Table of Contents

Lecture 3

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Today we continue going over Tao's sequence of Lemma 7.4.2 - 7.4.11

Lemma 7.4.2

I will prove the easy case (1-dim), you will do the general case in HW.

Let $E = (0,+\infty)$ be the open half space in $\R$. For any subset $A \In \R$, we need to prove that $$ m^*(A) = m^*(A \cap E^c) + m^*(A \cap E). $$ Let $A_+ = A \cap E$ and $A_- = A \cap E^c$, note that if $0 \in A$, then $0 \in A_-$. First, we note that $$ m^*(A) \leq m^*(A_-) + m^*(A_+)$$ by finite sub-additivity of outer-measure. (This is because any open cover of $A_-$ and open cover of $A_+$, union together form an open cover of $A$.) Next, we need to show that, for any $\epsilon>0$, we have $$ m^*(A) + 2\epsilon \geq m^*(A_-) + m^*(A_+).$$ The plan is the following. Given an open covering of $A$ by intervals $\{B_j\}_{j=1}^\infty$, such that $m^*(A) + \epsilon > \sum_j |B_j|$. We define $$ B_j^+ = B_j \cap (0,\infty) , \quad B_j^- = B_j \cap (-\infty, \epsilon / 2^j) $$ then $\{B_j^+\}$ forms a cover of open interval of $A_+$, similarly $\{B_j^+\}$ forms a cover of open interval of $A_-$. $|B_j^+| + |B_j^-| \leq |B_j| + \epsilon / 2^j$. Thus, we have $$ m^*(A) + 2\epsilon \geq \sum_j |B_j| + \epsilon \geq \sum_j |B_j^+| + \sum_j |B_j^-| \geq m^*(A_+) + m^*(A_-). $$

That finishes the proof for $n=1$ case. How to generalize to higher dimension? In the above discussion, we used the trick of $1 = \sum_j 1/2^j$, which is same trick as we prove outer-measure has countable sub-additivity. See the hint in Tao's Exercise for a two-step prove, that utilizes many results that we have proven.

Lemma 7.4.4

We only prove some part here.

Proof:

Lemma 7.4.5

finite additivity: if $\{E_j\}_{j=1}^N$ is a finite collection of disjoint measurable sets, then for any subset $A$, we have $$ m^*(A \cap (\cup_j E_j)) = \sum_j m^*(A\cap E_j) $$

Proof: again, we try the case of $N=2$ first, to get intuition. Let $B = A \cap (E_1 \cup E_2)$, and $B_1 = A \cap E_1 = B \cap E_1$, and $B_2 = A \cap E_2 = B\cap E_2$, so, we are trying to prove that $$ m^*(B) = m^*(B_1) + m^*(B_2) $$ since $E_1$ and $E_2$ are disjoint, we notice that $B_2 = B \RM E_1$, thus the above holds by measurability of $E_1$ applied to the test set $B$

Discussion Time

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