diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex
index 2153570..80bedbe 100644
--- a/inputs/lecture_10.tex
+++ b/inputs/lecture_10.tex
@@ -9,7 +9,7 @@ We consider $(\R, \cB(\R))$.
    By $M_1 (\R)$ we denote the set of all probability measures on $\left( \R, \cB(\R) \right)$.
 \end{notation}
 
-For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}\dif\bP(x)$.
+For all $\bP \in M_1(\R)$ we define $\phi_{\bP}(t) = \int_{\R} e^{\i t x}\bP(\dif x)$.
 If $X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable, we write
 $\phi_X(t) \coloneqq  \bE[e^{\i t X}] = \phi_{\mu}(t)$,
 where $\mu = \bP X^{-1}$.
@@ -20,22 +20,22 @@ where $\mu = \bP X^{-1}$.
     exists and is equal to the LHS.
     Note that the term on the RHS is integrable, as
     \[
-    \lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \pi(t) = a - b
+    \lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \phi(t) = a - b
     \]
     and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
     % TODO think about this
 
     We have
     \begin{IEEEeqnarray*}{rCl}
-        &&\lim_{T \to  \infty} \frac{1}{2 \pi} \int_{-T}^T \int_{\R} \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} dt \dif\bP(x)\\
-        &\overset{\text{Fubini for  $L^1$}}{=}& \lim_{T \to  \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} dt \dif\bP(x)\\
-        &=& \lim_{T \to  \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{\i t (b-x)}- e^{\i t (x-a)}}{-\i t} dt \dif\bP(x)\\
-        &=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \underbrace{\int_{-T}^T \left[ \frac{\cos(t (x-b)) - \cos(t(x-a))}{-\i t}\right] dt \dif\bP(x)}_{=0 \text{, as the function is odd}}
+        &&\lim_{T \to  \infty} \frac{1}{2 \pi} \int_{-T}^T \int_{\R} \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} \dif t \bP(\dif x)\\
+        &\overset{\text{Fubini for  $L^1$}}{=}& \lim_{T \to  \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{-\i t b}- e^{-\i t a}}{-\i t} e^{\i t x} \dif t \bP(\dif x)\\
+        &=& \lim_{T \to  \infty} \frac{1}{2 \pi} \int_{\R} \int_{-T}^T \frac{e^{\i t (b-x)}- e^{\i t (x-a)}}{-\i t} \dif t \bP(\dif x)\\
+        &=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \underbrace{\int_{-T}^T \left[ \frac{\cos(t (x-b)) - \cos(t(x-a))}{-\i t}\right] \dif t \bP(\dif x)}_{=0 \text{, as the function is odd}}
       \\&&
-        + \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} dt \dif\bP(x)\\
-        &=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} dt \dif\bP(x)\\
+        + \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} \dif t \bP(\dif x)\\
+        &=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} \dif t \bP(\dif x)\\
         &\overset{\substack{\text{\autoref{fact:intsinxx},}\\\text{dominated convergence}}}{=}& \frac{1}{\pi} \int -\frac{\pi}{2} \One_{x < a} + \frac{\pi}{2} \One_{x > a }
-                                                                - (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) \dif\bP(x)\\
+                                                                - (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) \bP(\dif x)\\
         &=&  \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
         &=& \frac{F(b) + F(b-)}{2} - \frac{F(a) - F(a-)}{2}
     \end{IEEEeqnarray*}
@@ -44,13 +44,13 @@ where $\mu = \bP X^{-1}$.
 \begin{fact}
     \label{fact:intsinxx}
     \[
-    \int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2}
+    \int_0^\infty \frac{\sin x}{x} \dif x = \frac{\pi}{2}
     \]
     where the LHS is an improper Riemann-integral.
     Note that the LHS is not Lebesgue-integrable.
     It follows that
     \begin{IEEEeqnarray*}{rCl}
-        \lim_{T \to  \infty} \int_0^T \frac{\sin(t(x-a))}{x} dt &=&
+        \lim_{T \to  \infty} \int_0^T \frac{\sin(t(x-a))}{x} \dif t &=&
     \begin{cases}
         - \frac{\pi}{2}, &x < a,\\
         0, &x = a,\\
@@ -64,7 +64,7 @@ where $\mu = \bP X^{-1}$.
     Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
     Then $\bP$ has a continuous probability density given by
     \[
-    f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) dt.
+    f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) \dif t.
     \]
 \end{theorem}
 
@@ -83,7 +83,7 @@ where $\mu = \bP X^{-1}$.
     \end{itemize}
 \end{example}
 \begin{refproof}{thm:lec10_3}
-    Let $f(x) \coloneqq  \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) dt$.
+    Let $f(x) \coloneqq  \frac{1}{2 \pi} \int_{\R} e^{ - \i t x} \phi(t) \dif t$.
     \begin{claim}
         If $x_n \to  x$, then $f(x_n) \to  f(x)$.
     \end{claim}
@@ -101,17 +101,17 @@ where $\mu = \bP X^{-1}$.
 
