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content/Math4121/Math4121_L37.md

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