Maximums/Minimums of Finite Sets of Lebesgue Measurable Functions

# Maximums/Minimums of Finite Sets of Lebesgue Measurable Functions

Recall that if $f$ is an extended real-valued function with Lebesgue measurable domain $D(f)$, then $f$ is said to be Lebesgue measurable if for all $\alpha \in \mathbb{R}$ we have that the following set is a Lebesgue measurable set:

(1)
\begin{align} \quad \{ x \in D(f) : f(x) < \alpha \} \end{align}

On the Linearity Properties of Lebesgue Measurable Functions page we saw that if $f$ and $g$ are Lebesgue measurable functions on a common domain then for all $\alpha, \beta \in \mathbb{R}$ we have that the function $\alpha f + \beta g$ is also a Lebesgue measurable function.

On the Products of Lebesgue Measurable Functions page we saw that if $f$ and $g$ are Lebesgue measurable functions on a common domain then $f^2$ and $fg$ are also Lebesgue measurable functions.

We now prove that if $f$ and $g$ are Lebesgue measurable functions on a common domain then the maximum and minimum of these functions are also Lebesgue measurable functions.

 Theorem 1: Let $f$ and $g$ be Lebesgue measurable functions defined on a common Lebesgue measurable domain $D(f) = D(g)$. Then $\max \{ f, g \}$ and $\min \{ f, g \}$ are Lebesgue measurable functions.
• Proof: Since $f$ and $g$ are Lebesgue measurable functions on the Lebesgue measurable set $D(f) = D(g)$ then for all $\alpha \in \mathbb{R}$ we have that the following sets are Lebesgue measurable sets:
(2)
\begin{align} \quad \{ x \in D(f) : f(x) > \alpha \} \quad , \quad \{ x \in D(g) : g(x) > \alpha \} \\ \quad \{ x \in D(f) : f(x) < \alpha \} \quad , \quad \{ x \in D(g) : g(x) < \alpha \} \end{align}
• Let $\alpha \in \mathbb{R}$ and consider the following sets:
(3)
\begin{align} \quad \{ x \in D(f) = D(g) : \max \{ f(x), g(x) \} > \alpha \} = \{ x \in D(f) : f(x) > \alpha \} \cup \{ x \in D(g) : g(x) > \alpha \} \end{align}
(4)
\begin{align} \quad \{ x \in D(f) = D(g) : \min \{ f(x), g(x) \} < \alpha \} = \{ x \in D(f) : f(x) < \alpha \} \cup \{ x \in D(g) : g(x) < \alpha \} \end{align}
• Therefore $\{ x \in D(f) = D(g) : \max \{ f(x), g(x) \} > \alpha \}$ and $\{ x \in D(f) = D(g) : \min \{ f(x), g(x) \} < \alpha \}$ are Lebesgue measurable sets. Hence $\max \{ f, g \}$ and $\min \{ f, g \}$ are Lebesgue measurable functions.
 Corollary 2: Let $f_1$, $f_2$, …, $f_n$ be Lebesgue measurable functions defined on a common Lebesgue measurable domain. Then $\max \{ f_1, f_2, ..., f_n \}$ and $\min \{ f_1, f_2, ..., f_n \}$ are Lebesgue measurable functions.

Corollary 2 can easily be proven by mathematical induction.

 Corollary 3: Let $f$ be a Lebesgue measurable function defined on a Lebesgue measurable domain $D(f)$. Then $|f|$ is a Lebesgue measurable function.
• Proof: Define two new functions $f^+$ and $f^-$ on $D(f)$ by:
(5)
\begin{align} \quad f^+ = \max \{ f, 0 \} \quad , \quad f^- = \max \{ -f, 0 \} \end{align}
• From Theorem 1 we have that $f^+$ and $f^-$ are Lebesgue measurable functions. Furthermore it is easy to show that $|f| = f^+ + f^-$. So by the linearity property of Lebesgue measurable functions we have that $|f|$ is a Lebesgue measurable function. $\blacksquare$