# 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:

- 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:

- 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:

- 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:

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:

- Hence $\{ x \in D(f) : f(x) = \alpha \}$ is Lebesgue measurable.

- Furthermore we have that:

- The intersection of a countable collection of Lebesgue measurable sets is Lebesgue measurable, so $\{ x \in D(f) : f(x) = \infty \}$ is Lebesgue measurable. $\blacksquare$