Search Results

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 …

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 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 …

The criterion suggested in the question works fine for $\sigma$-finite spaces, and Michael Greinecker's answer is correct under this assumption. However, the suggested criterion is not (provably) suf …

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 …

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 …

You got a wrong answer on Math Stackexchange from Daron. The free Boolean algebra on countably many generators is the Boolean algebra of clopens of $2^\omega$ (topologized with the product topology), …

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 …

The answer is no. Jiří Nedoma proved that if $(X,\Sigma)$ is a measurable space $|X| > 2^{\aleph_0}$, then the diagonal is not a measurable subset of $(X\times X, \Sigma \otimes \Sigma)$. (The articl …

The fact you are probably looking for is that, for any Baire space $X$ (e.g. a completely metrizable space or a compact Hausdorff space) the inclusion map $\mathfrak{R}(X) \rightarrow \mathfrak{B}o(X) …

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 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 …

Iosif Pinelis has given an answer to question 1 and partial answers to 2 and 3. Since he advised me to turn my comments into an answer, here it is. I will deal with the case where the axiom of choice …

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 …