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\begin{document}
\begin{enumerate}

\item Show 7 $\rightarrow$ 9

\begin{equation}
F_n \overbrace{\rightarrow}^{d} F \Rightarrow E[g_n(x_n)] = \int gdF_n \rightarrow \int gdF=E[g(x)]
\end{equation}

By \textbf{Hally-Bray Theorem}, consider some $(\Omega, \mathscr{A}, P)$.  Suppose that $g$ is bdd and is continuous a.s. $F$. $(g\in C_b)$, then

\begin{equation}
E[g_n(x_n)] = \int gdF_n \rightarrow \int gdF=E[g(x)]
\end{equation}

By converse, since $E[g_n(x_n)] \rightarrow E[g(x)]$ $\Rightarrow$ $F_n \rightarrow F$

\newpage

\item Show 9 $\rightarrow$ 12

Show that $E[g_n(x_n)] = \int gdF_n \rightarrow \int gdF=E[g(x)]$ $\forall$ $g\in C_b \Rightarrow \liminf P_n(B) \geq P(B)$ $\forall$ open sets $B$

Define a function, $g$, such that satisfies first part, then will show because of the way the function is setup, the function will imply the second part.

Let $B$ be an open set and $F=B^c$.  Consider a sequence of functions, $g_n \in C_b$, 
\begin{equation}
g_n(x) = \min{1,nd(x,F)}
\end{equation}

Show that $g_n(x) \uparrow 1(x)$
\begin{description}
\item[a.]  When $x\notin F \Rightarrow$ let $\epsilon=d(x,F)>0$, then $g_n(x) = \min(1,n\epsilon)$, where $n\epsilon \geq 1$.  For $n$ large, $\Rightarrow g_n(x) =1$
\item[b.]  When $x \in F$, then $d(x,F) =0 \Rightarrow g_n(x)=0$ 
\end{description}

Therefore, 
\begin{equation*}
g_n(x) = \left\{\begin{array}{rl} 0 & x \notin F=B^c \\ 1 & x \in F=B^c \end{array}\right.
\end{equation*}
as $n\rightarrow \infty$

By 9, 
$P_n(x) \geq \int g_k dF_n \rightarrow \int g_kdF$ as $n \rightarrow \infty$

and

$\liminf P_n(B) \geq \int g_kdF$

by letting $k \rightarrow \infty$ and MCT, 

\begin{eqnarray}
P_n(B) & \geq & \int g_kdF_n \overbrace{\rightarrow}^{9} \int g_k \uparrow P(B), n \rightarrow \infty \nonumber
\\
\liminf_n P_n(B) & \geq & \lim_{k\rightarrow \infty} \int g_kdF \overbrace{=}^{MCT} \int \lim_{k\rightarrow \infty} g_kdF = \int gdF = P(B) \nonumber
\\
\liminf_n P_n(B) & \geq & \int 1_BdF = P(B)
\end{eqnarray}

\newpage
\item Show 9 $\rightarrow$ 11

Show that $E[g_n(x_n)] = \int gdF_n \rightarrow \int gdF=E[g(x)]$ $\forall$ $g\in C_b \Rightarrow \limsup P_n(B) \leq P(B) \forall$ closed sets $B$

\begin{eqnarray}
P(B) &=& \int 1_BdF_n \leq \int g_kdF_n \overbrace{\rightarrow}^{9} \int g_k dF \downarrow P(B^c) \nonumber
\\
\limsup_n P_n(B) & \leq & \lim_{k\rightarrow \infty} \int g_k dF_n \overbrace{=}^{MCT} = \int \lim_{k\rightarrow \infty} g_kdF = \int gDF = P(B)
\end{eqnarray}

\newpage
\item Show 11 $\rightarrow$ 12

\begin{eqnarray}
\limsup P_n(B) & \leq & P(B)	\forall \hspace{.15cm} \textrm{closed sets} \hspace{.15cm} B  \nonumber
\\
1-\limsup P_n(B) & \geq & P(B^c)	\nonumber
\\
1+\liminf(-P_n(B)) & \geq & P(B^c)	\nonumber
\\
\liminf(1-P_n(B)) & \geq & P(B^c)	\nonumber
\\
\liminf(P_n(B^c)) & \geq & P(B^C)
\end{eqnarray}

\newpage
\item Show 12 $\rightarrow$ 11

\begin{eqnarray}
\liminf P_n(B) & \geq & P(B)	\forall \hspace{.15cm} \textrm{open sets} \hspace{.15cm} B \nonumber
\\
1-\liminf P_n(B) & \leq & P(B^c)	\nonumber
\\
1+\limsup(-P_n(B)) & \leq & P(B^c)	\nonumber
\\
\limsup(1-P_n(B)) & \leq & P(B^c)	\nonumber
\\
\limsup P_n(B^c) & \leq & P(B^c)
\end{eqnarray}

\newpage
\item Show 11 and 12 $\rightarrow$ 13

Define the following:

\begin{description}
\item[a.]  $B^0$ is the interior of $B$, where $B^0 =  \cup\{U \subset B, \hspace{.15cm} \textrm{s.t.} \hspace{.15cm} U \hspace{.15cm} \textrm{is open}\}$, (Union of all open subsets of $B$.
\item[b.]  $\bar{B}$ is the closure of $B$, where $\bar{B} =  \bigcap \{B \subset C, \hspace{.15cm} \textrm{s.t.} \hspace{.15cm} C \hspace{.15cm} \textrm{is closed}\}$, (Union of all open subsets of $B$.
\end{description}

\begin{equation*}
P(B)  \leq  \liminf P_n(B) \leq \limsup P_n(B) \leq P(B) 
\end{equation*}

$\forall$ $P-$continuity sets $B$, $(P(\delta B)=0)$, then the boundary $B$, $\delta B = \bar{B}/B^0$

Since $P(\delta B)=0$, then $P(\bar{B})=P(B^0) = P(B) \rightarrow \lim P_n(B) = P(B)$

\newpage
\item Show 13 $\rightarrow$ 9

For $\lim P_n(B) = P(B)$ $\forall$ $P-$ continuity sets $B$, , $(P(\delta B)=0)$, where $B \subset \Omega=\mathbb{R}$.

Let $B=(-\infty, x] \rightarrow \delta(-\infty,x] = \{x\}$.  Then $P(-\infty,x])=0$ $\forall$ continuity points $x \in \Omega$

\begin{equation*}
F_n \rightarrow F \Rightarrow F_n((-\infty,x]) \rightarrow F((-\infty,x])
\end{equation*}



\end{enumerate}
\end{document}