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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 $f: \R^n \to \R$, then for almost all $p$, the density $\delta(p,f)$ of $f$ at $p$ 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 $f: \R^n \to \R$, then for almost all $p$, the density $\delta(p,f)$ of $f$ at $p$ exists and equals to the value $f(p)$. 
  
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 If $f$ is not bounded, then we can split $f$ to be a bounded part and an unbounded part. For any $n>0$, we can define $f = f_n + g_n$, where $f_n = \max (f, n)$. Then, as $n \to \infty$,  $g_n \to 0$ a.e., and by DCT, $\int g_n \to 0$. Pick $N$ large enough, such that $\int g_N < \epsilon /2$. Then, let $\delta = (\epsilon/2) / N$. Then, for any countable disjoint open intervals $\{(a_i, b_i)\}$ with length $< \delta$, we have  If $f$ is not bounded, then we can split $f$ to be a bounded part and an unbounded part. For any $n>0$, we can define $f = f_n + g_n$, where $f_n = \max (f, n)$. Then, as $n \to \infty$,  $g_n \to 0$ a.e., and by DCT, $\int g_n \to 0$. Pick $N$ large enough, such that $\int g_N < \epsilon /2$. Then, let $\delta = (\epsilon/2) / N$. Then, for any countable disjoint open intervals $\{(a_i, b_i)\}$ 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 $\epsilon$ is arbitrary, we do get $H(a)=H(b)$.  Since $\epsilon$ is arbitrary, we do get $H(a)=H(b)$. 
  
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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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