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--- math105-s22:s:hexokinase:start [2022/02/09 15:16]
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-==== Feb 31 ====
+==== Jan 31 ====
 === Definitions of measurability ===
-Lebesgue's original definition of measurability(?): https://hsm.stackexchange.com/questions/7282/what-was-lebesgues-original-definition-of-a-measurable-set
+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
+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 $m_*(\R\setminus\Q) = 0$).
+==== Feb 2 ====
+=== Equivalence of definitions of measurability ===
+//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.
+==== Feb 3 ====
+=== Boundaries ===
+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?
+=== Measurability questions ===
+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?
+==== Feb 5 ====
+=== Half-open intervals and semirings ===
+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.
+==== Feb 7 ====
+=== Borel sets ===
+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?
+==== Feb 18 ====
+=== Question 0 ===
+{{ :math105-s22:s:hexokinase:4.0.pdf |}}
+==== Feb 21 ====
+=== Conjectures on products ===
+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.
+==== Mar 20 ====
+=== Littlewood's three principles ===
+== Principle 1 ==
+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.
+== Principle 2 ==
+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.
+== Principle 3 ==
+Not much to say on this one, since it came up already in HW 5.
 ===== Homework =====
-{{ :math105-s22:s:hexokinase:1.pdf |}}
+{{ :math105-s22:s:hexokinase:1.pdf |}} \\
+{{ :math105-s22:s:hexokinase:2.pdf |}} \\
+{{ :math105-s22:s:hexokinase:3.pdf |}} (slightly updated since Gradescope feedback) \\
+{{ :math105-s22:s:hexokinase:4.pdf |}} (slightly updated since Gradescope feedback) \\
+{{ :math105-s22:s:hexokinase:5.pdf |}} \\
+{{ :math105-s22:s:hexokinase:6.pdf |}} \\
+{{ :math105-s22:s:hexokinase:7.pdf |}} \\
+{{ :math105-s22:s:hexokinase:8.pdf |}} \\
+{{ :math105-s22:s:hexokinase:9.pdf |}} \\
+{{ :math105-s22:s:hexokinase:10.pdf |}}

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