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The answer is yes, following from Klee's extension of Keller's theorem on homeomorphisms of infinite dimensional compact convex sets: Klee, V. L., Some topological properties of convex sets, Trans. Am …

It seems that the part you're having trouble with is proving that the restriction of a bounded $\tau$-additive Borel measure to a Borel subset is $\tau$-additive (and you can follow the rest of Bogach …

The first time I taught myself rigorous measure theory, I used the "linear functionals on $C(X)$" approach for compact Hausdorff spaces, so I got first-hand knowledge of where it doesn't work for prob …

It is not true that $B$ is necessarily a measure algebra. The counterexample is due to Michel Talagrand, who constructed a Maharam algebra that is not a measure algebra. Maharam, D., An algebraic cha …

I decided not to "comment-answer" this question, so other people have duplicated some of my answer in the comments. Since there are several questions, I will separate them and answer them individually …

There are no such locally compact groups, because if $G$ is a locally compact group under the topology $\tau$, then the Weil topology $\tau_\mu$ defined by the Haar measure $\mu$ is the same as the or …

$\newcommand{\N}{\mathbb{N}}\newcommand{\R}{\mathbb{R}}$There are examples on $\R$ with the Borel $\sigma$-algebra $\mathcal{B}$. We take the null ideal to be the meagre Borel sets $\mathcal{M}$ (the …

The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire prope …

This can be proved without introducing ultrafilters by name, by doing "finitary measure theory" and using Zorn's lemma. An algebra $A$ on a set $X$ is just a $\sigma$-algebra without the $\sigma$, i …

Separability is not necessary. In fact, tightness of a family of Borel probability measures implies relative compactness in the vague/weak-* topology on any completely regular space. For instance, thi …

The answer is no. Assume that such a measure $\mu$ exists. First, since every singleton in $[0,1]$ is closed with empty interior, $\mu(\{x\}) = 0$ for all $x \in [0,1]$. Write $B_{x,\epsilon}$ for th …

The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the …

The problem is that you have to take uncountable unions of sets of the form $[a,b) \times [c,d)$ to get every open set in the Sorgenfrey plane, so the $\sigma$-algebra generated by $[a,b) \times [c,d) …

It is also possible to show that the category of measurable spaces, $\newcommand{\Mble}{\mathbf{Mble}}\Mble$, is not cartesian closed by using more category theory and less measure theory (though stil …

Fremlin's measure theory textbook is a good reference for these things. I am splitting things up into the Boolean algebra part and the real-valued functions part. Complete Boolean algebras: The way …