Quartz sync: Mar 10, 2025, 12:16 PM

content/Definitions/Measure Theory/Fatou's Lemma.md

@@ -0,0 +1,11 @@
+# Definition
+Have [[Measure|measure]] $\mu$ on $X$, and $f_{n} : X \to [0, \infty]$ [[Measurable|measurable]]. Then $\int \lim_{ n \to \infty } \inf f_{n} \, d\mu \le \lim_{ n \to \infty } \inf \int f_{n} \, d\mu$
+
+> [!info] What is $\lim\inf$?
+> Definition of [[Infimum|infimum]] (it is basically the opposite of a [[Supremum|supremum]]).
+> 
+> $\{ x_{n} \} \subset [0, \infty]$
+> $\lim_{ n \to \infty }\inf x_{n} = \sup_{m}\inf_{n \geq m} x_{n}$
+> 
+> $\inf_{n \geq m} = y_{m} \leq y_{m+1} \leq \dots$
+

content/Definitions/Vector Spaces/Vector Space.md

@@ -0,0 +1,7 @@
+# Definition
+A **vector space** is a set with elements, often called vectors, that can be added together and multiplied by numbers called scalars. These vector spaces must hold [[Properties of a Vector Space|these properties]].
+
+There are different types of **vector spaces** that can exists, some of them being
+- Real Vector Spaces
+- [[Complex Vector Space|Complex Vector Spaces]]
+- [[Normed Vector Space|Normed Vector Spaces]]

content/Lectures/Lecture 15.md

@@ -39,6 +39,7 @@ Then $A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]]
 > By the previous two lemmas, we have
 > $b \geq \lim_{ n \to \infty } \int_{A_{n}} f_{n} \, d\mu \geq \lim_{ n \to \infty } c \times \int_{A_{n}} s \, d\mu$
 > > [!note]- Note on the $A_{n}$
+> > 
 > > $$\int_{A} \subset \int_{X}$$
 > > $$A \subset X$$
 > 
@@ -50,15 +51,9 @@ Then $A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]]
 
 QED.
 # Corollary - Fatou's Lemma
-Have [[Measure|measure]] $\mu$ on $X$, and $f_{n} : X \to [0, \infty]$ [[Measurable|measurable]]. Then $\int \lim_{ n \to \infty } \inf f_{n} \, d\mu \le \lim_{ n \to \infty } \inf \int f_{n} \, d\mu$
-> [!info] What is $\lim\inf$?
-> Definition of [[Infimum|infimum]] (it is basically the opposite of a [[Supremum|supremum]]).
-> 
-> $\{ x_{n} \} \subset [0, \infty]$
-> $\lim_{ n \to \infty }\inf x_{n} = \sup_{m}\inf_{n \geq m} x_{n}$
-> 
-> $\inf_{n \geq m} = y_{m} \leq y_{m+1} \leq \dots$
-
+Defined in the lecture here: [[Fatou's Lemma]]
+> [!info] Definition
+> Have [[Measure|measure]] $\mu$ on $X$, and $f_{n} : X \to [0, \infty]$ [[Measurable|measurable]]. Then $\int \lim_{ n \to \infty } \inf f_{n} \, d\mu \le \lim_{ n \to \infty } \inf \int f_{n} \, d\mu$
 ## Proof
 Use [[Lebesgue's Monotone Convergence Theorem]] on $g_{m} = \inf_{n \geq m} f_{n}$.
 $g_{1} \leq g_{2} \leq \dots$ are [[Measurable|measurable]] functions.

