fix bug container may not eixsts.

Jenkinsfile

@@ -31,9 +31,9 @@ pipeline {
             steps {
                 echo "Deploying docker image ${registry}:v${version}.${env.BUILD_ID}"
                 echo "Stopping existing container"
-                sh 'docker stop notenextra'
+                sh 'docker stop notenextra || true'
                 echo "Removing existing container"
-                sh 'docker rm notenextra'
+                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}'
             }

pages/Math4121/Math4121_L35.md

@@ -48,18 +48,25 @@ Let $F=\{x\in E: f(x)>g(x)\}$. Then $m(F)=0$.
 $$
 \begin{aligned}
 \int_E f \, dm &= \int_{E\setminus F} f \, dm + \int_F f \, dm\\
-&\leq \int_{E\setminus F} g \, dm
+&\leq \int_{E\setminus F} g \, dm+\int_E g \, dm\\
+&= \int_E g \, dm
 \end{aligned}
 $$
 
-
-
-
-
-
-
 QED
 
+#### Proposition 6.13
 
+If $f$ is non-negative and $\int_E f \, dm =0$, then $f=0$ almost everywhere on $E$, $f(x)=0$ $\forall x\in E\setminus F$, where $m(F)=0$.
 
+Proof:
 
+Let $E_n=\{x\in E: f(x)\geq \frac{1}{n}\}$. Then $\frac{1}{n}\chi_{E_n}(x)\leq f(x)$ for all $x\in E$.
+
+By definition $\frac{1}{n}m(E_n)=\int_E \frac{1}{n}\chi_{E_n} \, dm \leq \int_E f \, dm =0$.
+
+Therefore, $m(E_n)=0$ for all $n$.
+
+Now $U=\{x\in E: f(x)>0\}=\bigcup_{n=1}^{\infty} E_n$.
+
+QED