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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.

1 vote

Completeness of Borel measure

No, it is not possible for $\mu$ to be complete. There exists a closed subset $K$ of $X$ with $\mu(K)=0$ and a continuous onto map $f\colon K\to2^\omega$. With $K,f$ as above, if $A\subseteq …
George Lowther's user avatar
4 votes

Borel kernel over an analytic set implies existence of a Borel map

The following is a construction of an analytic set $A\subseteq\omega^\omega\times\omega^\omega$ whose vertical sections exclude at most a single point, and which does not admit a Borel uniformization. …
George Lowther's user avatar
3 votes
Accepted

Existence of dominating measure for weak*-compact set of measures

There always exists a dominating measure. First, given two finite measures $\mu,\nu$ on $(\Omega,\mathcal{F})$, the Lebesgue decomposition theorem says that there is an $A\in\mathcal{F}$ such that $1 …
George Lowther's user avatar
13 votes
Accepted

What is a Gaussian measure?

You could alternatively try defining Gaussian measures as $2$-stable distributions. This does remove any reliance on finite dimensional projections, and even removes reference to topology. Let $V$ be …
George Lowther's user avatar
17 votes
Accepted

Bochner integral of stochastic process = path by path Lebesgue integral?

Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question. For a p …
George Lowther's user avatar
10 votes

Does there exist an event independent of a given sigma-algebra?

No. If the probability space contains atoms then you can easily construct sub-$\sigma$-algebras which are not independent to any nontrivial events. However, we can still construct examples, even for …
George Lowther's user avatar
23 votes
Accepted

Is arbitrary union of closed balls in $\mathbb{R}^n$ Lebesgue measurable?

No, in dimension $N>1$, it does not have to be Borel measurable. E.g., in 2 dimensions, consider, a non Borel measurable subset of the reals $S$, and let $A$ be the union of closed unit balls centered …
George Lowther's user avatar
29 votes

Is there a measure zero set which isn't meagre?

There's already been some good answers to this. However, this is something that I have also thought about recently, because I happen to have come across several meagre sets of full Lebesgue measure in …
George Lowther's user avatar
9 votes
Accepted

Is every probability space a factor space of the Haar Measure on some group?

It is possible to find the following: A compact abelian group G with Haar measure $\mu_G$, a subset $S\subseteq G$ of full outer Haar measure and a measurable function $f\colon S\to X$ with $\mu_P(E)= …
George Lowther's user avatar
26 votes

Non-Borel sets without axiom of choice

Measure theory without the Axiom of Choice (not even countable choice) is discussed in Fremlin, Measure Theory, Volume 5, Chapter 56. This is freely available online. Thanks to MO and ex-falso-quodlib …
George Lowther's user avatar
5 votes
Accepted

Non-existence of integral with respect to Poisson Random Measure

As I mentioned in my comment, you can prove the statement and its converse by looking at the moment generating function. Supposing that f ≥ 0 and λ > 0 is a real number, the following is true for a Po …
George Lowther's user avatar