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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.

5 votes
Accepted

Summability issues of measure when we decompose a measurable set into two non-measurable parts

Suppose $E\subseteq[0,1]$, $\ F=[0,1]\setminus E$, $\ \lambda^*(E)=a$, $\ \lambda^*(F)=b$, $\ a+b=1$. There are Lebesgue measurable sets ($G_\delta$ sets) $A,B\subseteq[0,1]$ such that $E\subseteq A$, …
bof's user avatar
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1 vote

Covering families of sets by small-measure partitions

Let $X=[0,1]$, let $\epsilon\gt0$, and let $\{A_i:i\in\mathbb N\}$ be the set of all $A\subseteq[0,1]$ such that $A$ is a finite union of rational intervals and $\mu(A)\lt\epsilon$. Let $\{B_i:i\in\ma …
bof's user avatar
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14 votes

Countably representing all closed sets of positive measure

Suppose $\{C_m\}_{m=1}^\infty$ is a sequence of nonempty subsets of $[0,1]$. Choose $x_m\in C_m$ and let $X=\{x_m:m\in\mathbb N\}$. The set $X$, being countable, has measure zero, so $[0,1]\setminus X …
bof's user avatar
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6 votes
Accepted

A balancing property of infinite subsets of $\mathbb{N}$

By the strong law of large numbers, if $S$ is an infinite subset of $\omega$, a random subset of $\omega$ will be well-balanced with respect to $S$ with probability one. By the countable additivity o …
bof's user avatar
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3 votes

Graph on $\mathbb{N}$ where almost every vertex is shy

Take any locally finite countable graph with infinitely many shy vertices, e.g., the disjoint union of $\aleph_0$ copies of $K_{1,2}$. Identify the vertex set with $\mathbb N$ in such a way that the s …
bof's user avatar
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11 votes
Accepted

Uncountable families of measurable sets with pairwise positive intersections

By the Erdős–Dushnik–Miller theorem, if the index set $I$ is infinite, then either there is a subset $J\subseteq I$ of the same cardinality as $I$ such that $A_j\cap A_{j'}$ has positive measure for a …
bof's user avatar
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3 votes
Accepted

An uncountable measurable subset of $\Bbb R$ containing no nonempty perfect set

A set $B\subseteq\mathbb R$ is a Bernstein set if $B$ contains no nonempty perfect set while having nonempty intersection with every nonempty perfect set; in other words, if neither $B$ nor $\mathbb R …
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0 votes

Why should one still teach Riemann integration?

Why whould one still teach Riemann integration? Because of this theorem: If the Riemann integral of $f$ on $[a,b]$ exists and is equal to $c$, then the Lebesgue integral of $f$ on $[a,b]$ likewise ex …
8 votes

Axiom of choice, Banach-Tarski and reality

Physical applications of the Banach–Tarski theorem were explored by Henry Kuttner, The Time Axis, Startling Stories 18:3 (January 1949), 13–82. Some excerpts from pp. 66–67: "'Professor Raphael M. …
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