content/Lectures/Lecture 16.md

@@ -0,0 +1,87 @@
+---
+lecture: 16
+date: 2025-03-10
+---
+# Recap
+Recap from the [[Lecture 15|previous]].
+$L_{1}(\mu) = \left\{  \text{measure}\ f : \underbrace{X}_{\mu} \to \mathbb{C} \, | \, \underbrace{\int |f| \, d\mu}_{= \, \sup_{0 \leq s \leq |f|} \int s \, d\mu} \lt \infty \right\}$
+$s = \Sigma_{i = 1}^{n} a_{i} \times X_{A_{i}}$
+$\int s \, d\mu = \Sigma_{i=1}^{n} a_{i} \times \mu(A_{i})$
+
+$f = \mathrm{Re} f + i \mathrm{Im} f = (\mathrm{Re} f)^{+} - (\mathrm{Re} f)^{-} + i((\mathrm{Im} f)^{+} - (\mathrm{Im} f)^{-})$
+$0 \leq \int (\mathrm{Re} f)^{\pm} \, d\mu \leq \int |f| \, d\mu$
+# Proposition
+$L^{1} (\mu)$ is a [[Vector Space|vector space]] under [[Pointwise|pointwise]] operations, and the $\int : L^{1}(\mu) \to \mathbb{C} \, d\mu$ is linear, and $| \int f \, d\mu | \leq \int |f| \, d\mu$.
+## Proof
+Let $f, g \in L^{1}(\mu)$, $a \in \mathbb{C}$.
+
+Assume first that $f, g \geq 0$ [[Simple Function|simple]] with distinct values $a_{i}$ and $b_{i}$, let $A_{i} = \{  x \in X \, | \, f(x) = a_{i} \}$ and $B_{j} = \{ x \in X \, | \, g(x) = b_{j} \}$
+
+Then $\int_{A_{i} \cap B_{j}} (f + g) \, d\mu = (a_{i} + b_{j}) \times \mu (A_{i} \cap B_{j})$
+$= a_{i} \times \mu (A_{i} \cap B_{j}) + b_{j} \times \mu(A_{i} + B_{j})$
+$= \int_{A_{i} + B_{j}} f \, d\mu + \int_{A_{i} + B_{j}} g \, d\mu$
+> [!note]
+> $X \underbrace{=}_{\text{disjoint union}} \cup(A_{i} \cap B_{j})$
+> $A \mapsto \int_{A} s \, d\mu$
+> [[Measure]] $L1$.
+
+$\implies \int_{X} (f + g) \, d\mu = \int f \, d\mu + \int g \, d\mu$
+LHS $= \Sigma_{i j} \int (f + g) \, d\mu = \Sigma_{i j} \left( \int_{A_{i} \cap B_{j}} f \, d\mu + \int_{A_{i} \cap B_{j}} g \, d\mu\right)$
+$= \Sigma_{i j} \int_{A_{i} \cap B_{j}} f \, d\mu + \Sigma_{i j} \int_{A_{i} \cap B_{j}} g \, d\mu$
+$=^{L_{1}} \int_{X} f \, d\mu + \int_{X} g \, d\mu$
+
+$L2$:
+$0 \leq s_{1} \leq s_{2} \leq \dots \leq f$
+$f=\lim_{ n \to \infty } s_{n}$
+
+By $L_{2}$ and [[Lebesgue's Monotone Convergence Theorem]] (LMCT) $\implies \int$ is additive for all non-negative [[Measurable|measurable]] functions $f$ and $g$.
+$\int  (f + g) \, d\mu =^{L_{2}} \int \lim_{ n \to \infty } (s_{n} + s_{n}') \, d\mu =^{\text{LMCT}} \lim_{ n \to \infty } \int (s_{n} + s_{n}') \, d\mu$
+(from above) $= \lim_{ n \to \infty } (\int s_{n} \, d\mu + \int s_{n}' \, d\mu$
+($\lim$ splits) $\lim_{ n \to \infty } \int s_{n} \, d\mu + \lim_{ n \to \infty } \int s_{n}' \, d\mu$
+$=^{\text{LMCT}} \int f \, d\mu + \int g \, d\mu$
+
+For the complex case we have $\int \underbrace{| af + g |}_{\leq |a| \times |f| + |g|} \, d\mu \leq \int (| a | \times |f| + |g|) \, d\mu$
+$= \int |a| \times |f| \, d\mu + \int |g| \, d\mu = |a| \times \underbrace{\int |f|}_{\lt \infty} \, d\mu + \underbrace{\int |g| \, d\mu}_{\lt \infty} < \infty$, so $af + g \in L^{1}(\mu)$ when $f, g \in L^{1}(\mu)$.
+So $L^{1}(\mu)$ is a [[Vector Space|vector space]].
+> [!info]- Why it's $af + g$ and not $af + bg$
+> Usually you would use $af + bg$ and have to prove for the $bg$ part as well, but $af + g$ is enough for the proof as you can do:
