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Hi! I'm Max.
Major: Pure math.
Math classes taken:
Reasons for taking 105:
Some resources that seem relevant to this course:
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:
$$ l = \sup([0,x] \setminus \bigcup C) $$ $$ r = \inf([x,1] \setminus \bigcup C) $$ 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:
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 $m_*(\R\setminus\Q) = 0$).