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math105-s22:notes:lecture_15

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math105-s22:notes:lecture_15 [2022/03/08 00:27]
pzhou 		math105-s22:notes:lecture_15 [2022/03/09 12:37] (current)
pzhou
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 ====== Lecture 15 ====== ====== Lecture 15 ======
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 +[[https://berkeley.zoom.us/rec/share/OvE6Kwx30h9tqRwyGp3PCtUcEp97b98ahuWVfO29jmfOGLRiEiJpwEFrfAPWnC2p.RD0hzsRPKYEE009Z | video ]]
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 Last time, we considered the (long and hard) Lebesgue density theorem, which says, given any Lebesgue locally integrable function f:Rn→Rf: \R^n \to \R, then for almost all pp, the density $\delta(p,f)of of fat at pexistsandequalstothevalue exists and equals to the value f(p)$.  Last time, we considered the (long and hard) Lebesgue density theorem, which says, given any Lebesgue locally integrable function f:Rn→Rf: \R^n \to \R, then for almost all pp, the density $\delta(p,f)of of fat at pexistsandequalstothevalue exists and equals to the value f(p)$. 
  
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 If ff is not bounded, then we can split ff to be a bounded part and an unbounded part. For any n>0n>0, we can define f=fn+gnf = f_n + g_n, where fn=max⁡(f,n)f_n = \max (f, n). Then, as n→∞n \to \infty,  gn→0g_n \to 0 a.e., and by DCT, ∫gn→0\int g_n \to 0. Pick NN large enough, such that ∫gN<ϵ/2\int g_N < \epsilon /2. Then, let $\delta = (\epsilon/2) / N.Then,foranycountabledisjointopenintervals. Then, for any countable disjoint open intervals \{(a_i, b_i)\}withlength with length < \delta$, we have  If ff is not bounded, then we can split ff to be a bounded part and an unbounded part. For any n>0n>0, we can define f=fn+gnf = f_n + g_n, where fn=max⁡(f,n)f_n = \max (f, n). Then, as n→∞n \to \infty,  gn→0g_n \to 0 a.e., and by DCT, ∫gn→0\int g_n \to 0. Pick NN large enough, such that ∫gN<ϵ/2\int g_N < \epsilon /2. Then, let $\delta = (\epsilon/2) / N.Then,foranycountabledisjointopenintervals. Then, for any countable disjoint open intervals \{(a_i, b_i)\}withlength with length < \delta$, we have 
-$$ \sum_i \int_{a_i}^b_i f(t) dt \leq \sum_i \int_{a_i}^b_i f_N(t) dt +  \sum_i \int_{a_i}^b_i g_N(t) dt \leq N \delta + \int_a^b g_N \leq \epsilon/2 + \epsilon/2 = \epsilon $$+ \sum_i \int_{a_i}^{b_i} f(t) dt \leq \sum_i \int_{a_i}^{b_i} f_N(t) dt +  \sum_i \int_{a_i}^{b_i} g_N(t) dt \leq N \delta + \int_a^b g_N \leq \epsilon/2 + \epsilon/2 = \epsilon
 Done for (a).  Done for (a). 
  
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 Since ϵ\epsilon is arbitrary, we do get H(a)=H(b)H(a)=H(b).  Since ϵ\epsilon is arbitrary, we do get H(a)=H(b)H(a)=H(b). 
  
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 +I will leave Pugh section 10 for presentation project. 
math105-s22/notes/lecture_15.1646728055.txt.gz · Last modified: 2022/03/08 00:27 by pzhou

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