Consider the space $(\Omega,\mathcal{F},P)$. Show that $E(c)=c,\forall c\in \mathbb{R}$.
I thought about two distinct ways to show that.
$E(c)=\int_\Omega c\ dP=c\int_\Omega dP=cP(\Omega)=c$;
Let $f=c$ a.e.. Then $f=c.\chi_1 (\{f=c\})+b.\chi_2(\{f\neq c\})$ as a simple function. Then $E(f)=\int_\Omega f dP=c .P(\{f=c\})$. Clearly, $P(\{f=c\})=1$, otherwise the sum of this probability with the probability of its complement wouldn't be 1.
Are both acceptable?
Thanks!
\cdot
in the $\TeX$ markup, as in $c\cdot \Pr(f=c).$ $\endgroup$