4
\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}

I'm getting the error in the second to last \end{enumerate}

Im doing this for the first time, kindly help, please.

2
\*${\sum\limits_{n=1}^\infty}$$\delta_a$ $(E_n)$=${\sum\limits_{n=1 n\neq i}^\infty}$$\de

here you are going into and out of math mode for each term which makes the source impossible to follow (it is not surprising that you missed a $ but also makes a very poor layout. this is a single displayed expression and should be set as such.

I probably missed some cases, but this uses slightly more reasonable markup and runs without error.

\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}
1

The last $ is missing:

    \item Let E = $\{n_1 \dots n_k\ \Leftrightarrow$E = 
           ${\bigcup\limits_{n=1}^k}$ $\{n_k\}$\\*
 %%%
                                            ^^^^^

and this one looks not correct:

      $\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)

      $

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