\documentclass [12pt,letterpaper]{exam}
\usepackage{amsmath, amsthm, amsfonts, amssymb, amscd, latexsym}
\usepackage{type1cm}
\usepackage{simplemath}
\oddsidemargin 0.0in
\evensidemargin 0.0in
\textwidth 6.0in
\headheight 0.0in
\topmargin 1.0in
\textheight 8.5in
\header{ECSE-500-01-1}{Assignment 8 Answers}{19.11.2014}
%\begin{math}
\newcounter{count}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\printanswers
\begin{enumerate}
\item If E=$\phi$ then it is obvious that a $\notin$ E $\Rightarrow$ $\delta_a$ (E)=0. Now let $\ra E_n$ be a sequence
of subsets of X such that:
\\*n$\neq$m$\Rightarrow$ $E_n$$\cap$$E_m$=$\Phi$
This is not a trivial statement as it also implies that "a" could belong to at most one term of the sequence.
\\*Hence we have:
\begin{enumerate}
\item'a' belongs to a term $E_i$ of the sequence
\item'a' does not belong to any of the terms.
\end{enumerate}
For (a):
\\* We have a $\in$ ${\bigcup\limits_{n=1}^\infty}$ $E_n$=${\bigcup\limits_{n=1 \\* n\neq i}^\infty}$ $E_n$ $\cup$ $E_i$$\Rightarrow$ $\delta_a$(${\bigcup\limits_{n=1}^\infty}$$E_n$)=1
\\*But,
\\*${\sum\limits_{n=1}^\infty}$$\delta_a$ $(E_n)$=${\sum\limits_{n=1 n\neq i}^\infty}$$\delta_a$ $(E_n)$ + $\delta_a$ $(E_i)$=1 (Since $\delta_a$ $(E_n)$=$\forall$ n$\neq$ i)
\\*Hence,
\\*$\delta_a$(${\bigcup\limits_{n=1}^\infty}$$E_n)$= ${\sum\limits_{n=1}^\infty}$ $\delta_a$ $(E_n)$\\*\\*
For (b):\\*
a $\notin$ ${\bigcup\limits_{n=1}^\infty}$ $E_n$ $\Leftrightarrow$a $\notin$ $E_n$ $\forall$ n $\Rightarrow$${\bigcup\limits_{n=1}^\infty}$$\delta_a$($E_n$)
Thus from the above we conclude that $\delta_a$ is a measure on X (the Dirac Measure in a)
\item \begin{enumerate}
\item $\int_E$f d$\mu$ = $\int_E$f d$\delta_a$
\\*Assuming f is measurable, we know that if a $\notin$E then \\*
$\delta_a$(E)=0 and thus $\int_E$f d$\delta_a$=0 \\*
If a$\in$E, then we can define the set B=\{a\} and C=E-B \\*
We get, $\mu$(e)=$\delta_a$(e)=0. Since a$\notin$e.
Also, we have B$\subset$E, thus, assuming E$\in$m,
by corollary of theorm 11.2 (lecture 10) we have \\*
$\int_E$f d$\delta_a$ = $\int_a$f d$\delta_a$ = f(a) \\*
Thus, we have $\int_E$f d
$
\delta_a =
\begin{cases}
0 & \text{ if } a\in E\\
f(a) & \text{ if } a\notin E
\end{cases}
$\\*
\item Let E = $\{n_1 ... n_k\ \Leftrightarrow$E = ${\bigcup\limits_{n=1}^k}$ $\{n_k\}\\*
Then,\\*
\int_E f d\mu = {\int_{\bigcup\limits_{n=1}^k{n}} f d\mu_c}=
{\sum\limits_{n=1}^k} {\int_{\{n\}} f d\mu_c =
{\sum\limits_{n=1}^k} f(n_i) =
{\sum\limits_{x_i\in E}}f(n_i)
$\end{enumerate}
\end{enumerate}
%\end{math}
\end{document}
我在倒数第二个中遇到了错误\end{enumerate}
这是我第一次这样做，请帮帮我。
答案1
\*${\sum\limits_{n=1}^\infty}$$\delta_a$ $(E_n)$=${\sum\limits_{n=1 n\neq i}^\infty}$$\de
在这里，您将进入和退出每个术语的数学模式，这使得源代码无法跟踪（您错过了一个并且使得布局很差也就不足为奇了$。这是一个单独显示的表达式，应该这样设置。
我可能错过了一些情况，但这使用了稍微更合理的标记并且运行时没有错误。
\documentclass [12pt,letterpaper]{exam}
\usepackage{amsmath, amsthm, amsfonts, amssymb, amscd, latexsym}
\usepackage{type1cm}
%\usepackage{simplemath}
\newcommand\ra{arightarrow}
\oddsidemargin 0.0in
\evensidemargin 0.0in
\textwidth 6.0in
\headheight 0.0in
\topmargin 1.0in
\textheight 8.5in
\header{ECSE-500-01-1}{Assignment 8 Answers}{19.11.2014}
%\begin{math}
\newcounter{count}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\printanswers
\begin{enumerate}
\item If $E=\phi$ then it is obvious that
$a \notin E \Rightarrow \delta_a (E)=0$. Now let $\ra E_n$ be a sequence
of subsets of $X$ such that:
\[
\neq m\Rightarrow E_n \cap E_m=\Phi
\]
This is not a trivial statement as it also implies that $a$
could belong to at most one term of the sequence.
