Search Results

Call a function $F$ nice, if $$\DeclareMathOperator{\Dm}{d\!} \begin{align} &\zeta(2)\sum_{\frac{a}{b}\in\mathbb{Q}_n} \frac{F(\log \frac{a}{b})}{\sqrt{abn}}\xrightarrow[n\to \infty]{}\int F(x)\Dm x …

Yes, by Luzin's theorem. Fix $\varepsilon>0$ and take a compact subset $K$ of measure at least $1-\varepsilon$ such that $F$ is continuous on $K$. Then $F(K)$ is a compact set of at least the same me …

No even in dimension 1 (and multiplying the example for $\mathbb{R}$ by the small segment you get a counterexample in $\mathbb{R}^2$). Take the set $A_n\subset \mathbb{R}$ defined as $\bigcup_{k\in \ …

You may think the following way. Ignore the sharp constants and study when $xy\le C(x^p+y^q) $ for some $C>0$ and arbitrary positive $x, y$. This is equivalent to asking when $xy\le C\max(x^p, y^q)$ ( …

No. Denote $T_k=T\cap [k,k+1)$. Then $\sum |T_k|<\infty$ (where $|X|$ stands for the measure of $X\subset \mathbb{R}$). Choose a segment $[a,b]\subset (0,\epsilon)$. Note that if $r\in [a,b]$ and $nr\ …

Yes. For fixed $n$, we approximate our set $E$ from above by an open set $U=\sqcup \Delta_i$ ($\Delta_i$ are disjoint intervals) with such accuracy that one of intervals $\Delta_i$ satisfies $|E\cap \ …

Let $E:=\cup_{n=1}^\infty [n^2+n\sqrt{2},n^2+n\sqrt{2}+1]$. Clearly it has Banach density 0. Let $f$ be a $T$-periodic function, $T>0$, $f\in L^1[0,T]$. It is convenient to suppose that $T>1$ (that ma …

All irrational numbers in $(0,1)$ have unique representation as $\sum_{k=2}^\infty c_k/k!$, where $c_k\in \{0,1,\dots,k-1\}$. 'Digits' $c_k$ are independent, so you may choose the events $E_n$ as '$c_ …

A set of positive measure contains a closed subset of positive measure. It is known that closed subset of reals may have cardinality either continuum or at most countable (http://en.wikipedia.org/wiki …

I think, the reason is that if the ground space has infinite measure, you can not define the measurable sets as those for which inner measure equals the outer measure: it may happen that both are infi …

Without loss of generality your atom $A$ is compact (by inner regularity). Call a point $a\in A$ negligible if it has a neighborhood with zero measure inside $A$. Clearly the set $B$ of negligible poi …

Consider the 1-periodic function $f$ with Fourier series $$\sum n^{-1}\cos (2\pi n! x).$$ Note that it satisfies your property, since all but finitely many summands are $h$-periodic for any rational $ …

Yes, it is true. Since $2\min(a,b)=a+b-|b-a|$ for positive numbers $a,b$, and $F(z)=e^{g(z)}-1$ is differentiable, it suffices to prove that $|F|$ is differentiable except at a countable set of points …

Yes, by Kantorovich--Rubinstein duality $W(\mu_1,\mu_2)=\sup_{f\,\text{is 1-Lip}} \int f d(\mu_1-\mu_2)$.

If $F(x)=\int_a^x f(x)dx$ (for some fixed $a$), then $x$ being a Lebesgue point of $f$ yields $F'(x)=f(x)$; and the derivatives enjoy the intermediate value property by Darboux theorem.