diff --git a/docker-compose.yaml b/docker-compose.yaml
index 8e1975c..c7e5cbb 100644
--- a/docker-compose.yaml
+++ b/docker-compose.yaml
@@ -3,7 +3,7 @@ services:
     build:
       context: ./
       dockerfile: ./Dockerfile
-    image: trance0/notenextra:v1.1.10
+    image: trance0/notenextra:v1.1.11
     restart: on-failure:5
     ports:
       - 13000:3000
diff --git a/pages/Math4121/Math4121_L28.md b/pages/Math4121/Math4121_L28.md
index bd1212a..50f4e0b 100644
--- a/pages/Math4121/Math4121_L28.md
+++ b/pages/Math4121/Math4121_L28.md
@@ -118,3 +118,14 @@ So $S$ is measurable.
 
 QED
 
+#### Proposition 5.9 (Preview)
+
+Any finite union (and intersection) of measurable sets is measurable.
+
+Proof:
+
+Let $S_1, S_2$ be measurable sets.
+
+We prove by verifying the Caratheodory's criteria for $S_1\cup S_2$.
+
+QED
diff --git a/pages/Math416/Math416_L20.md b/pages/Math416/Math416_L20.md
new file mode 100644
index 0000000..4ef9f96
--- /dev/null
+++ b/pages/Math416/Math416_L20.md
@@ -0,0 +1,143 @@
+# Math416 Lecture 20
+
+## Laurent Series and Isolated Singularities
+
+### Isolated Singularities
+
+$f$ has an isolated singularity at $z_0$ if $f$ is analytic everywhere in some punctured disk $0 < |z - z_0| < R$ except at $z_0$ itself.
+
+#### Removable Singularities
+
+We call $z_0$ a removable singularity if there exists $g\in O(B_r(z_0))$ such that $f(z) = g(z)$ for all $z\in B_r(z_0) \setminus \{z_0\}$.
+
+#### Poles
+
+We call $z_0$ a pole if there are finitely many terms with negative powers in the Laurent series expansion of $f$ about $z_0$.
+
+#### Essential Singularities
+
+We call $z_0$ an essential singularity if there are infinitely many terms with negative powers in the Laurent series expansion of $f$ about $z_0$.
+
+#### Theorem: Criterion for a removable singularity (Riemann removable singularity theorem)
+
+Suppose $f$ has an isolated singularity at $z_0$. Then it is removable if and only if $f$ is bounded on a punctured disk centered at $z_0$.
+
+#### Theorem 8.10 (Casorati-Weierstrass Theorem)
+
+If $z_0$ is an essential singularity of $f$, then $\forall r>0$, $\overline{f(B_r(z_0)\setminus\{z_0\})}= \mathbb{C}$.
+
+Proof:
+
+Suppose $w\notin$ closure range fo $f$ on $B_r(z_0)\setminus\{z_0\}$, then $\exists \epsilon > 0$ such that $B_\epsilon(w)\cap f(B_r(z_0)\setminus\{z_0\})=\emptyset$.
+
+$g(z)=1/(f(z)-w)$ and $|g(z)|\leq \frac{1}{\epsilon}$, which is bounded. By Riemann removable singularity theorem, $g$ has a removable singularity. So $f(z)=\frac{1}{g(z)}+w$ is holomorphic on $B_r(z_0)\setminus\{z_0\}$.
+
+Suppose $g(z_0)\neq 0$, then $f$ has a removable singularity at $z_0$.
+
+Suppose $g(z_0)=0$, then $f$ has a pole at $z_0$.
+
+This contradicts the assumption that $z_0$ is an essential singularity.
+
+QED
+
+#### Theorem 8.11 (Picard's Theorem)
+
+If $z_0$ is an essential singularity of $f$, then $\forall r>0$, $f(B_r(z_0)\setminus\{z_0\})$ contains every point in $\mathbb{C}$ except possibly one.
+
+#### Definition: Residue
+
+Suppose $f$ has an isolated singularity at $z_0$. The residue of $f$ at $z_0$, write $res_{z_0}(f)$, is the coefficient of $(z-z_0)^{-1}$ in the Laurent series expansion of $f$ about $z_0$.
+
+> Preview:
+>
+> Residue Theorem:
+>
+> Suppose $G$ is a simply connected domain, $F$ is a finite set in $G$ $h$ is holomorphic on $G\setminus F$. Let $\gamma$ be a simple closed curve in $G$, containing $\lambda_1, \lambda_2, \cdots, \lambda_N$ from $F$. Then
+>
+> $$\int_\gamma h(z) dz = 2\pi i \sum_{j=1}^N res_{\lambda_j}(h)$$
+
+Special case:
+
+When $\gamma=\partial B_r(z_0)$, $f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$ converges on $A(z_0;0,R)$, then
+
+$$
+\int_{\gamma} f(z) dz = 2\pi i \sum_{n=-\infty}^\infty a_n \int_{\gamma} (z-z_0)^n dz=2\pi i a_{-1}
+$$
+
+Example:
+
+1. Find residue of $f(z)=\frac{\sin z}{z^4}$ at $z=0$.