+> $$af + bg = af + (bg + 0)$$
+> as the parts in the bracket would belong in $L1$ as well.
+
+Then use the definition of the integral for complex functions to see that it is additive. Clearly $\int  a f \, d\mu = a \int f \, d\mu$ when $a \geq 0$. For $a \lt 0$ use $(-g)^{\pm} = g^{\mp}$ etc. Check for $a = i$, but this is okay as $\int i f \, d\mu = \int i (\mathrm{Re} f + i \mathrm{Im} f) \, d\mu = \int (-\mathrm{Im} f + i \mathrm{Re} f) \, d\mu = - \int \mathrm{Im}f \, d\mu + i \int \mathrm{Re} f \, d\mu = i \times \int f \, d\mu$.
+So $\int a f \, d\mu = a \int f \, d\mu, \; \forall a \in \mathbb{C}$, and the integral is a linear function.
+
+Finally, pick $b \in \mathbb{C}$ such that $| \int f \, d\mu | = b \times \underbrace{\int f \, d\mu}_{z}$
+> [!note]- What is happening at $z$
+> Remember the rules of complex numbers:
+> $$|z| = b \times z$$
+> $$b = \frac{|z|}{z}$$
+
+(linear) $= \int b f \, d\mu = \int \underbrace{\mathrm{Re}(b f)}_{\leq |b f|} \, d\mu \leq \int  |b f| \, d\mu = \int \underbrace{|b|}_{=1} \times |f| \, d\mu = \int |f| \, d\mu$
+QED.
+# Lebesgue's Dominated Convergence Theorem
+It is a theorem for complex valued functions.
+Also see the [[Lebesgue's Dominated Convergence Theorem|definition]].
+
+Let $\{ f_{n} \}$ be [[Measurable|measurable]] functions $f_{n} : X \to \mathbb{C}$, and $\mu$ [[Measure|measure]] on $X$. Assume $\exists g \in L^{1}$ such that all $f_{n} \leq g$.
+Then $\lim_{ n \to \infty } \int  f_{n} \, d\mu = \int \underbrace{\lim_{ n \to \infty } f_{n}}_{f} \, d\mu$.
+If you have a measure so that the whole space is finite, $\mu(x) < \infty$:
+$\implies \int 1 \, d\mu = \mu(x) < \infty$ (can use $g$ instead of the $1$)
+## Proof
+By [[Fatou's Lemma]]
+$\int 2g \, d\mu \underbrace{=}_{f = \lim f_{n}} \int \lim_{ n \to \infty } \inf(\underbrace{2g - |f_{n} - f|}_{\geq 0}) \, d\mu$
+(by [[Fatou's Lemma]]) $\leq \lim_{ n \to \infty } \inf \int (2g - |f_{n} - f|) \, d\mu$
+(by [[#Proposition|previous proposition]]) $\int 2g \, d\mu + \lim_{ n \to \infty }\inf\left( -\int |f_{n} - f| \, d\mu \right)$
+$= \int 2g \, d\mu - \lim_{ n \to \infty }\sup \int |f_{n} - f | \, d\mu$ so $\lim_{ n \to \infty } \sup \int |f_{n} - f \, d\mu \leq 0$, and $\lim_{ n \to \infty } \underbrace{\int |f_{n} - f| \, d\mu}_{\geq 0} \leq 0$, so $\lim_{ n \to \infty } \int |f_{n} - f| \, d\mu = 0$.
+
+Then $\lim_{ n \to \infty } | \int f_{n} \, d\mu - \int f \, d\mu |$
+(by [[#Proposition|previous proposition]]) $= \lim_{ n \to \infty } | \int (f_{n} - f) \, d\mu |$
+(again, by [[#Proposition|previous proposition]]) $\leq \lim_{ n \to \infty } \int |f_{n} - f \, d\mu = 0$.
+
+QED.
+> [!example] Another example
+> 
+> $$\lim_{ n \to \infty } \lim_{ m \to \infty } \frac{n}{m} = \lim_{ n \to \infty } 0 = 0$$
+> Then swap the limits
+> $$\lim_{ m \to \infty } \lim_{ n \to \infty } \frac{n}{m} = \lim_{ m \to \infty } \infty = \infty$$
+> Which are two different numbers as $0 \neq \infty$.

content/index.md

@@ -23,3 +23,4 @@ title: ACIT4330 Lecture Notes
 - [[Lecture 13 - Measure Theory]]
 - [[Lecture 14]]
 - [[Lecture 15]]
+- [[Lecture 16]]