fix bad substitution

Jenkinsfile

@@ -29,13 +29,16 @@ pipeline {
         }
         stage('Deploy') {
             steps {
-                echo "Deploying docker image ${registry}:v${version}.${env.BUILD_ID}"
-                echo "Stopping existing container"
-                sh 'docker stop notenextra || true'
-                echo "Removing existing container"
-                sh 'docker rm notenextra || true'
-                echo "Running new docker container"
-                sh 'docker run -d -p 13000:3000 --name notenextra ${registry}:v${version}.${env.BUILD_ID}'
+                script {
+                    def imageTag = "${registry}:v${version}.${env.BUILD_ID}"
+                    echo "Deploying docker image ${imageTag}"
+                    echo "Stopping existing container"
+                    sh 'docker stop notenextra-jenkins || true'
+                    echo "Removing existing container"
+                    sh 'docker rm notenextra-jenkins || true'
+                    echo "Running new docker container"
+                    sh "docker run -d -p 13000:3000 --name notenextra-jenkins ${imageTag}"
+                }
             }
         }
     }

pages/Math4121/Math4121_L35.md

@@ -67,6 +67,27 @@ By definition $\frac{1}{n}m(E_n)=\int_E \frac{1}{n}\chi_{E_n} \, dm \leq \int_E
 
 Therefore, $m(E_n)=0$ for all $n$.
 
-Now $U=\{x\in E: f(x)>0\}=\bigcup_{n=1}^{\infty} E_n$.
+Now $U=\{x\in E: f(x)>0\}=\bigcup_{n=1}^{\infty} E_n$, and $E_n\subseteq E_{n+1}$ for all $n$.
+
+Therefore, $m(U)=m(\bigcup_{n=1}^{\infty} E_n)=\lim_{n\to\infty} m(E_n)=0$.
 
 QED
+
+### Convergence Theorems
+
+When does $\lim_{n\to\infty} \int_E f_n \, dm = \int_E \lim_{n\to\infty} f_n \, dm$?
+
+#### Theorem 6.14 Monotone Convergence Theorem
+
+Let $\{f_n\}$ be a monotone increasing sequence of measurable functions on $E$ and $f_n\to f$ almost everywhere on $E$. ($f_n(x)\leq f_{n+1}(x)$ for all $x\in E$ and $n$)
+
+If there exists $A>0$ such that $\left|\int_E f_n \, dm\right|\leq A$ for all $n\in \mathbb{N}$, then $f(x)=\lim_{n\to\infty} f_n(x)$ exists for almost every $x\in E$ and it is integrable on $E$ and
+
+$$
+\int_E f \, dm = \lim_{n\to\infty} \int_E f_n \, dm
+$$
+
+Proof:
+
+QED
+