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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.
8
votes
Accepted
Sufficient condition for the graph of a measurable map to be measurable
The assumption that $Y$ is countably separated cannot be meaningfully weakened. The following is Proposition 2.1 of [Musial, Kazimierz. "Projective limits of perfect measure spaces." Fund. Math 110.16 …
6
votes
Accepted
Are measurable maps with countably separated image in a Banach space always strongly measura...
If $(\Omega,\Sigma_\Omega)$ is standard Borel, $E$ metrizable, and $X:\Omega_\Sigma\to E$ measurable, then $X(\Omega)$ is separable. This is Proposition 1.11 in "Probability Distributions on Banach Sp …
5
votes
Accepted
How probability-rich is the $\sigma$-algebra generated by a sequence of sets? (Sierpiński's ...
Here is a proof that probability measures have closed range under DC, summarized from the discussion with Iosif Pinelis. I try to be as explicit as possible about any choice arguments used.
For any pr …
2
votes
Accepted
Pushforward of measure under surjective map
The answer to the first part is no. The point is that surjectivity is not as strong a condition in this context as one might wish for.
Let $X=Y=Z=[0,1]$ with the Borel $\sigma$-algebra and $\mu_X=\mu_ …
5
votes
Accepted
Pointwise limit of a "net" of measurable functions is measurable?
No. Let $N\subseteq[0,1]$ be a non-measurable set. Let $\mathcal{F}$ be the family of indicator functions of finite subsets of $N$. Then $\mathcal{F}$ is a net under the pointwise ordering with limit …
1
vote
Accepted
Sufficient conditions for the graph measurability of a multivalued function
Graph measurability of $P$ is not sufficient. Let $E\subseteq[0,1]^2$ be a Borel set whose projection $\pi(E)$ onto the first coordinate is not Borel. Let $X=\mathbb{R}$ and let $B$ have the constant …
3
votes
Accepted
Relationship between Baire sigma algebra and Borel sigma algebra of an uncountable product
Let all factor spaces be nontrivial compact Hausdorff spaces. Then every continuous function is determined by countably many coordinates, and so is, consequently, every Baire measurable set.
It follow …
2
votes
Accepted
Conditional expectation: commuting integration and supremum
Note: This answer used to be a counterexample that missed the mark.
The way to get around he definitional issues with conditional expectations is to work with regular conditional probabilities in prod …
10
votes
Measure without measurable sets
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 …
13
votes
Product of Borel sigma algebras
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 …
2
votes
A limit definition of regular conditional probability
Essentially yes. Only essentially, because the conditional probability on the right hand has to be defined in the first place.
The limit exists almost everywhere in the topology of weak convergence of …
4
votes
Measurability of random function with values in $C(K,E)$
The Borel $\sigma$-algebra on the space $C(K,E)$ of continuous functions from a compact metrizable space $K$ to a separable metric space with the induced uniform metric is generated by the evaluation …
2
votes
Accepted
Let $c: X \times Y \to \overline{\mathbb R}$ be $\gamma$-measurable. Is $c_x:Y \to \overline...
No to both. Let $\gamma$ be the uniform distribution on $\Delta=\{(x,x)\mid x\in [0,1]\}$, the diagonal of $[0,1]^2$. The marginals are simply the uniform distribution on $[0,1]$. Fix some function $g …
1
vote
Accepted
Is the conditional expectation of a Caratheodory function a Caratheodory function?
Here is a positive answer for the case that $\Sigma_0$ is generated by a random variable with values in a Polish space, so that we can use regular conditional probabilities and for some kernel $\kappa …
9
votes
Is a random subset of the real numbers non-measurable? Is the set of measurable sets measur...
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 …