# Measure theory

• Apr 7th 2010, 07:48 AM
Chandru1
Measure theory
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.
• Apr 7th 2010, 10:55 AM
southprkfan1
Quote:

Originally Posted by Chandru1
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