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header{ECSE-500-01-1}{Assignment 8 Answers}{19.11.2014}
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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 $in$ ${bigcuplimits_{n=1}^infty}$ $E_n$=${bigcuplimits_{n=1 \* nneq i}^infty}$ $E_n$ $cup$ $E_i$$Rightarrow$ $delta_a$(${bigcuplimits_{n=1}^infty}$$E_n$)=1
        \*But,
        \*${sumlimits_{n=1}^infty}$$delta_a$ $(E_n)$=${sumlimits_{n=1 nneq i}^infty}$$delta_a$ $(E_n)$ + $delta_a$ $(E_i)$=1 (Since $delta_a$ $(E_n)$=$forall$ n$neq$ i)
        \*Hence,
        \*$delta_a$(${bigcuplimits_{n=1}^infty}$$E_n)$= ${sumlimits_{n=1}^infty}$ $delta_a$ $(E_n)$\*\*
        For (b):\*
        a $notin$  ${bigcuplimits_{n=1}^infty}$ $E_n$ $Leftrightarrow$a $notin$ $E_n$ $forall$ n $Rightarrow$${bigcuplimits_{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 } ain E\
          f(a) & text{ if } anotin E
          end{cases}
          $\*
          item Let E = ${n_1 ... n_k Leftrightarrow$E = ${bigcuplimits_{n=1}^k}$ ${n_k}\*
          Then,\*
          int_E f dmu = {int_{bigcuplimits_{n=1}^k{n}} f dmu_c}=
          {sumlimits_{n=1}^k} {int_{{n}}  f dmu_c =
          {sumlimits_{n=1}^k} f(n_i) =
          {sumlimits_{x_iin E}}f(n_i)
          $end{enumerate}
end{enumerate}
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