+
+$\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$
+
+$\frac{\sin z}{z^4} = \frac{1}{z^4} - \frac{1}{3!z} + \frac{z}{5!} - \cdots$
+
+$res_{z=0}(\frac{\sin z}{z^4}) = \frac{1}{3!}$
+
+2. Find residue of $f(z)=\frac{1}{(z+2)(z-5)}$ at $z=5$ and $z=-2$.
+
+$res_{z=5}(f(z))=(z-5)^{-1}\cdot\frac{1}{z+2}|_{z=5}=\frac{1}{7}$
+
+$res_{z=-2}(f(z))=(z+2)^{-1}\cdot\frac{1}{z-5}|_{z=-2}=-\frac{1}{7}$
+
+#### Corollary of residue
+
+Suppose $f$ has an simple pole at $z_0$. Then $res_{z_0}(f)=\lim_{z\to z_0}(z-z_0)f(z)$.
+
+Proof:
+
+$f(z)=a_{-1}(z-z_0)^{-1}+\sum_{n=0}^\infty a_n(z-z_0)^n$, $(z-z_0)f(z)=a_{-1}+\sum_{n=0}^\infty a_n(z-z_0)^{n+1}$
+
+$\lim_{z\to z_0}(z-z_0)f(z)=\lim_{z\to z_0}(z-z_0)\cdot a_{-1}+\lim_{z\to z_0}(z-z_0)\cdot\sum_{n=0}^\infty a_n(z-z_0)^n=a_{-1}$
+
+QED
+
+#### Find residue for poles with higher order
+
+Suppose $f$ has a pole of order 2 at $z_0$. Then $f(z)=\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}+\sum_{n=0}^\infty a_n(z-z_0)^n$, $(z-z_0)^2f(z)=a_{-2}+(z-z_0)a_{-1}+\sum_{n=0}^\infty a_n(z-z_0)^{n+2}$
+
+Method 1:
+
+$res_{z_0}(f)=a_{-1}=\lim_{z\to z_0}\frac{f(z)-\lim_{z\to z_0}(z-z_0)^2f(z)}{(z-z_0)}$
+
+Method 2:
+
+$res_{z_0}(f)=\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)|_{z=z_0}$
+
+So suppose $f$ has a pole of order $n$ at $z_0$. Then $res_{z_0}(f)=\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)|_{z=z_0}$
+
+Proof:
+
+$f(z)=\frac{a_{-n}}{(z-z_0)^n}+\frac{a_{-n+1}}{(z-z_0)^{n-1}}+\cdots+\frac{a_{-1}}{z-z_0}+\sum_{m=0}^\infty a_m(z-z_0)^m$
+
+$(z-z_0)^nf(z)=a_{-n}+(z-z_0)a_{-n+1}+\cdots+(z-z_0)^{n-1}a_{-1}+\sum_{m=0}^\infty a_m(z-z_0)^{m+n}$
+
+$\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)=(n-1)!a_{-1}+\sum_{m=0}^\infty a_m(m+n)(m+n-1)\cdots(m+1)(z-z_0)^{m-1}$
+
+$\lim_{z\to z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)=a_{-1}$
+
+QED
+
+## Chapter 9: Generalized Cauchy's Theorem
+
+### Simple connectedness
+
+#### Proposition 9.1
+
+Let $\phi$ be a continuous nowhere vanishing function from $[a,b]\subset\mathbb{R}$ to $\mathbb{C}\setminus\{0\}$. Then there exists a continuous function $\psi:[a,b]\to\mathbb{C}$ such that $e^{\psi(t)}=\phi(t)$ for all $t\in[a,b]$.
+
+Moreover, $\psi$ is uniquely determined up to an additive integer multiple of $2\pi i \mathbb{Z}$.
+
+Proof:
+
+Uniqueness:
+
+Suppose $\phi_1$ and $\phi_2$ are both continuous functions so that $e^{\phi_1(t)}=\phi(t)=e^{\phi_2(t)}$ for all $t\in[a,b]$.
+
+Then $e^{\phi_1(t)-\phi_2(t)}=1$ for all $t\in[a,b]$. So $\phi_1(t)-\phi_2(t)=2k\pi i$ for some $k\in\mathbb{Z}$.
+
+Existence:
+
+Continue on Thursday.
+
+QED
diff --git a/pages/Math416/_meta.js b/pages/Math416/_meta.js
index 0aeecca..6bcf6cd 100644
--- a/pages/Math416/_meta.js
+++ b/pages/Math416/_meta.js
@@ -23,4 +23,5 @@ export default {
     Math416_L17: "Complex Variables (Lecture 17)",
     Math416_L18: "Complex Variables (Lecture 18)",
     Math416_L19: "Complex Variables (Lecture 19)",
+    Math416_L20: "Complex Variables (Lecture 20)",
 }