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math105-s22:notes:lecture_15 [2022/03/08 00:27] pzhou created |
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====== Lecture 15 ====== | ====== Lecture 15 ====== | ||
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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, | 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, | ||
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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 $n \to \infty$, | 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 $n \to \infty$, | ||
- | $$ \sum_i \int_{a_i}^b_i f(t) dt \leq \sum_i \int_{a_i}^b_i f_N(t) dt + \sum_i \int_{a_i}^b_i g_N(t) dt \leq N \delta + \int_a^b g_N \leq \epsilon/2 + \epsilon/2 = \epsilon $$ | + | $$ \sum_i \int_{a_i}^{b_i} f(t) dt \leq \sum_i \int_{a_i}^{b_i} f_N(t) dt + \sum_i \int_{a_i}^{b_i} g_N(t) dt \leq N \delta + \int_a^b g_N \leq \epsilon/2 + \epsilon/2 = \epsilon $$ |
Done for (a). | Done for (a). | ||
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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' | 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' | ||
$$ \sum_i |H(b_i) - H(a_i)| \leq \epsilon / 2(b-a) \sum_i (b_i-a_i) \leq \epsilon /2 $$ | $$ \sum_i |H(b_i) - H(a_i)| \leq \epsilon / 2(b-a) \sum_i (b_i-a_i) \leq \epsilon /2 $$ | ||
- | Since $H$ is absolutely continuous, thus for $\epsilon/ | + | Since $H$ is absolutely continuous, thus for $\epsilon/ |
$$ |H(b) - H(a)| \leq \sum_{n=1}^N |H(b_n) - H(a_n)| + \sum_{n=1}^{N-1} |H(b_{n+1}) - H(a_n)| + |H(b) - H(b_N)| + |H(a_1) - H(a)| \leq \epsilon/2 + \epsilon/2 = \epsilon $$ | $$ |H(b) - H(a)| \leq \sum_{n=1}^N |H(b_n) - H(a_n)| + \sum_{n=1}^{N-1} |H(b_{n+1}) - H(a_n)| + |H(b) - H(b_N)| + |H(a_1) - H(a)| \leq \epsilon/2 + \epsilon/2 = \epsilon $$ | ||
Since $\epsilon$ is arbitrary, we do get $H(a)=H(b)$. | Since $\epsilon$ is arbitrary, we do get $H(a)=H(b)$. | ||
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+ | I will leave Pugh section 10 for presentation project. |