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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.
3
votes
Reference for Function-Valued Random Variables?
Brownian motion, i.e. Wiener measure, is a good source of ideas and examples here.
For instance if $W_t$ is 1-dimensional standard Brownian motion at time $t$ and $$P(\forall x\,F(x)=x^2)=1$$ and $Y=W …
1
vote
Continuity of Brownian motion constructed from Kolmogorov extension theorem?
The process $X$ you mention is uniformly continuous on the rationals* in the compact interval $[0,n]$, with probability 1. So you define Brownian motion $B$ to be the unique continuous extension of: $ …
4
votes
Accepted
When are events in tail $\sigma$-algebra the limsup of some sequence of events?
No, let $\mathcal S_0=\{\{0,1\},\emptyset\}$ and let
$\mathcal S_1$ be the powerset of $\{0,1\}$. Let
$\mathcal F_n=\mathcal S_0^{n-1}\times\mathcal S_1
\times\mathcal S_0^{\infty}$.
Let us write $X …
2
votes
1
answer
201
views
Non-uniqueness in Krylov-Bogoliubov theorem
So apparently the Krylov-Bogoliubov theorem says that every continuous function $f:X\to X$ on a compact metrizable space $X$ has an invariant probability measure $\mu$.
Of course, if $X$ is just a si …
3
votes
Accepted
Binarily universal members of $[0,1]$
$U$ is a $G_\delta $ set, hence Borel at level 2 of the Borel hierarchy. It has Lebesgue measure 1 and is also comeager, so it is large both in the sense of measure and of Baire category.
1
vote
Design measure, which cannot be factorized as a product of measures
I suppose you can just randomly perturb the independent distribution. That is, let
$$\mu(x_i,y_j)=\mu_x(x_i)\mu_y(y_j)+\epsilon_{i,j}$$
where the $\epsilon_{i,j}$ form a random matrix (sufficiently ra …
2
votes
Intuitional feeling of harmonic measure on one-third Cantor set
Let's say we start a 1-dimensional Brownian motion somewhere in the unit interval $[0,1]$, and look at where it hits the middle-third Cantor set. Certain points have higher probability than others.
I …
2
votes
1
answer
269
views
Lower bounds from Fourier dimension?
According to Mattila, Geometry of sets and measures in Euclidean spaces, p. 168, the Fourier dimension $\text{dim}_F(A)$ of $A\subseteq \mathbb R^n$ is the unique number in $[0,n]$ such that for any $ …
2
votes
Accuracy of the truncated Hausdorff moment problem
For $p=\infty$, there is a best possible lower bound of
$$\inf_{|s| = n} \alpha_p(s)=\infty$$
since you can change the maximum of a (continuous, say) function without changing its integrals much.
Tha …
4
votes
Unusual augmentation of a filtration
They're the same, $\mathcal G_t=\mathcal F_t$.
Indeed, suppose $A\in\mathcal G_t$.
So in particular $A\in\bigcap_{n=1}^\infty(\mathcal F_{t+1/n}\vee\mathcal N)$.
Note that for any $\sigma$-algebra $ …
3
votes
Is this generating family of a measurable space of point measures a pi-system?
It's not a $\pi $-system. For a counter example let $ A=\{0, 1\} $, $ B=\{2, 3\}$, $ u=v=1$. Show there is no $ C $ and $ w $ with $$ A_u\cap B_v= C_w $$ by first finding out what $ w $ would have to …
7
votes
Which distributions can you sample if you can sample a Gaussian?
Just want to observe that if $X$ is $\mathcal N(0,1)$ and $Y$ has any distribution with a strictly increasing* cdf $F_Y$ such that $F_Y^{-1} \circ F_X$ is computable (in whatever model of computabilit …
1
vote
Measures of disjoint unions and complements of a collection of sets
Yes, and this is a big part of why Dynkin systems are of interest.
$\pi$-systems, on the other hand, have a simpler definition, but $\mu(A\cap B)$ is not determined by $\mu(A)$ and $\mu(B)$.
4
votes
Accepted
Measurability of $\{ x \in X ; H_0 x \subset A \}$
No, it is not true.
It suffices to show that the complement is not necessarily Borel.
Let $B=X\setminus A$, which is a general Borel set. We have
$$
\{x:H_0x\not\subseteq A\} = \{x:H_0x\cap B\ne\varn …
1
vote
Function from a compact metric space to the subsets of the naturals
To answer the second part (Can $b$ be chosen so that it depends only on $a$ and not the function $f$):
No, it cannot.
First note that there exist infinitely many upper density one subsets $S_n\subse …