General Measurable Functions

Recall that an extended real-valued function $f$ is said to be Lebesgue measurable on its domain (assumed to be a Lebesgue measurable set) if for every $\alpha \in \mathbb{R}$ the set $\{ x \in D(f) : f(x) < \alpha \}$ is a Lebesgue measurable set.

Given any measurable space $(X, \mathcal A)$, we can define general measurable functions in the same way.

Theorem 1: Let $(X, \mathcal A)$ be a measurable space and let $f$ be an extended real-valued function defined on a measurable set $E$. Then $f$ is said to be a Measurable Function on $E$ if for all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) < \alpha \}$ is a measurable set.

Similarly as to Lebesgue measurable functions, we can list equivalent statements for the measurability of a function.

Theorem 1: Let $(X, \mathcal A)$ be a measurable space and let $f$ be an extended real-valued function defined on a measurable set $E$. Then the following are equivalent: a) $f$ is measurable on $E$. b) For all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) < \alpha \}$ is measurable. c) For all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) \geq \alpha \}$ is measurable. d) For all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) > \alpha \}$ is measurable. e) For all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) \leq \alpha \}$ is measurable.

• Proof: Let $(X, \mathcal A)$ be a measurable space.

• $a) \Leftrightarrow b)$ This follows from the definition.

• $b) \Rightarrow c)$ Suppose that for all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) < \alpha \}$ is measurable. Let $\alpha \in \mathbb{R}$. Then:

(1)

\begin{align} \quad \{ x \in E : f(x) \geq \alpha \} = \{ x \in E : f(x) < \alpha \}^c \end{align}

• Since $\{ x \in E : f(x) < \alpha \} \in \mathcal A$ we have that $\{ x \in E : f(x) < \alpha \}^c \in \mathcal A$ (since $\mathcal A$ is a $\sigma$-algebra). This shows that $\{ x \in E : f(x) \geq \alpha \}$ is measurable.

• $c) \Rightarrow d)$ Suppose that for all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) \geq \alpha \}$ is measurable. Then:

(2)

\begin{align} \quad \{ x \in E : f(x) > \alpha \} = \bigcup_{n=1}^{\infty} \underbrace{\left \{ x \in E : f(x) \geq \alpha + \frac{1}{n} \right \}}_{\mathrm{measurable}} \end{align}

• Since $\{ x \in E : f(x) > \alpha \}$ is a countable union of measurable sets this implies that $\{ x \in E : f(x) > \alpha \}$ is measurable.

• $d) \Rightarrow e)$ Suppose that for all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) > \alpha \}$ is measurable. Then:

(3)

\begin{align} \quad \{ x \in E : f(x) \leq \alpha \} = \{ x \in E : f(x) > \alpha \}^c \end{align}

• So $\{ x \in E : f(x) \leq \alpha \}$ is measurable.

• $e) \Rightarrow a)$ Suppose that for all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) \leq \alpha$ is measurable. Then:

(4)

\begin{align} \quad \{ x \in E : f(x) < \alpha \} = \bigcup_{n=1}^{\infty} \underbrace{\left \{ x \in E : f(x) \leq \alpha + \frac{1}{n} \right \}}_{\mathrm{measurable}} \end{align}

• Since $\{ x \in E : f(x) < \alpha \}$ is a countable union of measurable sets this implies that $\{ x \in E : f(x) < \alpha \}$ is measurable. $\blacksquare$

Theorem 2: Let $(X, \mathcal A)$ be a measurable space and let $f$ be an extended real-valued function defined on a measurable set $E$. If $f$ is a measurable function then for all $\alpha \in \mathbb{R}$, the set $\{ x \in E : f(x) = \alpha \}$ is measurable.

The converse of Theorem 2 does not hold in general. That is, it is possible for all $\alpha \in \mathbb{R}$ the set $\{ x \in E : f(x) = \alpha \}$ is measurable and $f$ is not a measurable function.

• Proof: Since $f$ is a measurable function on $E$, for all $\alpha \in \mathbb{R}$ the following sets are measurable:

(5)

\begin{align} \quad \{ x \in E : f(x) \leq \alpha \} \quad \mathrm{and} \quad \{ x \in E : f(x) \geq \alpha \} \end{align}

• The intersection of two measurable sets is measurable, and:

(6)

\begin{align} \quad \{ x \in E : f(x) = \alpha \} = \{ x \in E : f(x) \leq \alpha \} \cap \{ x \in E : f(x) \geq \alpha \} \end{align}

• So $\{ x \in E : f(x) = \alpha \}$ is measurable. $\blacksquare$