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No, consider the characteristic function $ f $ of a union $ A$ of infinitely many disjoint closed annuli centered on the origin in $\mathbb R^2$. If the annuli have sufficiently fast-shrinking radii t …

Let $$ f(x,y)=x^{f(x,y-1)}, $$ $$ f(x,1)=x. $$ Then your function is given by $g(n)=f(n,n)$.

If you allow $A$ to contain an interval, let $$A = (\mathbb Q\cap [0,1])\cup [0,1/2].$$ If not, consider $q_i$ the $i$th rational number in some enumeration of $\mathbb Q\cap [0,1]$, and let $$A=\bigc …

Inspired by Hausdorff dimension, we can try to let $X$ consist of numbers whose decimal expansion is of the form $$ 0.x_100x_4x_5x_60000\dots $$ and $Y$ consist of the "complementary" numbers: $$ 0.0x …

Sure, consider the set of pairs $(q,r)$ where $q$ is rational and $r$ is a piecewise linear continuous function with rational break-points (and rational values at the breakpoints) with finitely many p …

This is $\mathbb E (X_1^{n-1} X_2) $ where $ X_i $ is the $ i $ th order statistic of a sample from the uniform distribution on $[0, t] $. To evaluate it can try using the joint pdf of these order sta …

Just to show that $F:R^1\to R^1$ can be fairly nice without your relation holding, let $$F(x)=\begin{cases}1&\text{if }x=0,\\ x&\text{otherwise}. \end{cases}$$ Then $$\overline{\text{co}}\left\{\bigca …

$U$ 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.

We may assume $n\le m$ throughout. Let $y>x>0$. There are only finitely many $n$ with $$ 1/n\ge x/2. $$ For each such $n$, the set of $1/n+1/m$ as $m$ varies is not dense in any subinterval of $(x,y …

Yes, if $X$ is an analytic set. See: Besicovitch, On existence of subsets of finite measure of sets of infinite measure, 1952 R. O. Davies, Subsets of finite measure in analytic sets, 1951. For di …

This is a refined version of my earlier question Convex extensibility of combination of two lines. Is there a smooth function $f:[0,1]\times [0,1]\rightarrow\mathbb R$ such that for all $x\in [0 …

Not what you're really looking for, but here's a "word answer" in terms of a probabilistic interpretation. Suppose Alice and Bob are playing a game where they are presented with $2n+1$ indistinguisha …

Do there exist "nice" (maybe analytic?) functions $f_0,f_1:\mathbb R \to \mathbb R$ such that $\forall n\in\mathbb N,\forall \sigma\in\{0,1\}^n,\exists x\in\mathbb R, \forall \tau\in\{0,1\}^n, (\tau\ …

Yes. See Exercise 1.13(a) of Mörters and Peres, Brownian motion. http://www.stat.berkeley.edu/~peres/bmbook.pdf

Infinite, complete, separable linear order with at most countably many jumps and not both a greatest and least element. See http://www.math.uni-hamburg.de/home/geschke/papers/SeparableLinearOrders2. …