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--- math105-s22:notes:lecture_4 [2022/01/22 19:03]
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+++ math105-s22:notes:lecture_4 [2026/02/21 14:41] (current)
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 ====== Lecture 4 ======
+{{ :math105-s22:notes:note_jan_27_2022_math105_4.pdf |note}}, [[https://berkeley.zoom.us/rec/share/FpG9GNaABIipr523cZa_36y6w6O9We_zWhb-llmVIusvAyLlFmVqtiBUpjp9Q2hZ.-dvCEBgnfAdGXV7e | video ]]
-Lemma 7.4.8 - Lemma 7.4.11
+==== Cor 7.4.7 ====
+If $A \In B$ are both measurable, then $B \RM A$ is measurable, and $m(B \RM A) = m(B) - m(A)$.
+We need to show that for any subset $E \In \R^n$... wait a second, do we really need to go by definitions again? After all these preparations, we should be able to exploit our sweat. Hint
+  * $B \RM A = B \cap A^c$.
+  * $B = A \sqcup (B \RM A)$, a decomposition into measurable subsets.
+==== Lemma 7.4.8: Countable addivitivty ====
+Let $\{E_j\}_{j=1}^\infty$ be a countable collection of **disjoint** subsets. Then, their union $E$ is measurable, and we have
+$$ m^*(E) = \sum_{j=1}^\infty m^*(E_j) $$
+Proof: Tao's proof is really clever, let's first try to go through the proof, then discuss how we can come up with it ourselves.
+First, we start by showing $E$ is measurable from the definition: we want to show for any subset $A$, we have
+$$ m^*(A) = m^*(A \cap E) + m^*(A \RM E) $$
+Suffice to show $\geq$ direction. Let $F_N = \sum_{j=1}^N E_j$. We have two expressions
+  * $m^*(A \cap E) \leq \sum_{j=1}^\infty m(A \cap E_j) = \sup_{N > 1} \sum_{j=1}^N m^*(A \cap E_j) = \sup_{N > 1} m^*(A \cap F_N) $
+  * $m^*(A \RM E) \leq m^*(A \RM F_N)$ for all $N$
+Hence,
+$$ m^*(A \RM E) + m^*(A \cap E) \leq \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM E)) = \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM F_N)) = \sup_{N > 1}  m^*(A) = m^*(A) $$
+OK, that shows $E$ is measurable. To finish off, we need to show countable addivity
+$$ m^*(E) = \sum_{j=1}^\infty m^*(E_j) $$
+Since $E = \cup E_j$, we have $\leq$ from countable sub-addivity. Then, by monotonicity, we have
+$$ m^*(E) \geq m^*(F_N) = \sum_{j=1}^N m^*(E_j) $$
+since this is true for all $N$, we can sup over $N$, and get
+$$ m^*(E) \geq \sup_N \sum_{j=1}^N m^*(E_j) = \sum_{j=1}^\infty m^*(E_j) $$
+----
+One slogan is to approximate $E$ by $F_N$. We want to prove
+ $$ m^*(A) \geq  m^*(A \cap E) + m^*(A \RM E) $$
+If we have
+  * (1) $m^*(A \cap E) \leq m^*(A \cap F_N)$ and
+  * (2) $m^*(A \RM E) \leq m^*(A \RM F_N)$,
+then we can write
+$$ m^*(A \cap E) + m^*(A \RM E) \leq m^*(A \cap F_N) + m^*(A \RM F_N) = m^*(A) $$
+but unfortunately, (1) is wrong. One way to remedy this, is to show that (assuming $m^*(A)< \infty$),  for any $\epsilon > 0$, there exists an $N$, such that $m^*(A \cap E) \leq m^*(A \cap F_N) + \epsilon$ holds. (see if you can make this approach work). Another more elegant approach is done as above, using countable subadditivity to get $\leq$, then introduce a $\sup$ to get to finite $N$.
+Try to forget this proof, and come up with your own. It might be fun.
+==== Lemma 7.4.9 ====
+The $\sigma$-algebra property.
+Given a countable collection of measurable set $\Omega_j$, one need to prove that $\cup \Omega_j$ and $\cap \Omega_j$ are measurable.
+We only need to prove the case of $\Omega = \cup_j \Omega_j$, since the $\cap$ operation can be obtained by taking complement and $\cup$. The hint is to define
+$$ \Omega_N = \cup_{j=1}^N \Omega_j$$
+and $E_N = \Omega_N \RM \Omega_{N-1}$, then $\{E_j\}$ are measurable, mutually disjoint, and $\cup_j E_j = \cup_j \Omega_j = \Omega$.
+==== Lemma 7.4.10 ====
+Every open set can be written as a finite or countable union of open boxes.
+I will leave this as discussion problem.
+  * A subset $U$ is open, if for every point $x \in U$, there exists an open ball $B(x,r) \In U$.
+  * claim: a subset $U$ is open, iff $\forall x\in U$, there exists an open box $B$, such that $x \in B \In U$.
+  * claim: a subset $U$ is open, iff $\forall x\in U$, there exists an open box $B$ with rational boundary coordinates, such that $x \in B \In U$.
+  * There are countably many open boxes with rational boundary coordinates.
+==== Lemma 7.4.11 ====
+All open sets are measurable.
+Since open boxes are measurable, and countable union of measurable sets are measurable.
+==== Discussion Problem ====
+An alternative definition for measurable set is the following:
+=== Def 2 ===
+A subset $E$ is measurable, if for any $\epsilon>0$, there exists an open set $U\supset E$, such that $m^*(U \RM E) < \epsilon$.
+Can you show that this definition is equivalent to the Caratheodory criterion (the one we had been using)?

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