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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 …

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 $\mu(B)$.

$U$ 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.

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 …

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 …

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 …

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 …

If we have a point of increase at $t$, witnessed by $\epsilon$, the question is threefold: why can we assume $t-\epsilon=0$, $B_t(\omega)\le 1$, and $B_{t+\epsilon}(w)-B_t(\omega)\ge 2$. These conditi …

The basic step is to construct a nowhere dense set of positive, and controlled, measure. Then iteratively in each interval where the set is empty, you replace by another such set. See for example a pa …

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 …

Yes, it requires the axiom of choice (and so No, it's not possible without), in the sense that Zermelo-Fraenkel set theory without choice (ZF) is not enough. ZF together with the Axiom of Determinacy …

Let $p_i$ be a list of the rational numbers. Let $U_{i,n}$ be an open interval centered on $p_i$ of length $2^{-i}/n$. Then $V_n=\cup_i U_{i,n}$ is an open cover of the rationals, of measure at most $ …

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 …

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 $ …

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 $ …