation given by $e(f,x)\ma …

It is shown in "Linear Operators, Part I" 1988 by Dunford and Schwarz as IV.8.16 on page 296 that the dual of $L_\infty(\mu)$ can be identified with the space of finitely additive bounded signed measu …

To close a gap: From the answer of Gerald Edgar, we know that the answer to the second question is no if the spaces involved have cardinality larger than $\mathfrak{c}$. This leaves open what happens …

The KET fails for general measurable spaces, the classical example can be found in a paper by Andersen and Jessen. Topological assumptions are necessary so that the resulting measure is not only finit …

Yes, that is the approach of Daniell and Stone. To see how the approach works and how general it is, you can take a look at the four-part series on "Notes on Integration" by M. H. Stone. See here, her …

It is a consequence of Riesz' Lemma that every open ball in an infinite dimensional normed space contains a disjoint sequence of smaller open balls. They all have the same measure under a translation …

The answer is yes. First, it follows from the following result and the Riesz–Markov–Kakutani representation theorem that we can always find a suitable Baire measure representing a positive linear fu …

The problem of choosing subsets at random has been studied in a rather different context in mathematical economics. Suppose we choose a subset of $[0,1]$ by independently throwing a fair coin for each …

Yes. Every uncountable Polish space is isomorphic as a measurable space to the unit interval by Kuratowski's isomorphism theorem and admits, therefore, a nonatomic probability measure. On a Polish spa …

No. Let $\Omega=[0,1]$. If $x\in[0,1]$, let $\delta_x$ be the point mass at $x$. They are all Radon measures. It is not that hard to show that $\|\delta_x-\delta_{x'}\|=2$. So you can construct an unc …

The standard reference is probably still the book Vector Measures by Diestel and Uhl. If $f$ is the pointwise limit of Borel measurable functions, then $f$ is Borel measurable. To see this, note that …