Lecture 01: Introduction: a non-measurable set

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Strong rec Jose Avilez · 2021-06 · Mathematics Stack Exchange · on “Advanced online courses for different subjects”

A few years ago, IMPA recorded lectures on Measure Theory, which you can find here [this video] . You can find the auxiliary notes here: [LINK] While I have not seen the measure theory lectures myself, I did follow the probability theory lecture series they recorded, which were of superb quality. I expect the same for this lecture series.

Recommends Santiago Arango-Piñeros · 2020-08 · Mathematics Stack Exchange · on “Volume of a bounded open set in $\Bbb{R}^n$”

To understand the notion of boundedness in higher dimensions, the first thing you need to understand is the Euclidean norm in $\mathbb{R}^n$ , defined by $||\mathbf{x}|| := (x_1^2 + \dots + x_n^2)^{1/2}$ . Note that when $n=1$ this is the usual absolute value! A subset $S \subseteq \mathbb{R}^n$ is bounded if there exists some positive constant $B>0$ such that $||\mathbf{x}||< B$ for every $x\in S$ . Geometrically, a set is bounded if it fits inside some $n-$ dimensional ball centered at the origin. (Think about the cases $n=1,2,3$ for intuition). Your second question is actually very profound. First of all, the volume of a set is always a non negative quantity (so positive means $\neq 0$ ),

Cites Proko · 2020-12 · Mathematics Stack Exchange · on “Show that $\mathbb{R}|_{\sim}$ is uncountable for $\sim (x, y) : y - x \in \mathbb{Q}$”

I'm self-learning some measure theory. During the following lecture ( I had included the appropriate starting time inside the link, but timespan is 7:50-9:00): [this video] the Professor Claudio Landim states that if we take $\Lambda = \mathbb{R}|_{\sim}$ where $\sim (x, y) : y - x \in \mathbb{Q}$ then $\Lambda$ is uncountable. The lecturer explains that if indeed $\Lambda$ was countable, then every element in $\mathbb{R}$ could be represented by element of $\Lambda$ (some equivalence class) and by element of this equivalence class. Couldn't it? Why not? I cannot understand it.

Cites stochastic-conch · 2021-04 · Mathematics Stack Exchange · on “What is the significance of a Borel $\sigma$-algebra?”

The Borel- $\sigma$ -algebra is important, because some very natural measures simply cannot be defined on all the subsets of $\mathbb{R}^d$ . Let me show you the standard example: take $d=1$ . We would like to define a "length"-function, i.e. a function on all the subsets of $\mathbb{R}$ (denoted $\wp(\mathbb{R})$ ) which to each subset $A\subseteq\mathbb{R}$ assigns the length of $A$ (whatever that means). Call this function $\lambda$ . For us to think of $\lambda(A)$ as the length of $A$ , $\lambda$ must satisfy the following properties: $\lambda:\wp(\mathbb{R})\to[0,\infty)\cup\{+\infty\}$ (lengths should be non-negative, but may be infinite), $\lambda((a,b))=b-a$ for all $a<b$ (we know w

Cites Booker ML · 2026-01 · Mathematics Stack Exchange · on “Is there a measure on $\mathbb{R}$ that satisfies these properties?”

I am taking a course in measure theory: [this video] In the first lecture, the teacher proposed a "perfect" measure and proved that such a measure does not exist.This measure includes the following properties: $m((a, b]) = b - a$ , $m(A) = m(A + x)$ , $x \in \mathbb{R}$ , $A$ is a set, and $m(\cup_i A_i) = \sum_i m(A_i)$ , $i \neq j$ , $A_i \cap A_j = \emptyset$ But when I came across Lebesgue measure, I found that Lebesgue measure also satisfies the above 3 properties. Does that mean Lebesgue measure is not a perfect measure? So where did I go wrong?