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--- math105-s22:notes:lecture_15 [2022/03/08 00:27]
pzhou created
+++ math105-s22:notes:lecture_15 [2022/03/09 12:37] (current)
pzhou
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 ====== Lecture 15 ======
+[[https://berkeley.zoom.us/rec/share/OvE6Kwx30h9tqRwyGp3PCtUcEp97b98ahuWVfO29jmfOGLRiEiJpwEFrfAPWnC2p.RD0hzsRPKYEE009Z | video ]]
 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
-$$ \sum_i \int_{a_i}^b_i f(t) dt \leq \sum_i \int_{a_i}^b_i f_N(t) dt +  \sum_i \int_{a_i}^b_i g_N(t) dt \leq N \delta + \int_a^b g_N \leq \epsilon/2 + \epsilon/2 = \epsilon $$
+ \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).
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 How to use the condition  $H'=0$  a.e.? Let  $Z\In [a,b]$  be a nullset, such that if  $x \in A = [a,b] \RM Z$ , we have $H'(x)=0 $. By definition, it means that$ \lim_{h \to 0} [H(x+h) - H(x)] / h = 0 $. Then, we claim, for any$ x \in A $,  for any$ \epsilon $, there exists$ \delta $, such that if$ x \in [a, b ] $with$ 0<b - a< \delta $, we have$ |H(b_i) - H(a_i)| / (b_i - a_i) < \epsilon / 2(b-a) $(why?). Thus, we have a Vitali covering of$ A $by closed interval. Thus, there is a countable collection of disjoint intervals$ [a_i, b_i]$, such that
  $\sum_i |H(b_i) - H(a_i)| \leq \epsilon / 2(b-a) \sum_i (b_i-a_i) \leq \epsilon /2$
-Since  $H$  is absolutely continuous, thus for $\epsilon/2$, there is a $\delta>0$, such that  $\sum_{i=1}^n |b_i - a_i| < \delta$  implies $\sum_i |H(b_i) - H(a_i)|<\epsilon/2 $. Let$ N $be large enough, such that$ m([a,b] - \sqcup_{i=1}^N [a_i,b_i]) < \delta $. And relabel these$ N $intervals, so that$ a \leq a_1 < b_1 < a_2 < b_2 \cdots < a_N < b_N \leq b$, then we have
+Since  $H$  is absolutely continuous, thus for $\epsilon/2$, there is a $\delta>0$, such that  $\sum_{i=1}^n |b_i - a_i| < \delta$  implies $\sum_i |H(b_i) - H(a_i)|<\epsilon/2 $. Let$ N $be large enough, such that$ m ( [a,b] - \sqcup_{i=1}^N [a_i,b_i]) < \delta $. And relabel these$ N $intervals, so that$ a \leq a_1 < b_1 < a_2 < b_2 \cdots < a_N < b_N \leq b$, then we have
  $|H(b) - H(a)| \leq  \sum_{n=1}^N |H(b_n) - H(a_n)| + \sum_{n=1}^{N-1} |H(b_{n+1}) - H(a_n)| + |H(b) - H(b_N)| + |H(a_1) - H(a)|  \leq \epsilon/2 + \epsilon/2 = \epsilon$
 Since  $\epsilon$  is arbitrary, we do get  $H(a)=H(b)$ .
+-------
+I will leave Pugh section 10 for presentation project.

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