Lebesgue Measurable Functions and Open Sets

# Lebesgue Measurable Functions and Open Sets

 Theorem 1: Let $f$ be defined on a Lebesgue measurable set $E$. Then $f$ is a Lebesgue measurable function if and only if for every open set $O \subseteq \mathbb{R}$ we have that $f^{-1}(O)$ is a Lebesgue measurable set.
• Proof: $\Rightarrow$ Suppose that $f$ is Lebesgue measurable and let $O \subseteq \mathbb{R}$ be an open set. Every open set in $\mathbb{R}$ is a countable union of bounded open intervals, say:
(1)
\begin{align} \quad O = \bigcup_{n=1}^{\infty} I_n \end{align}
• For each open interval $I_n = (a_n, b_n)$ let $A_n = (a_n, \infty)$ and let $B_n = (-\infty, b_n)$. Then:
(2)
\begin{align} \quad I_n = A_n \cap B_n \end{align}
• Since $f$ is a Lebesgue measurable function, for each $a_n, b_n \in \mathbb{R}$ we have that the following sets are Lebesgue measurable:
(3)
\begin{align} \quad \{ x \in E : f(x) > a_n \} \quad \mathrm{and} \quad \{ x \in E : f(x) < b_n \} \end{align}
• But these sets are simply $f^{-1}(A_n)$ and $f^{-1}(B_n)$ respectively. Hence:
(4)
\begin{align} \quad f^{-1}(O) &= f^{-1} \left ( \bigcup_{n=1}^{\infty} I_n \right ) \\ &= f^{-1} \left ( \bigcup_{n=1}^{\infty} [A_n \cap B_n] \right ) \\ &= \bigcup_{n=1}^{\infty} f^{-1} (A_n \cap B_n) \\ &= \bigcup_{n=1}^{\infty} [f^{-1}(A_n) \cap f^{-1}(B_n)] \end{align}
• Since $f^{-1}(A_n)$ and $f^{-1} (B_n)$ are Lebesgue measurable sets, so is their intersection $f^{-1}(A_n) \cap f^{-1} (B_n)$. And a countable union of Lebesgue measurable sets is Lebesgue measurable, so $f^{-1} (O)$ is a Lebesgue measurable set.
• $\Leftarrow$. Suppose that for every open set $O \subseteq \mathbb{R}$ we have that $f^{-1}(O)$ is open. In particular, for each $\alpha \in \mathbb{R}$ the open interval $(-\infty, \alpha)$ is an open set, and so sets of the form below are Lebesgue measurable by assumption:
(5)
\begin{align} \quad f^{-1}((\alpha, \infty)) = \{ x \in E : f(x) < \alpha \} \end{align}
• Therefore $f$ is a Lebesgue measurable function.