diff --git a/inputs/a_1_counterexamples.tex b/inputs/a_1_counterexamples.tex
index ebe6368..02e59ca 100644
--- a/inputs/a_1_counterexamples.tex
+++ b/inputs/a_1_counterexamples.tex
@@ -8,10 +8,21 @@ Exercise 4.3
 10.2
 
 
-Martingales converging a.s.~but not in $L^1$.
+\begin{example}[Martingale not converging in $L^1$]
+    Let $\Omega = [0,1]$, $\bP = \lambda\upharpoonright [0,1]$.
+    Define $X_n \coloneqq 2^n \cdot \One_{[0,2^n]}$,
+    and let $(\cF_n)_n$ be the canonical filtration.
+    Then $(X_n)_{n}$ is a Martingale
+    with $\bE[X_0] = 1$,
+    but $X_n \xrightarrow{a.s.} 0$.
+\end{example}
+
 
 Stopping times
 
-
+\begin{example}[{Martingale such that $\bE[X_T] \neq \bE[X_0]$}]
+    Consider the simple random walk and $T = \inf \{n : X_n \ge 1\}$.
+    Obviously $X_T = 1$.
+\end{example}
 
 
diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex
index 3db4926..4195844 100644
--- a/inputs/lecture_16.tex
+++ b/inputs/lecture_16.tex
@@ -193,7 +193,9 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
 \end{definition}
 \begin{corollary}
     Suppose that $f: \R \to  \R$ is a convex function such that $f(X_n) \in L^1(\bP)$.
-    Suppose that $(X_n)_n$ is a martingale\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}.
+    Suppose that $(X_n)_n$ is a martingale%
+    \footnote{In this form it means, that there is some filtration,
+    that we don't explicitly specify}.
     Then $(f(X_n))_n$ is a sub-martingale.
 \end{corollary}
 \begin{proof}
diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex
index 16dcb16..faf98bb 100644
--- a/inputs/lecture_18.tex
+++ b/inputs/lecture_18.tex
@@ -140,7 +140,7 @@ First, we need a very important inequality:
     Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
     denote the \vocab{running maximum}.
     \begin{enumerate}[(1)]
-        \item Then \[ \forall  \ell > 0 .~\bP[X_n^\ast \ge  \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ell\}} |X_n| \dif \bP \le  \frac{1}{\ell} \bE[|X_n|].  \]
+        \item Then \[ \forall  \ell > 0 .~\bP[X_n^\ast \ge  \ell] \le \frac{1}{\ell} \int_{\{X_n^\ast \ge \ell\}} |X_n| \dif \bP \le  \frac{1}{\ell} \bE[|X_n|].  \]
                 (Doob's $L^1$ inequality).
         \item Fix $p > 1$. Then \[
         \bE[(X_n^\ast)^p] \le  \left( \frac{p}{p-1} \right)^p \bE[|X_n|^p].