     We'll show that for all  $a < b$ we have
     \[
-    \bP\left( (a,b] \right)  = \int_a^b (x) dx.\label{thm10_3eq1}
+    \bP\left( (a,b] \right)  = \int_a^b (x) \dif x.\label{thm10_3eq1}
     \]
     Let $F$ be the distribution function of $\bP$.
     It is enough to prove \autoref{thm10_3eq1}
     for all continuity points $a $ and $ b$ of $F$.
     We have
     \begin{IEEEeqnarray*}{rCl}
-        RHS &\overset{\text{Fubini}}{=}& \frac{1}{2 \pi} \int_{\R} \int_{a}^b e^{-\i t x} \phi(t) dx dt\\
-            &=& \frac{1}{2 \pi} \int_\R \phi(t) \int_a^b e^{-\i t x} dx dt\\
-            &=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) dt\\
-            &\overset{\text{dominated convergence}}{=}& \lim_{T \to \infty} \frac{1}{2\pi} \int_{-T}^{T} \phi(t) \left( \frac{e^{-\i t b} - e^{- \i t a}}{- \i t} \right) dt
+        RHS &\overset{\text{Fubini}}{=}& \frac{1}{2 \pi} \int_{\R} \int_{a}^b e^{-\i t x} \phi(t) \dif x \dif t\\
+            &=& \frac{1}{2 \pi} \int_\R \phi(t) \int_a^b e^{-\i t x} \dif x \dif t\\
+            &=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) \dif t\\
+            &\overset{\text{dominated convergence}}{=}& \lim_{T \to \infty} \frac{1}{2\pi} \int_{-T}^{T} \phi(t) \left( \frac{e^{-\i t b} - e^{- \i t a}}{- \i t} \right) \dif t
     \end{IEEEeqnarray*}
     By \autoref{inversionformula}, the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
 \end{refproof}
@@ -122,17 +122,17 @@ However, Fourier analysis is not only useful for continuous probability density
     Let $\bP \in M_1(\lambda)$.
     Then
     \[
-    \forall x \in  \R ~ \bP\left( \{x\}  \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) dt.
+    \forall x \in  \R ~ \bP\left( \{x\}  \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) \dif t.
     \]
 
 \end{theorem}
 \begin{refproof}{bochnersformula}
     We have
     \begin{IEEEeqnarray*}{rCl}
-        RHS &=& \lim_{T \to  \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \dif\bP(y) \\
-            &\overset{\text{Fubini}}{=}& \lim_{T \to  \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} dt\\
-            &=& \lim_{T \to  \infty} \frac{1}{2T} \int_{\R} \dif\bP(y) \int_{-T}^T \cos(t(y - x)) dt\\
-            &=& \lim_{T \to  \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} \dif\bP(y)\\
+        RHS &=& \lim_{T \to  \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \bP(\dif y) \\
+            &\overset{\text{Fubini}}{=}& \lim_{T \to  \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} \dif t\\
+            &=& \lim_{T \to  \infty} \frac{1}{2T} \int_{\R} \bP(\dif y) \int_{-T}^T \cos(t(y - x)) \dif t\\
+            &=& \lim_{T \to  \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} \bP(\dif y)\\
     \end{IEEEeqnarray*}
     Furthermore
     \[
@@ -143,7 +143,7 @@ However, Fourier analysis is not only useful for continuous probability density
     \]
     Hence
     \begin{IEEEeqnarray*}{rCl}
-            \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} \dif\bP(y) &=& \bP\left( \{x\}\right)
+            \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} \bP(\dif y) &=& \bP\left( \{x\}\right)
     \end{IEEEeqnarray*}
     % TODO by dominated convergence?
 \end{refproof}
@@ -168,9 +168,9 @@ However, Fourier analysis is not only useful for continuous probability density
 
     For part (b) we have:
     \begin{IEEEeqnarray*}{rCl}
-        \sum_{j,k}  c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k}  c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} \dif\bP(x)\\
-                                                       &=& \int_{\R} \sum_{j,k}  c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} \dif\bP(x)\\
-                                                       &=& \int_{\R}\sum_{j,k}  c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} \dif\bP(x)\\
+        \sum_{j,k}  c_j \overline{c_k} \phi(t_j - t_k) &=& \sum_{j,k}  c_j \overline{c_k} \int_\R e^{\i (t_j - t_k) x} \bP(\dif x)\\
+                                                       &=& \int_{\R} \sum_{j,k}  c_j \overline{c_k} e^{\i t_j x} \overline{e^{\i t_k x}} \bP(\dif x)\\
+                                                       &=& \int_{\R}\sum_{j,k}  c_j e^{\i t_j x} \overline{c_k e^{\i t_k x}} \bP(\dif x)\\
                                                        &=& \int_{\R} \left| \sum_{l}  c_l e^{\i t_l x}\right|^2 \ge  0
     \end{IEEEeqnarray*}
 \end{refproof}