Hi! I'm Max.

Major: Pure math.

Math classes taken:

• 55
• 104
• 114
• 250A

Reasons for taking 105:

• 104 was my favorite math class.
• Prof. Zhou seems like a good instructor.
• The setup of this class is interesting, unusual, and potentially very good.
• Prof. Zhou made a class Discord server. (An instructor-made Discord server is a good sign, in my view.)

Some resources that seem relevant to this course:

• Lebesgue measure/integration
  • Rieffel's notes: https://math.berkeley.edu/~rieffel/measinteg.html
  • Wodzicki's notes and problems: https://math.berkeley.edu/~wodzicki/105.S12/index.html
• Multi-dimensional calculus
  • Honda's notes on differential geometry: https://www.math.ucla.edu/~honda/math225a/revised%20course%20notes.pdf
  • Leoni's notes on vector analysis: https://www.math.cmu.edu/~gautam/sj/teaching/2016-17/269-vector-analysis/pdfs/giovanni-2015.pdf
  • (also Wodzicki I think, same link as above)
• Fourier analysis
  • Leoni's notes on harmonic analysis: http://giovannileoni.weebly.com/uploads/3/1/0/5/31054371/singulars15-lectures-2015-05-01.pdf

I might just use the course textbooks rather than any of the above resources.

At the top of page 384 Pugh specifies (after defining the Lebesgue outer measure on $\R$) that the collection of open intervals must be countable. I think (at least for this particular outer measure) the countability condition is unnecessary. Specifically, I claim that for any collection $C$ of open intervals in $\R$ there's a countable collection of open intervals $C'$ such that $$\bigcup C = \bigcup C'$$ To prove this, first note:

• It suffices to prove this for the case where $\bigcup C \subseteq [0,1]$
• Each point in $x\in C$ can be mapped to $(l,r)$, the biggest interval contained in $\bigcup C$ that includes $x$. We construct this by letting

$$ l = \sup([0,x] \setminus \bigcup C) $$ $$ r = \inf([x,1] \setminus \bigcup C) $$ Clearly the $(l,r)$ are disjoint and their union is $\bigcup C$. Letting $C'$ be the set of all these $(l,r)$, we see that $\bigcup C = \bigcup C'$, so now we just need to show that $C'$ is countable.

To do this, pick a positive integer $n$ and break the real line into pieces of the form $[\frac{k}{n}, \frac{k+1}{n}]$ for natural numbers $k<n$. Now we construct a partial function $$\{0, 1, 2, \dots n-1\} \to C'$$ by mapping each $k$ to the unique interval $(l,r) \in C'$ such that $$[\frac{k}{n}, \frac{k+1}{n}] \subseteq (l,r)$$ (if such an interval exists). Let $a_n$ be the range of this partial function. Clearly, for each $(l,r) \in C'$, there is some $n$ such that $(l,r) \in a_n$. This gives $$C' = \bigcup_{n=1}^\infty a_n$$ Since $C'$ is a union of countably many finite sets, it's countable. QED.

Prof. Zhou mentioned translation invariance. I think it intuitively makes sense to strengthen this condition to congruence invariance: if $A,B \subseteq \R^n$ and $A,B$ are isometric, then we would want $m(A) = m(B)$.

Also, in order for the properties he posited to be incompatible, I think we also need to specify that:

• every set can be measured (he did this implicitly, or maybe explicitly)
• some set has nonzero measure

I recently realized that every open set in $\R$ can be expressed as a countable union of disjoint open intervals (assuming you allow $\pm\infty$ to be endpoints of intervals).
Whereas, if I remember correctly, the Cantor set is closed and no positive-length interval is contained in the Cantor set. So it's just an uncountable union of zero-length closed intervals (i.e. singletons).
I think open sets are, in this sense, generally much simpler than closed ones, which is surprising given that they are each others' complements. E.g. if you take the complement of a complicated closed set like the Cantor set with uncountably many points that aren't part of intervals, you get a nice simple countable union of disjoint open intervals.

Lebesgue's original definition of measurability: https://hsm.stackexchange.com/questions/7282/what-was-lebesgues-original-definition-of-a-measurable-set

1.pdf