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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 …
Bjørn Kjos-Hanssen's user avatar
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: $ …
Bjørn Kjos-Hanssen's user avatar
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 …
Bjørn Kjos-Hanssen's user avatar
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.
Bjørn Kjos-Hanssen's user avatar
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 …
Bjørn Kjos-Hanssen's user avatar
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 …
Bjørn Kjos-Hanssen's user avatar
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 …
Bjørn Kjos-Hanssen's user avatar
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 $ …
Bjørn Kjos-Hanssen's user avatar
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 …
Bjørn Kjos-Hanssen's user avatar
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 …
Community's user avatar
  • 1
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 $\mu(B)$.
Bjørn Kjos-Hanssen's user avatar
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 …
Community's user avatar
  • 1
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 …
Bjørn Kjos-Hanssen's user avatar

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