math105-s22:notes:lecture_4
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math105-s22:notes:lecture_4 [2022/01/27 08:37] pzhou |
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| ====== Lecture 4 ====== | ====== Lecture 4 ====== | ||
| + | {{ : | ||
| ==== Cor 7.4.7 ==== | ==== Cor 7.4.7 ==== | ||
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| ==== Lemma 7.4.8: Countable addivitivty ==== | ==== 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 | Let $\{E_j\}_{j=1}^\infty$ be a countable collection of **disjoint** subsets. Then, their union $E$ is measurable, and we have | ||
| - | $$ m^*(E) = \sum_{j=1}^\infty m^*(E_j} $$ | + | |
| + | $$ m^*(E) = \sum_{j=1}^\infty m^*(E_j) $$ | ||
| 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. | 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 | First, we start by showing $E$ is measurable from the definition: we want to show for any subset $A$, we have | ||
| - | $$ m^*(A) = m^*(A \cap E) + m^(A \RM E) $$ | + | $$ m^*(A) = m^*(A \cap E) + m^*(A \RM E) $$ |
| Suffice to show $\geq$ direction. Let $F_N = \sum_{j=1}^N E_j$. We have two expressions | 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 \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) $ | ||
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| $$ m^*(A \RM E) + m^*(A \cap E) \leq \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM E)) = \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM F_N)) = \sup_{N > 1} m^*(A) = m^*(A) $$ | $$ m^*(A \RM E) + m^*(A \cap E) \leq \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM E)) = \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM F_N)) = \sup_{N > 1} m^*(A) = m^*(A) $$ | ||
| + | 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, | ||
| + | $$ m^*(E) \geq m^*(F_N) = \sum_{j=1}^N m^*(E_j) $$ | ||
| + | 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, | ||
| + | |||
| + | 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 $E_N = \Omega_N \RM \Omega_{N-1}$, | ||
| + | |||
| + | ==== 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, | ||
| + | * 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> | ||
| + | |||
| + | Can you show that this definition is equivalent to the Caratheodory criterion (the one we had been using)? | ||
math105-s22/notes/lecture_4.1643272639.txt.gz · Last modified: 2026/02/21 14:43 (external edit)
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