From 65388d60b231db12b9995c0dbdf0740dce27d88d Mon Sep 17 00:00:00 2001
From: Josia Pietsch <josia.pietsch@uni-bonn.de>
Date: Sat, 15 Jul 2023 15:08:17 +0200
Subject: [PATCH] fixed error

---
 Makefile              | 2 +-
 inputs/lecture_20.tex | 2 +-
 wtheo.sty             | 3 +++
 3 files changed, 5 insertions(+), 2 deletions(-)

diff --git a/Makefile b/Makefile
index f42a265..2b611ef 100644
--- a/Makefile
+++ b/Makefile
@@ -1,5 +1,5 @@
 pdf: init
-	latexmk < /dev/null
+	latexmk -halt-on-error < /dev/null
 
 clean:
 	latexmk -c
diff --git a/inputs/lecture_20.tex b/inputs/lecture_20.tex
index ade9eff..00827ad 100644
--- a/inputs/lecture_20.tex
+++ b/inputs/lecture_20.tex
@@ -208,7 +208,7 @@ we need the following theorem, which we won't prove here:
     then $X^T$ is also a supermartingale,
     and we have $\bE[X_{T \wedge n}] \le  \bE[X_0]$ for all $n$.
     If $(X_n)_n$ is a martingale, then so is $X^T$
-    and $\bE[X_{T \wedge n}] \le  \bE[X_0]$.
+    and $\bE[X_{T \wedge n}] = \bE[X_0]$.
 \end{theorem}
 \begin{proof}
     First, we need to show that $X^T$ is adapted.
diff --git a/wtheo.sty b/wtheo.sty
index 9ab9e63..852d865 100644
--- a/wtheo.sty
+++ b/wtheo.sty
@@ -24,6 +24,9 @@
 \usepackage{float}
 %\usepackage{algorithmicx}
 
+
+% \font\nullfont=cmr10
+
 \usepackage{pgfplots}
 \pgfplotsset{compat = newest}