Linearity Properties of Lebesgue Measurable Functions

Linearity Properties of Lebesgue Measurable Functions

Recall from the Lebesgue Measurable Functions page that an extended real-valued function $f$ on a Lebesgue measurable domain $D(f)$ is a Lebesgue measurable function if for all $\alpha \in \mathbb{R}$ the set $\{ x \in D(f) : f(x) < \alpha \}$ is a Lebesgue measurable set.

We now show that the collection of Lebesgue measurable functions has somewhat of a linear structure to it. That is, if $f$ and $g$ are Lebesgue measurable functions with common domain then $f + g$ is a Lebesgue measurable function, and for any $c \in \mathbb{R}$ we have that $cf$ is a Lebesgue measurable function.

 Theorem 1 (Linearity of Lebesgue Measurable Functions): Let $f$ and $g$ be Lebesgue measurable functions on the same domain ($D(f) = D(g)$) and let $c \in \mathbb{R}$. Then: a) $f + g$ is a Lebesgue measurable function. b) $cf$ is a Lebesgue measurable function.
• Proof of a): Let $\alpha \in \mathbb{R}$. Consider the following set:
(1)
\begin{align} \quad \{ x \in D(f + g) : f(x) + g(x) < \alpha \} = \{ x \in D(f + g) : f(x) < \alpha - g(x) \} \end{align}
• Consider the inquality $f(x) < \alpha - g(x)$. By the density of the rational numbers there must exist a rational number $r_n \in \mathbb{Q}$ such that $f(x) < r_n < \alpha - g(x)$. Let $\{ r_n \}_{n=1}^{\infty} = \mathbb{Q}$ be an enumeration of the rational numbers. The set above can be rewritten as the following union
(2)
\begin{align} \quad \{ x \in D(f + g) : f(x) + g(x) < \alpha \} &= \bigcup_{n=1}^{\infty} \{ x \in D(f + g) : f(x) < r_n < \alpha - g(x) \}\\ &= \bigcup_{n=1}^{\infty} \left [ \{ x \in f(x) : f(x) < r_n \} \cap \{ x \in g(x) : r_n < \alpha - g(x) \} \right ] \\ &= \bigcup_{n=1}^{\infty} \left [ \underbrace{\{ x \in f(x) : f(x) < r_n \}}_{\mathrm{Lebesgue \: measurable}} \cap \underbrace{\{ x \in g(x) : g(x) < \alpha - r_n \}}_{\mathrm{Lebesgue \: measurable}} \right ] \end{align}
• The intersection of Lebesgue measurable sets is Lebesgue measurable and a countable union of Lebesgue measurable sets is Lebesgue measurable since the collection of Lebesgue measurable sets is a $\sigma$-algebra. Therefore $\{ x \in D(f + g) : f(x) + g(x) < \alpha \}$ is a Lebesgue measurable set. So $f + g$ is a Lebesgue measurable function. $\blacksquare$
• Proof of b): Let $\alpha \in \mathbb{R}$. There are three cases to consider.
• Case 1: If $c > 0$ then $\displaystyle{\{ x \in D(f) : cf(x) < \alpha \} = \left \{ x \in D(f) : f(x) < \frac{\alpha}{c} \right \}}$. So $\{ x \in D(f) : cf(x) < \alpha \}$ is a Lebesgue measurable set.
• Case 2: If $c < 0$ then $\displaystyle{\{ x \in D(f) : cf(x) < \alpha \} = \left \{ x \in D(f) : f(x) > \frac{\alpha}{c} \right \}}$. So $\{ x \in D(f) : cf(x) < \alpha \}$ is a Lebesgue measurable set.
• Case 3: If $c = 0$ then $\displaystyle{\{ x \in D(f) : cf(x) < \alpha \} = \{ x \in D(f) : 0 < \alpha \}}$. If $\alpha > 0$ then this set is $D(f)$ which is Lebesgue measurable. If $\alpha \leq 0$ then this set is $\emptyset$ which is Lebesgue measurable.
• In all three cases we see that $\{ x \in D(f) : cf(x) < \alpha \}$ is a Lebesgue measurable set. So $cf$ is a Lebesgue measurable function. $\blacksquare$