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
Slide 25 of this link gives a different definition and also calls it Lebesgue's: https://people.math.harvard.edu/~shlomo/212a/11.pdf
In light of the second definition (outer measure equals inner measure), I question my earlier claims about inner measure (specifically $m_*(\R\setminus\Q) = 0$).

Caratheodory's criterion: for any $A \subseteq \R^n$: $m^*(A) = m^*(A \cap E) + m^*(A \setminus E)$.
New criterion: for any $\varepsilon > 0$, there's an open set $U \supseteq E$ with $m^*(U\setminus E) < \varepsilon$.
Say a set is Caratheodory-measurable if it meets Caratheodory's criterion, and new-measurable if it meets the new criterion.
We will prove the equivalence of these criteria.
Thanks to Griffin (Shuqi Ke) for pointing out that his proof of Proposition 1 does not apply to all unbounded sets. My initial proof of Claim 1 was essentially identical to his proof of Proposition 1, and so it failed to prove the claim for sets with infinite outer measure.
Griffin influenced the preceding description. His Proposition 1 (on page 2 of https://drive.google.com/file/d/1XKo2wFOQ-k7hpbxgwRjgDzhCUq5YfUlz/view) inspired me to write up these proofs, however I proved the results independently (aside from the aforementioned correction).

Claim 1: Any Caratheodory-measurable set is new-measurable.
Let $E \subseteq \R^n$ be Caratheodory-measurable and suppose that $m^*(E) < \infty$. Let $\varepsilon > 0$, and let $(B_j)_{j\in J}$ a countable open box cover of E such that $$ \sum_{j \in J} |B_j| < m^*(E) + \varepsilon $$ We define $$ U = \bigcup_{j \in J} B_j $$ so that $U$ is open and $m^*(U) < m^*(E) + \varepsilon$.
We apply Caratheodory's criterion: $$ m^*(U) = m^*(E) + m^*(U \setminus E) $$ $$ m^*(U \setminus E) < \varepsilon $$ Since $\varepsilon > 0$ was arbitrary, this proves that $E$ is new-measurable.
Now we prove the general claim.
Suppose that $E$ is measurable and do not assume that it has finite outer measure.
Then we have $$ E = \bigcup_{n=1}^\infty (B_n(0) \cap E) $$ where $B_n(0)$ is the open ball with radius $n$ and center $0$.
Each $B_n(0)$ is Caratheodory-measurable with finite outer measure, so each $B_n(0) \cap E$ is Caratheodory-measurable with finite outer measure and is therefore new-measurable.
Applying Lemma 2 of Homework 2, we find that $E$ is new-measurable.

Claim 2: Any new-measurable set is Caratheodory-measurable.
Let $E \subseteq \R^n$ be new-measurable, $A \subseteq \R^n$, $\varepsilon > 0$, and $U \supseteq E$ open with $m^*(U \setminus E) < \varepsilon$. $$ m^*(A \setminus E) \leq m^*(A \setminus U) + m^*(U \setminus E) \leq m^*(A \setminus U) + \varepsilon $$ $$ m^*(A \setminus E) + m^*(A \cap E) \leq m^*(A \setminus E) + m^*(A \cap U) \leq m^*(A \setminus U) + m^*(A \cap U) + \varepsilon = m^*(A) + \varepsilon $$ (Note that for the $=$, we use the Caratheodory-measurability of open sets.)
The rest of the proof is clear.

A neighborhood is a set containing an open set.
A boundary is a closed non-neighborhood.
Is every boundary the boundary of a closed neighborhood? (I think not.)
Is every boundary the boundary of a neighborhood?
Does every boundary have measure 0? Which boundaries have measure 0?

Why does Tao define measurability using open sets instead of measurable ones? His definition isn't equivalent to ours (it's stricter).
What are some Lebesgue-measurable sets that aren't Borel sets?

Concept from Rieffel's notes (which are linked above): Let $P \subseteq \mathscr{P}(X)$. $P$ is a semiring iff

• For any $E,F \in P:\ \ E\cap F \in P$
• For any $E, F \in P$ there exist disjoint sets $F_1, \dots F_k \in P$ such that $E\setminus F = \bigsqcup_{i=1}^k F_i$

Examples:

• the set of left-closed, right-open intervals $[a,b) \subseteq \R$ ($a \leq b$).
• the set of all $\prod_{i=1}^n [a_i, b_i) \subseteq \R^n$.

On each of these we can define a premeasure (see Rieffel's notes for details) which maps each set to its volume. This can be extended to an outer measure (the Lebesgue outer measure) in the usual way. Note: I have not verified that the second example yields a premeasure. I think it does.

Is any $G_\delta$ not an $F_\sigma$?
Is any Borel set not a $G_\delta$ or an $F_\sigma$?
Can we characterize/classify the Borel sets?

4.0.pdf

Let $E \subseteq \R^m\times\R^n$.
I conjecture that $E$ is measurable if and only if $E_x \subseteq \R^n$ is measurable for a.e. $x\in\R^m$.
Furthermore, supposing $E$ is measurable and $\Omega \subseteq\R^m$ is the full-measure set on which $E_x$ is measurable, I conjecture that $x \mapsto m_n(E_x)$ is a measurable function and $$ \int_{\R^m} (x \mapsto m_n(E_x)) = m_{m+n}(E) $$ I think the second conjecture has a hint of Fubini.

This made me wonder what exactly “regularity” means, so I looked at Wikipedia's definition, which I now provide. Given a measure space $(X, \Sigma, \mu)$ and a topological space $(X, \tau)$ (with the same underlying set), a set $E\in\Sigma$ is inner regular if $$ \mu(E) = \sup\{ \mu(K) \vert K\in\Sigma \textrm{ compact } \} $$ and outer regular if $$ \mu(E) = \inf\{ \mu(U) \vert U\in\Sigma \textrm{ open } \} $$ It is regular if both of these hold, and $\mu$ is regular if every $E\in\Sigma$ is regular.

I was initially confused by this one; using the preimage definition of continuity, I believed that $\xi_\Q$ was a counterexample since $\Q^c$ has empty interior. However, when we restrict our domain, the meaning of “open” in our domain changes, which accounts for this.

Not much to say on this one, since it came up already in HW 5.

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