Lebesgue Measurable Functions
 Definition: Let $f$ be an extended real-valued function and let $D(f)$ be a Lebesgue measurable set. $f$ is said to be a Lebesgue Measurable Function if for all $\alpha \in \mathbb{R}$ the set $\{ x \in D(f) : f(x) < \alpha \}$ is Lebesgue measurable.
In the definition above, there are two requirements for an extended real-valued function $f$ to be Lebesgue measurable. We must have that $D(f)$ is a Lebesgue measurable set and that for all $\alpha \in \mathbb{R}$\}$the sets$\{ x \in D(f) : f(x) < \alpha \}$are Lebesgue measurable sets. The following theorem gives us different criteria for an extended real-valued function$f$with Lebsegue measurable domain$D(f)$to be a Lebesgue measurable function.  Theorem 1: Let$f$be an extended real-valued function and let$D(f)$be a Lebesgue measurable set. The following statements are equivalent: a) For all$\alpha \in \mathbb{R}$the set$\{ x \in D(f) : f(x) < \alpha \}$is Lebesgue measurable. b) For all$\alpha \in \mathbb{R}$the set$\{ x \in D(f) : f(x) \geq \alpha \}$is Lebesgue measurable. c) For all$\alpha \in \mathbb{R}$the set$\{ x \in D(f) : f(x) > \alpha \}$is Lebesgue measurable. d) For all$\alpha \in \mathbb{R}$then set$\{ x \in D(f) : f(x) \leq \alpha \}$is Lebesgue measurable. By definition, (a) simply states that$f$is a Lebesgue measurable function. We use the fact that the collection of Lebesgue measurable sets is a$\sigma$-algebra in the following proof. • Proof: Let$\alpha \in \mathbb{R}$. •$a) \Leftrightarrow b)$Suppose that (a) holds. Then$\{ x \in D(f) : f(x) < \alpha \}^cis a Lebesgue measurable set and: (1) \begin{align} \quad \{ x \in D(f) : f(x) < \alpha \}^c = \{ x \in D(f) : f(x) \geq \alpha \} \end{align} • So\{ x \in D(f) : f(x) \geq \alpha \}$is Lebesgue measurable. The converse is clear.$\blacksquare$. • **$c) \Leftrightarrow d)$Suppose that (c) holds. Then$\{ x \in D(f) : f(x) > \alpha \}^cis a Lebesgue measurable set and: (2) \begin{align} \quad \{ x \in D(f) : f(x) > \alpha \}^c = \{ x \in D(f) : f(x) \leq \alpha \} \end{align} • So\{ x \in D(f) : f(x) \leq \alpha \}$is Lebesgue measurable. The converse is clear.$\blacksquare$•$a) \Leftrightarrow d)$Suppose that (a) holds. Then for each$n \in \mathbb{N}$the set$\displaystyle{\left \{ x \in D(f) : f(x) < \alpha + \frac{1}{n} \right \}}is Lebesgue measurable and: (3) \begin{align} \quad \{ x \in D(f) : f(x) \leq \alpha \} = \underbrace{\bigcap_{n=1}^{\infty} \left \{ x \in D(f) : f(x) < \alpha + \frac{1}{n} \right \}}_{\mathrm{Lebesgue \: measurable}} \end{align} • Conversely, suppose that (d) holds. Then for eachn \in \mathbb{N}$then set$\displaystyle{\left \{ x \in D(f) : f(x) \leq \alpha - \frac{1}{n} \right \}}is Lebesgue measurable and: (4) \begin{align} \quad \{ x \in D(f) : f(x) < \alpha \} = \underbrace{\bigcup_{n=1}^{\infty} \left \{ x \in D(f) : f(x) \leq \alpha - \frac{1}{n} \right \}}_{\mathrm{Lebesgue \: measurable}} \quad \blacksquare \end{align}  Theorem 2: Letf$be an extended real-valued function and let$D(f)$be a Lebesgue measurable set. If$f$is a Lebesgue measurable function then for all$\alpha \in \mathbb{R}$the set$\{ x \in D(f) : f(x) = \alpha \}$is a Lebesgue measurable set. Furthermore, the set$\{ x \in D(f) : f(x) = \infty \}$is a Lebesgue measurable set. The converse of Theorem 2 above is not true in general. That is, if$f$is an extended real-valued function defined on a Lebesgue measurable set$D(f)$then if$\{ x \in D(f) : f(x) = \alpha \}$is Lebesgue measurable for every$\alpha \in \mathbb{R}$it may still be that$f$is not a Lebesgue measurable function. • Proof: Let$f$be an extended Lebesgue measurable function and let. • From the previous theorem we have that for all$\alpha \in \mathbb{R}$that$\{ x \in D(f) : f(x) \leq \alpha \}$and$\{ x \in D(f) : f(x) \geq \alpha \}$are Lebesgue measurable sets. The intersection of two Lebesgue measurable sets is Lebesgue measurable (since the set of Lebesgue measurable sets is a$\sigma-algebra) so: (5) \begin{align} \quad \{ x \in D(f) : f(x) \leq \alpha \} \cap \{ x \in D(f) : f(x) \geq \alpha \} = \{ x \in D(f) : f(x) = \alpha \} \end{align} • Hence\{ x \in D(f) : f(x) = \alpha \}is Lebesgue measurable. • Furthermore we have that: (6) \begin{align} \quad \{ x \in D(f) : f(x) = \infty \} = \bigcap_{n=1}^{\infty} \{ x \in D(f) : f(x) > n \} \end{align} • The intersection of a countable collection of Lebesgue measurable sets is Lebesgue measurable, so\{ x \in D(f) : f(x) = \infty \}$is Lebesgue measurable.$\blacksquare\$