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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 $ …

For reals $x<y$ denote $\rho(x,y)=1/2^n$ if integer $n$ is minimal such that there exist integer $i$ with $x<i/2^n\leqslant y$. We have $\rho(x,y)\geqslant |x-y|/2$ and ``in average'' $\rho(x,y)$ beha …

Yes, we need the boundedness of $|f|^p \ln|f|$ when both $x$ and $p$ vary, but important note is that $p$ may vary on a given segment $[p_1,p_2]$, $1<p_1<p_2$ (this would imply that the derivative $d/ …

Easier (well, I mean: requiring less deep knowledge) reason than in the answers by Goldstern and Will Brian. Your set $C$ may be approximated by a disjoint union of small dyadic intervals (simply dis …

I assume that by integrable you mean $\int |f|<\infty$. Assume that $f=f_1+\ldots+f_n$ where each $f_i$ is a continuous periodic function: $f_i(x+v_i)=f_i(x)$ for certain $v_i\in \mathbb{R}^n\setminu …

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 …

Yes, it is a part of Fubini theorem for the characteristic function of $Y$.

It suffices to prove that, given $\rho>t$, $g(t) \leqslant \rho$ for almost all $t$, where $g$ is your sum of squares. Assume the contrary. Then the $\phi$-preimage of the set $\Omega=\{(z_1,\dots,z_d …

1) Locally finite measure on the separable metric space $X$ is $\sigma$-finite. Indeed, fix a dense countable set $Z\subset X$ and call a ball with center in $Z$ and rational radius good, if it has fi …

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

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 \ …

Yes. It suffices to prove that for every rationals $p<q$ the set $A$ of those $x$ for which simultaneosly $\mu\text{-esssup}_{[0,x]} f<p$ and $q<f(x)$ satisfies $\mu(A)=0$. Note that if $x\in A$, the …

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 …

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.