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 $\{ x \in D(f) : f(x) < \alpha \}^c$ is 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 $\{ x \in D(f) : f(x) > \alpha \}^c$ is 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 each $n \in \mathbb{N}$ 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: Let $f$ 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. $\blacksquare$