122 lines
3.3 KiB
Markdown
122 lines
3.3 KiB
Markdown
# Math4121 Lecture 37
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## Extended fundamental theorem of calculus with Lebesgue integration
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### Density of continuous functions
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#### Lemma for density of continuous functions
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Let $K\subseteq U$ be bounded sets in $\mathbb{R}$, $K$ is closed and $U$ is open. Then there is a continuous function $g$ such that $\chi_K\leq g\leq \chi_U$.
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Proof in homework.
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Hint: Consider the basic intervals cases.
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#### Theorem for continuous functions
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Let $f$ be integrable. For each $\epsilon>0$, there is a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $\int_{\mathbb{R}}|f-g|dm<\epsilon$.
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Proof:
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First where $f=\chi_S$ for some bounded means set $S$. then extended to all $f$ integrable.
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First, assume $f=\chi_S$. Let $\epsilon>c$, we can find $K\subseteq S\subseteq U$. and $K$ is closed and $U$ is open such that (by definition of Lebesgue outer measure)
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$$
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m(K)+\frac{\epsilon}{2}>m(S)>m(U)-\frac{\epsilon}{2}
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$$
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In particular, $m(U\setminus K)=m(U)-m(K)<\epsilon$.
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By lemma, there is a continuous function $g$ such that $\chi_K\leq g\leq \chi_U$.
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So
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$$
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\int_E |\chi_S -g|dm=\int_{U\setminus K} |\chi_S -g|dm\leq m(U\setminus K)<\epsilon
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$$
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For the general case,
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By the Monotone Convergence Theorem (use $|f|\chi_{[-N,N]}$ to approximate $|f|$), we can find $N$ large such that
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$$
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\int_{E_N^c}|f|dm<\frac{\epsilon}{3}
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$$
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where $E_N=E\cap [-N,N]$.
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Notice that by the definition of Lebesgue integral, $\int f^+ dm=\sup\{\int \phi^+ dm:\phi\text{ is simple and } \phi\leq f^+\}$ and $\int f^- dm=\sup\{\int \phi^- dm:\phi\text{ is simple and } \phi\leq f^-\}$.
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By considering $f^+$ and $f^-$ separately, we can find a simple function $\phi$ such that
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$$
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\int_{E_N} |f-\phi|dm<\frac{\epsilon}{3}
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$$
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For each $i=1,2,\cdots,n$, we can find $g_i$ continuous such that
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$$
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\int_{E}|\chi_{S_i}-g_i|dm<\frac{\epsilon}{3M}
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$$
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where $M=\sum_{i=1}^n |\alpha_i|$.
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Take $g=\sum_{i=1}^n \alpha_i g_i$,
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$$
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\int_E |\phi-g|dm\leq \sum_{i=1}^n |\alpha_i|\int_E |g_i-\chi_{S_i}|dm<\frac{\epsilon}{3}
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$$
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$\phi-g=\sum_{i=1}^n \alpha_i (\chi_{S_i-g_i})$
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All in all,
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$$
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\begin{aligned}
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\int_E |f-g|dm&\leq \int_E|f-\phi|dm+\int_E |\phi-g|dm\\
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&=\int_{E_N^c}|f|dm+\int_E |f-\phi|dm+\int_E |\phi-g|dm\\
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&<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}\\
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&=\epsilon
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\end{aligned}
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$$
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QED
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### Road map for proving the fundamental theorem of calculus in Lebesgue integration
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Recall the Riemann-Stieltjes integral:
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If $g\in \mathscr{R}(\alpha)$ on $[a,b]$,
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$G(x)=\int_a^x g d\alpha$,
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then:
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1. $G$ is continuous on $[a,b]$
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2. If $g$ is continuous at $x\in [a,b]$, then $G$ is differentiable at $x$ with $G'(x)=g(x)$.
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To extend this to the case where $g$ is Lebesgue integrable, we use the Hardy-Littlewood maximal function.
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#### Definition of the Hardy-Littlewood maximal function
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Given an interval $I\subseteq \mathbb{R}$, define the averaging operator $A_I f(x)=\frac{\chi_I(x)}{m(I)}\int_I f(x)dm$.
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(This function takes the average of $f$ over the interval $I$.)
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The Hardy-Littlewood maximal function is defined as:
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$$
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f^*(x)=\sup_{I\text{ is open interval}}A_I f(x)
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$$
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We will show that $f^*$ is not that such worse than $f$. (Prove on Wednesday)
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Relates to the Fundamental Theorem of Calculus in Lebesgue integration.
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$$
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\frac{G(x+h)-G(x)}{h}=\frac{1}{h}\int_x^{x+h} g(t)dt=A_{[x,x+h]}g(x)
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$$
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If we can control all the averages, we can control the function.
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