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====== Lecture 3 ====== | ====== Lecture 3 ====== | ||
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Today we continue going over Tao's sequence of Lemma 7.4.2 - 7.4.11 | Today we continue going over Tao's sequence of Lemma 7.4.2 - 7.4.11 | ||
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Let $E = (0, | Let $E = (0, | ||
$$ m^*(A) = m^*(A \cap E^c) + m^*(A \cap E). $$ | $$ m^*(A) = m^*(A \cap E^c) + m^*(A \cap E). $$ | ||
- | Let $A_+ = A \cap E$ and $A_- = A \cap E^c$, note that if $0 \in A$, then $0 \in A_-$. First, we show that $$ m^*(A) \leq m^*(A_-) + m^*(A_+)$$. 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> | + | 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 $$ m^*(A) \leq m^*(A_-) + m^*(A_+)$$ |
$$ m^*(A) + 2\epsilon \geq m^*(A_-) + m^*(A_+).$$ | $$ m^*(A) + 2\epsilon \geq m^*(A_-) + m^*(A_+).$$ | ||
The plan is the following. Given an open covering of $A$ by intervals $\{B_j\}_{j=1}^\infty$, | The plan is the following. Given an open covering of $A$ by intervals $\{B_j\}_{j=1}^\infty$, | ||
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$$ m^*(A) + 2\epsilon \geq \sum_j |B_j| + \epsilon \geq \sum_j |B_j^+| + \sum_j |B_j^-| \geq m^*(A_+) + m^*(A_-). $$ | $$ m^*(A) + 2\epsilon \geq \sum_j |B_j| + \epsilon \geq \sum_j |B_j^+| + \sum_j |B_j^-| \geq m^*(A_+) + m^*(A_-). $$ | ||
- | 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. We can prove the general | + | 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. |
- | $$ m^*(A) | + | |
+ | ==== Lemma 7.4.4 ==== | ||
+ | We only prove some part here. | ||
+ | * (c) If $E_1, E_2$ are measurable, then $E_1 \cup E_2$ and $E_1 \cap E_2$ are measurable. | ||
+ | * (e) every open box, and every closed box, is measurable | ||
+ | * (f) if a set has outer-measure zero, then it is measurable. | ||
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+ | Proof: | ||
+ | * (c) Try the case first with $A = \{ x^2 + y^2 < 1 \} \In \R^2$ and $E_1 = \{x>0\}$ and $E_2 = \{y> | ||
+ | * (e) express open box as intersections of translated half-spaces. | ||
+ | * (f) Assume $E$ has outer-measure 0, then $0 \leq m^*(A \cap E) \leq m^*(E) = 0$, we can prove $m^*(A) \geq m^*(A \RM E) + m^*(A \cap E) = m^*(A \RM E)$ again by monotonicity. | ||
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+ | ==== Lemma 7.4.5 ==== | ||
+ | finite additivity: if $\{E_j\}_{j=1}^N$ is a finite collection of **disjoint** measurable sets, then for any subset $A$, we have | ||
+ | $$ m^*(A \cap (\cup_j E_j)) = \sum_j | ||
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+ | 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 $B$ | ||
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+ | ===== Discussion Time ===== | ||
+ | Let's fill in the details of the above sketches. | ||
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