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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 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 …

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 …

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 …

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 …

If you work in terms of topologies on M, then the way to go is to use the theory of dualities from functional analysis. The basic theory of these things can be found in books called "Topological Vecto …

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 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 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) …

$\newcommand{\N}{\mathbb{N}}\newcommand{\R}{\mathbb{R}}$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 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 …

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), …

In effect, you are asking if $\bigvee\limits_{i \in \omega}B_i = \bigcup\limits_{i \in \omega}B_i$, where the left hand side is the closure of the union, which is the join/supremum of the family $\{B_ …

The question has a trivial negative answer, as long as an atomlessly measurable cardinal exists (if one doesn't it is vacuously true, of course). Given an atomless probability measure $\mu$ on $(Y, \m …

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 …