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Problem. Let $\lambda$ be the standard product measure on the Hilbert cube $[-\frac12,\frac12]^\omega$ and $A,B$ be two $\lambda$-positive Borel subsets of $[-\frac12,\frac12]^\omega$. Is it true tha …

If $\mu^\star|P(X)$ would be a measure, then we could define a $\sigma$-additive measure $\lambda:P([0,1])\to[0,1]$ by the formula $\lambda(A)=\mu^\star(A\cap X)$ for $A\subset [0,1]$. This would impl …

This question has a negative answer (given by Gregorz Plebanek), which follows from the following theorem of Gitik and Shelah. Theorem (Gitik-Shelah, 1989): If a set $X$ admits an atomless probabilit …

There exists a counterexample under the Continuum Hypothesis, which implies that the unit interval $[0,1]$ admits a well-order $\preceq$ such that for every $x\in[0,1]$ the initial interval ${\downarr …

The equality $\mathcal M_\tau(X)=\mathcal M_t(X)$ for a complete metric space $X$ follows from three facts: 1) For any finitely additive measure $\mu$ on $X$ its support $supp(\mu)$ (i.e., the set of …

The answer is affirmative if $f$ is Borel-measurable. It is suffices to prove the equality for $f$ having finitely many values. Let $V=f(X\times Y)$ be the finite set of values of $f$. By the Borel-m …

The answer to this question is negative. To construct a counterexample, fix any free ultrafilter $\mathcal U$ on $X$ containing a discrete subspace $D$ of $X$. The characteristic function $u:\mathcal …

Theorem. If there exists a free ultrafilter $\mathcal U$ on $\omega$, then there exists a non-measurable subset of the real line. Proof. First observe that the free ultrafilter $\mathcal U$ is a non- …

A counterexample to this problem can be constructed as follows. Take a sequence $(K_n)_{n\in\omega}$ of pairwise disjoint nowhere dense compact sets $K_n\subset[0,1]$ of positive Lebesgue measure $\l …

The Borel $\sigma$-algebras generated by these two topologies seem to be equal. The idea of the proof is as follows. Let $\mathcal M_+$ be the subspace of $\mathcal M$ consisting of measures. It is kn …

A counterexample to this question (and its locally convex version) is any non-metrizable (locally convex) space $X$, which is hereditarily Lindelof. The hereditary Lindelofness of $X$ implies that any …

The affirmative answer to this question is given by the following theorem, proved by the technique of continuous selections. Theorem. For any compact metrizable space $X$ there exists a continuous …

A Borel subset $B$ of the unit interval $\mathbb I=(0,1)$ is defined to be a density neighborhood of a set $A\subseteq\mathbb I$ if for every $a\in A$ we have $$\lim_{\varepsilon\to0}\frac{\lambda(B\c …

Professor Wladyslaw Wilczynski kindly informed me that the answer to this problem is negative. Take any Lebesgue null dense $G_\delta$-set $A$ in the real line $\mathbb R$. Choose a countable dense su …

No, those sets are not open. Indeed, take any non-isolated point $x$ in $X$ and a sequence $(x_n)_{n\in\omega}$ that converges to $x$. For every $n$, consider the sign measure $\mu_n=\delta_{x}-\delta …