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pages/Math4121/Math4121_L37.md

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 ### Density of continuous functions
 
-#### Lemma: 
+#### 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 $\chi_K\leq g\leq \chi_U$.
 
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 $$
 
 If we can control all the averages, we can control the function.
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-
-
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pages/Math4121/Math4121_L38.md

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+# Math4121 Lecture 38
+
+## Extended fundamental theorem of calculus with Lebesgue integration
+
+### Hardy-Littlewood maximal function
+
+Given integrable $f$m and an interval $I$, look at the averaging operator $A_I f(x)=\frac{\chi_I(x)}{m(I)}\int_I f(y)dy$.
+
+The maximal function is defined as
+
+$$
+f^*(x)=\sup_{I \text{ is an open interval}} A_I f(x)
+$$
+
+#### Theorem Hardy-Littlewood Maximal Function Theorem
+
+Fix $f$ integrable. For each $\lambda>0$, we define
+
+$$
+E_\lambda^*=\{x\in\mathbb{R}: |f^*(x)|>\lambda\}
+$$
+
+Then
+
+$$
+m(E_\lambda^*)\leq \frac{2}{\lambda}\int_\mathbb{R} |f| dm
+$$
+
+To give context for the maximal estimate, for any $f$ integrable, $\lambda>0$,
+
+$$
+E_\lambda=\{x\in\mathbb{R}: |f(x)|>\lambda\}
+$$
+
+Then we have Marknov's inequality, $m(E_\lambda)\leq \frac{1}{\lambda}\int_\mathbb{R} |f| dm$. We know $|f(x)|>\lambda \chi_{E_\lambda}(x)$ so Markov's inequality follows by integrating.
+
+Proof:
+
+Let $f^*(x)=\sup_{I \text{ is an open interval such that } x\in I} \frac{1}{m(I)}\int_I f \, dm$.
+
+If $x\in E_\lambda^*$, then $\exists I$ open interval such that $x\in I$ and $\frac{1}{m(I(x))}\left|\int_{I(x)} f \, dm\right|>\lambda$.
+
+Take $K\subset E_\lambda^*$ compact. Then $K\subset \bigcup_{x\in K} I(x)$. Taking the finite subcover, we have $I_1, \ldots, I_n$ open intervals such that $K\subset \bigcup_{i=1}^n I_i$.
+
+If three intervals, $I,J,K$ have non-empty intersection, then one is contained in the union of the other two.
+
+In particular, we can find another subcover for $K$, $J_1, \ldots, J_N$ such that they have overlap of at most 2 (otherwise, we can remove the cover). We can state this as
+
+$$
+\sum_{j=1}^N \chi_{J_j}(x)\leq 2
+$$
+
+$$
+\begin{aligned}
+m(K)&\leq \sum_{j=1}^N m(J_j)\\
+&\leq \sum_{j=1}^N \frac{1}{\lambda}\left|\int_{J_j} f \, dm\right|\\
+&\leq \frac{1}{\lambda}\int_\mathbb{R} \sum_{j=1}^N \chi_{J_j}(x) |f(x)| \, dx\\
+&\leq \frac{2}{\lambda}\int_\mathbb{R} |f(x)| \, dx
+\end{aligned}
+$$
+
+Since $A_I f(x)$ is measurable, $f^*$ is measurable function and $E_\lambda$ is measurable, we have
+
+$$
+m(E_\lambda^*)\leq \frac{2}{\lambda}\int_\mathbb{R} |f(x)| \, dx
+$$
+
+QED
+
+## 3 Big Convergence Theorems
+
+### Theorem L.1 (Monotone Convergence Theorem)
+
+[Monotone convergence theorem](https://notenextra.trance-0.com/Math4121/Math4121_L36#theorem-614-monotone-convergence-theorem)
+
+### Theorem L.2 (Fatou's Lemma)
+
+Let $\{f_n\}_{n=1}^\infty$ be a sequence of non-negative measurable functions on $E$. Then
+
+$$
+\int_E \liminf_{n\to\infty} f_n \, dm\leq \liminf_{n\to\infty} \int_E f_n \, dm
+$$
+
+Proof:
+
+Let $g_n=\inf_{k\geq n} f_k$ is a monotone increasing nonnegative, and the following properties holds:
+
+$$
+\lim_{n\to\infty} g_n(x)=\sup_{n\geq 1} \inf_{k\geq n} f_k(x)=\liminf_{n\to\infty} f_n(x)
+$$
+
+$$
+\int_E g_n \, dm\leq \inf_{k\geq n} \int_E f_k\, dm
+$$
+
+So,
+
+$$
+ \int_E g_n \, dm\leq \inf_{k\geq n} \int_E f_k \, dm
+$$
+
+Apply the monotone convergence theorem to $g_n$, we have
+
+$$
+\lim_{n\to\infty} \int_E g_n \, dm\leq \liminf_{n\to\infty} \int_E f_n \, dm
+$$
+
+QED
+
+### Theorem L.3 (Dominated Convergence Theorem)
+
+Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on $\mathbb{R}$ converging to $f$ almost everywhere. If there exists integrable $g$ such that $|f_n|\leq |g|$ for all $n$, then
+
+$$
+\int_E f \, dm=\lim_{n\to\infty} \int_E f_n \, dm
+$$
+
+Proof:
+
+Consider the function $g+f_n$ and $g-f_n$, these are non-negative sequences of measurable functions. By Fatou's lemma, we have
+
+$$
+\int g\,dm+\int f\,dm=\int_E \liminf_{n\to\infty} (g+f_n) \, dm\leq \liminf_{n\to\infty} \int_E (g+f_n) \, dm=\int_E g\,dm+\liminf_{n\to\infty} \int_E f_n\,dm
+$$
+
+So, $\int_E f\,dm\leq \liminf_{n\to\infty} \int_E f_n\,dm$.
+
+Similarly, we have
+
+$$
+\int g\,dm-\int f\,dm=\int_E \liminf_{n\to\infty} (g-f_n) \, dm\leq \liminf_{n\to\infty} \int_E (g-f_n) \, dm=\int_E g\,dm-\limsup_{n\to\infty} \int_E f_n\,dm
+$$
+
+So, $\limsup_{n\to\infty} \int_E f_n\,dm\leq \int_E f\,dm$.
+
+So $\limsup_{n\to\infty} \int_E f_n\,dm\leq \int_E f\,dm\leq \liminf_{n\to\infty} \int_E f_n\,dm$.
+
+Since $\limsup_{n\to\infty} \int_E f_n\,dm\geq \liminf_{n\to\infty} \int_E f_n\,dm$, we have $\int_E f\,dm=\lim_{n\to\infty} \int_E f_n\,dm$.
+
+QED