Basic Theorems Regarding Lebesgue Measurable Functions

Basic Theorems Regarding Lebesgue Measurable Functions

Recall from the Lebesgue Measurable Functions page that an extended real-valued function $f$ with Lebesgue measurable domain $D(f)$ 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.

We will now look at some basic results regarding Lebesgue measurable functions.

 Theorem 1: Let $f$ be an extended real-valued function. If $m(D(f)) = 0$ then $f$ is a Lebesgue measurable function.
• Proof: Let $\alpha \in \mathbb{R}$. Then the set $\{ x \in D(f) : f(x) < \alpha \} \subset D(f)$. Therefore:
(1)
\begin{align} \quad m(\{ x \in D(f) : f(x) < \alpha \}) \leq m(D(f)) = 0 \end{align}
• Hence $m(\{ x \in D(f) : f(x) < \alpha \}) = 0$. But any set that has Lebesgue measure $0$ is Lebesgue measurable, so $\{ x \in D(f) : f(x) < \alpha \}$ is a Lebesgue measurable set. Thus $f$ is a Lebesgue measurable function. $\blacksquare$
 Theorem 2: Let $f$ be an extended real-valued function with Lebesgue measurable domain $D(f)$. If $f$ is a Lebesgue measurable function and $E$ is a Lebesgue measurable set such that $E \subseteq D(f)$ then the function $f$ restricted to $E$, $f \lvert_E$ is a Lebesgue measurable function.
• Proof: Let $\alpha \in \mathbb{R}$. We note that the intersection of Lebesgue measurable sets is a Lebesgue measurable set (since the collection of Lebesgue measurable sets is a $\sigma$-algebra) ]] and: (2) \begin{align} \quad \{ x \in E : f_E(x) < \alpha \} = \underbrace{\{ x \in D(f) : f(x) < \alpha \}}_{\mathrm{Lebesgue \: measurable}} \cap \underbrace{E}_{\mathrm{Lebesgue \: measurable}} \end{align} • Therefore\{ x \in E : f_E(x) < \alpha \}$is a Lebesgue measurable set. So$f \lvert_E$is a Lebesgue measurable function.$\blacksquare$ Theorem 3: Let$f$and$g$be extended real-valued functions with Lebesgue measurable domains$D(f)$and$D(g)$respectively with$D(f) \cap D(g) = \emptyset$. Let$h$be the extended real-valued function defined on$D(f) \cup D(g)$by$\displaystyle{h(x) = \left\{\begin{matrix} f(x) & \mathrm{if} \: x \in D(f) \\ g(x) & \mathrm{if} \: x \in D(g) \end{matrix}\right.}$. Then$h$is a Lebesgue measurable function. • Proof: Let$\alpha \in \mathbb{R}. Note that the union of two Lebesgue measurable sets is Lebesgue measurable and: (3) \begin{align} \quad \{ x \in D(f) \cup D(g) : h(x) < \alpha \} = \underbrace{\{ x \in D(f) : f(x) < \alpha \}}_{\mathrm{Lebesgue \: measurable}} \cup \underbrace{\{ x \in D(g) : g(x) < \alpha \}}_{\mathrm{Lebesgue \: measurable}} \end{align} • Hence\{ x \in D(f) \cup D(g) : h(x) < \alpha \}$is a Lebesgue measurable set. Thus$h$is a Lebesgue measurable function.$\blacksquare$ Theorem 4: Let$f$be an extended real-valued function with Lebesgue measurable domain$D(f)$. If$f$is continuous then$f$is a Lebesgue measurable function. • Proof: Let$\alpha \in \mathbb{R}$. Since$f$is a continuous function, the inverse image of any open subset of$\mathbb{R}$is open. In particular,$(-\infty, \alpha)$is an open interval and so$f^{-1} (-\infty, \alpha)is open. However: (4) \begin{align} \quad f^{-1} (-\infty, \alpha) = \{ x \in D(f) : f(x) < \alpha \} \end{align} • Every open set is a Lebesgue measurable set so\{ x \in D(f) : f(x) < \alpha \}$is a Lebesgue measurable set. Hence$f$is a Lebesgue measurable function.$\blacksquare\$