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:

\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:

\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:

\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:

\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. $\blacksquare$