62 lines
1.7 KiB
Markdown
62 lines
1.7 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:
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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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$$
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\int_{E_N^c}|f|dm<\frac{\epsilon}{2}
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$$
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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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QED
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