Lecture Notes

Contact Info: davideliahu@berkeley.edu

Hello! My name is David Alcalay. I'm a third year pure-math major here at UC Berkeley. I grew up in Davis, California, and transferred a year ago from community college. Analysis has always been my favorite area of mathematics, so I'm really excited to be taking Math 105. I'm really interested in Cantor sets, and also the topic of constructible numbers (think ruler-and-compass).

Outside of my education, I spend most of my time reading, hiking, doing art, and playing board games.

At UC Berkeley, I've taken the following math courses:

• Math 55 (Discrete Mathematics)
• Math 104 (Real Analysis)
• Math 113 (Abstract Algebra)
• I've also taken the equivalent of the lower-division calculus series, and the differential equations portion of Math 54, while at community college.

I'm currently enrolled in the following math courses:

• Math 54 (Linear Algebra and Differential Equations)
• Math 105 (A Second Course in Analysis)

My dream, upon finishing my education, is to become a research mathematician, specifically researching Cantor sets.

• Homework 1 (Incomplete)
• Homework 2 (Rough Draft)

• Lecture 1 - 18 January 2022
• Lecture 2 - 20 January 2022
• Lecture 3 - 25 January 2022
• Lecture 7 - 8 February 2022

• What modifications need to be made to the content covered in lecture so far so that the concepts of measure we've defined make sense in an arbitrary metric space?
  • Does the discrete metric on any space define an outer measure?
  • Does the supnorm metric on the space of bounded smooth functions define an outer measure?
  • When metrics define measures, are the measures unique?
  • How do isometric embeddings from one metric space to another affect measures that come from the metrics on the domain and codomain? (Does this question even make sense?)

The following proof is from Measure and Category, second edition, by John C. Oxtoby.

Definition: A subset $B \subseteq \mathbb{R}$ is called a Bernstein set if the following holds: For each subset $F \subseteq \mathbb{R}$ such that $F$ is an uncountable closed subset of $\mathbb{R}$, the intersections $B \cap F$ and $B^c \cap F$ is nonempty.

Theorem (F. Bernstein): There exists a Bernstein set. (pg. 23)

The proof relies on the Well-Ordering Principle, and uses transfinite induction to choose points from each uncountable closed subset of the reals.

Theorem: Bernstein sets are not Lebesgue-measurable.

proof. Let $B$ be a Bernstein set. Let $A \subseteq B$ be a Lebesgue-measurable subset. Suppose that $F \subseteq A$ is closed. If $F$ was uncountable, then $F$ would meet both $B$ and $B^c$ since both $B$ and $B^c$ are Bernstein sets by Bernstein's theorem, and $F$ meeting $B^c$ is a contradiction. Thus $F$ is at most countable, and so $m(F) = 0$.

Lemma: For a set $A$, we have $m^*(A) = \sup\{m(X) \mid X \subseteq A \text{ and } X \text{ is closed}\}$ (pg.15-16).

The by the lemma above, we thus have $m(A) = 0$. Hence if $B$ is Lebesgue measurable, then $m(B) = 0$. But $B^c$ is also a Bernstein set, and thus $m(B^c) = 0$. But $B \cup B^c = \mathbb{R}$, and so $m(B)$ and $m(B^c)$ can't both be zero. It follows that $B$ is not Lebesgue measurable. $\square$