\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

If $f$ and $g$ are measurable functions such that $\ds \int_E
f=\int_E g$ for all measurable sets $E$ then show that $f=g$ a.e.

${}$

(Hint.  Suppose $f=g$ a.e. is false and consider the sequence of
sets

$E_n=\{x|f(x)\geq g(x)+\frac{1}{n}\}$ or $F_n=\{x|g(x)\geq
f(x)+\frac{1}{n}\}$)


\vspace{0.25in}

{\bf Answer}

Let $A_1=\{x|f(x)>g(x)\}$ and $A_2=\{x|f(x)<g(x)\}$

$A-1\cup A_2=\{x|f(x)\not=g(x)\}$

Suppose $\ds m(A_1)>0, \,\,\, A_1=\bigcup_{n=1}^\infty E_n$

Therefore there exists $n, \,\,\, mE_n>0$

$\ds E_n=\{x|f(x)-g(x)\geq\frac{1}{n}\} \,\,\,\, f-g$ is
measurable therefore $E_n$ is measurable.

$\ds
\int_{E_n}f-\int_{E_n}g=\int_{E_n}f-g\geq\frac{1}{n}mE_n>0\,\,\,$
contradiction.

Similarly if $m(A_2)>0$ we get a contradiction.

\end{document}
