Let f be an extended real valued function defined on a measurable set E of R. Show that f is a measurable function on E iff for each rational number r the set $\displaystyle E(f<r)= \{ x \in E \ : \ f(x)<r\}$ is a measurable subset of E.
-->Suppose f is measurable
then $\displaystyle f^{-1} $ takes opens sets to measurable sets
Fix a rational number r
But $\displaystyle E(f<r)= \{ x \in E \ : \ f(x)<r\}$ = $\displaystyle f^{-1}$($\displaystyle -\infty$, r), which is measurable
<--Suppose for each rational number r the set $\displaystyle E(f<r)= \{ x \in E \ : \ f(x)<r\}$ is a measurable subset of E.
Fix b in $\displaystyle R $
We want to show $\displaystyle f^{-1}$($\displaystyle -\infty$, b) is measurable.
This is clearly true if b is rational (by assumption)
If b is irrational, then we can find a sequence of rationals, {$\displaystyle q_n $} that are increasing (or decreasing) and converge to b.
Let $\displaystyle S_k $ = E(f < $\displaystyle q_k $)
Let S = E(f < b)
Clearly, $\displaystyle S_1 \subset S_2 \subset ... $ and $\displaystyle S_k $ --> S, and since each $\displaystyle S_k $ is measurable then so is S