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math105-s22:notes:lecture_12 [2022/02/23 23:58]
pzhou |
math105-s22:notes:lecture_12 [2022/02/25 00:35] (current)
pzhou |
| ∫F+(x)dx=∫F−(x)dx | ∫F+(x)dx=∫F−(x)dx |
| hence F+(x)=F−(x) for almost all x. Thus, for a.e. x, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus ∫f(x,y)dy exists for a.e. x. | hence F+(x)=F−(x) for almost all x. Thus, for a.e. x, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus ∫f(x,y)dy exists for a.e. x. |
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| | ==== A Lemma ==== |
| | Suppose A is measurable, and B⊂A any subset, with Bc=A\B. Then |
| | m(A)=m∗(B)+m∗(Bc) |
| | Proof: |
| | m∗(B)=inf{m(C)∣A⊃C⊃B,Cmeasurable}=inf{m(A)−m(Cc)∣A⊃C⊃B,Cmeasurable} |
| | =m(A)−sup{m(Cc)∣A⊃C⊃B,Cmeasurable}=m(A)−sup{m(Cc)∣Cc⊂Bc,Ccmeasurable}=m(A)−m∗(Bc) |
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| ===== Pugh 6.8: Vitali Covering ===== | ===== Pugh 6.8: Vitali Covering ===== |
| **Vitali Covering Lemma:** Let $\vcal$ be a Vitali covering of a measurable bounded subset A by closed balls, then for any $\epsilon>0$, there is a countable disjoint subcollection $\vcal' = \{V_1, V_2, \cdots \},suchthatA \RM \cup_k V_kisanullset,and\sum_k m(V_k) \leq m(A) + \epsilon$. | **Vitali Covering Lemma:** Let $\vcal$ be a Vitali covering of a measurable bounded subset A by closed balls, then for any $\epsilon>0$, there is a countable disjoint subcollection $\vcal' = \{V_1, V_2, \cdots \},suchthatA \RM \cup_k V_kisanullset,and\sum_k m(V_k) \leq m(A) + \epsilon$. |
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| Proof: The construction is easy, like a 'greedy algorithm'. First, using the given ϵ, we find an open subset W⊃A, with m(W)≤m(A)+ϵ. Let $\vcal_1 = \{V \in \vcal: V \In W\}$, and $d_1 = \sup \{diam V: V \in \vcal_1\}$. We pick $V_1 \in \vcal_1$ where the diameter is sufficiently large, say diamV1>d1/2. Then, we delete V1 from W, let W2=W\V1, and consider $\vcal_2 = \{ V \in \vcal_1, V \In W_2\}, and define $d_2 = \sup \{diam V: V \in \vcal_2\},andpickV_2among\vcal_2sothatdiam V_2 > d_2 /2.Repeatthisprocess,wegetacollectionofdisjointclosedballs\{V_i\}.SufficetoshowthatA \RM \cup V_i$ is a null set. | Proof: The construction is easy, like a 'greedy algorithm'. First, using the given ϵ, we find an open subset W⊃A, with m(W)≤m(A)+ϵ. Let $\vcal_1 = \{V \in \vcal: V \In W\}$, and $d_1 = \sup \{diam V: V \in \vcal_1\}$. We pick $V_1 \in \vcal_1$ where the diameter is sufficiently large, say diamV1>d1/2. Then, we delete V1 from W, let W2=W\V1, and consider $\vcal_2 = \{ V \in \vcal_1, V \In W_2\}$, and define $d_2 = \sup \{diam V: V \in \vcal_2\}$, and pick V2 among $\vcal_2$ so that diamV2>d2/2. Repeat this process, we get a collection of disjoint closed balls {Vi}. Suffice to show that A\∪Vi is a null set. |
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| The crucial claim is the following, for any positive integer N, we have | The crucial claim is the following, for any positive integer N, we have |