Hence we have:
\begin{enumerate}
\item $a$ belongs to a term $E_i$ of the sequence
\item $a$ does not belong to any of the terms.
\end{enumerate}
For (a):
We have
\[
a\in {\bigcup\limits_{n=1}^\infty} E_n=
{\bigcup\limits_{n=1 n\neq i}^\infty}
E_n \cup E_i \Rightarrow \delta_a({\bigcup\limits_{n=1}^\infty}E_n)=1
\]
But,
\[
{\sum\limits_{n=1}^\infty}\delta_a (E_n)={\sum\limits_{n=1 n\neq i}^\infty}\delta_a (E_n) + \delta_a (E_i)=1 (Since \delta_a (E_n)=\forall n\neq i)\]
Hence,
\[
\delta_a({\bigcup\limits_{n=1}^\infty}E_n)= {\sum\limits_{n=1}^\infty} \delta_a (E_n)\]
For (b):
\[
a \notin {\bigcup\limits_{n=1}^\infty} E_n \Leftrightarrow a \notin E_n \forall n \Rightarrow{\bigcup\limits_{n=1}^\infty}\delta_a(E_n)
\]
Thus from the above we conclude that $\delta_a$ is a measure on $X$ (the Dirac Measure in $a$)
\item \begin{enumerate}
\item
\[\int_Ef d\mu = \int_Ef d\delta_a\]
Assuming $f$ is measurable, we know that if $a \notin $ then
\[\delta_a(E)=0 and thus \int_Ef d\delta_a=0 \]
If $a\in E$, then we can define the set $B=\{a\}$ and $C=E-B$.
We get, $\mu(e)=\delta_a(e)=0$. Since $a\notin e$.
Also, we have $B\subset E$, thus, assuming $E\in m$,
by corollary of theorm 11.2 (lecture 10) we have
\[\int_E f d\delta_a = \int_af d\delta_a = f(a) \]
Thus, we have
\[\int_Ef d
\delta_a =
\begin{cases}
0 & \text{ if } a\in E\\
f(a) & \text{ if } a\notin E
\end{cases}\]
\item Let
\[E = \{n_1 ... n_k\ \Leftrightarrow E = {\bigcup\limits_{n=1}^k} \{n_k\}\]
Then,
\begin{align}
\int_E f d\mu &= {\int_{\bigcup\limits_{n=1}^k{n}} f d\mu_c}\\
&=
{\sum\limits_{n=1}^k} \int_{\{n\}} f d\mu_c\\
& =
{\sum\limits_{n=1}^k} f(n_i)\\
& =
{\sum\limits_{x_i\in E}}f(n_i)
\end{align}
\end{enumerate}
\end{enumerate}
%\end{math}
\end{document}
答案2
最后$缺少的是：
\item Let E = $\{n_1 \dots n_k\ \Leftrightarrow$E =
${\bigcup\limits_{n=1}^k}$ $\{n_k\}$\\*
%%%
^^^^^
这看起来不正确：
$\int_E f d\mu = {\int_{\bigcup\limits_{n=1}^k{n}} f d\mu_c}=
{\sum\limits_{n=1}^k} {\int_\{n\}} f d\mu_c =
{\sum\limits_{n=1}^k} f(n_i) =
{\sum\limits_{x_i\in E}}f(n_i)
$