From c7901188d41f3bd1c34dab5e5139cd8677d374d1 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Wed, 16 Apr 2025 10:31:20 -0500
Subject: [PATCH] fix jenkinfile errors

---
 Jenkinsfile                    |  2 --
 pages/Math4121/Math4121_L35.md | 32 ++++++++++++++++++++++++++++++++
 2 files changed, 32 insertions(+), 2 deletions(-)

diff --git a/Jenkinsfile b/Jenkinsfile
index 04f7cfe..715851b 100644
--- a/Jenkinsfile
+++ b/Jenkinsfile
@@ -40,5 +40,3 @@ pipeline {
         }
     }
 }
-
-}
diff --git a/pages/Math4121/Math4121_L35.md b/pages/Math4121/Math4121_L35.md
index 08a6fd6..5ba91e9 100644
--- a/pages/Math4121/Math4121_L35.md
+++ b/pages/Math4121/Math4121_L35.md
@@ -31,3 +31,35 @@ $$
 
 We allow for $A-\infty = -\infty$ and $A+\infty = \infty$ for any $A\in \mathbb{R}$. But not $\infty-\infty$.
 
+#### Immediate Properties of Lebesgue Integral
+
+If $f$ is measurable and $m(E)=0$, then $\int_E f \, dm = 0$.
+
+If $E=E_1\cup E_2$ and $E_1\cap E_2=\emptyset$, then $\int_E f \, dm = \int_{E_1} f \, dm + \int_{E_2} f \, dm$.
+
+#### Corollary
+
+If $f\leq g$ almost everywhere, ($f\leq g$ except for a set of measure 0), then $\int_E f \, dm \leq \int_E g \, dm$.
+
+Proof:
+
+Let $F=\{x\in E: f(x)>g(x)\}$. Then $m(F)=0$.
+
+$$
+\begin{aligned}
+\int_E f \, dm &= \int_{E\setminus F} f \, dm + \int_F f \, dm\\
+&\leq \int_{E\setminus F} g \, dm
+\end{aligned}
+$$
+
+
+
+
+
+
+
+QED
+
+
+
+