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Let $\lambda\ne 0$ and both $f, g$ be non-negative measurable homogeneous functions of degrees $d_1,d_2$ respectively, and either $d_1d_2\ne 0$, or $d_1\ne 0$ and $f$ is strictly positive on the unit …

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 …

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 vertices can be shy. You may add edges to your graph recursively, on $n$-th step fixing all edges from $1,2,\ldots,n$ and possibly some other (finitely many) edges, so that $1,2,\ldots,n$ already …

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 $xy\le C\max(x^p, y^q)$ ( …

Concerning problem 2. I do not know whether this "elementary", but at least formally it does not use the notion of outer measure or Lebesgue measure. Replace the closed intervals to open intervals. No …

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.

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 …

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 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 …

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 \ …

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 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 …

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 …

An argument for Sierpiński's Intermediate Value Theorem for non-atomic measures without using Zorn's lemma (but possibly implicitly using countable axiom of choice, that I am not qualified to judge), …