Properties of Lebesgue Measurable Functions 2

# Properties of Lebesgue Measurable Functions 2

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 properties of Lebesgue measurable functions. More properties can be found on the Properties of Lebesgue Measurable Functions 1 page.

 Theorem 3: Let $f$ be a Lebesgue measurable function and let $c \in \mathbb{R}$. Then $f^2$ is a Lebesgue measurable function.
• Proof: Let $\alpha \in \mathbb{R}$. There are two cases to consider:
• Case 1: If $\alpha \geq 0$ then $\{ x \in D(f^2) : [f(x)]^2 < \alpha \} = \{ x \in D(f) : f(x) < \sqrt{\alpha} \}$ which is a Lebesgue measurable set.
• Case 2: If $\alpha < 0$ then $\{ x \in D(f^2) : [f(x)]^2 < \alpha \} = \emptyset$ which is a Lebesgue measurable set.
• In all two cases we see that $\{ x \in D(f^2) : [f(x)]^2 < \alpha \}$ is a Lebesgue measurable set. So $f^2$ is a Lebesgue measurable function. $\blacksquare$
 Theorem 4: Let $f$ and $g$ be Lebesgue measurable functions on the same domain ($D(f) = D(g)$). Then $fg$ is a Lebesgue measurable function.
• We use the following identity:
(1)
\begin{align} \quad f(x)g(x) = \frac{1}{4} \left [ \left ( f(x) + g(x) \right )^2 - \left ( f(x) - g(x) \right )^2 \right ] \end{align}
• Since $f$ and $g$ are Lebesgue measurable functions, the sum and difference $f + g$ and $f - g$ are Lebesgue measurable from Theorem 1. The squares $(f + g)^2$ and $(f - g)^2$ are Lebesgue measurable from Theorem 3. Then difference $(f + g)^2 - (f - g)^2$ is Lebesgue measurable from Theorem 1. And $\displaystyle{\frac{1}{4} [ (f + g)^2 - (f - g)^2 ]}$ is Lebesgue measurable from Theorem 2. So $fg$ is a Lebesgue measurable function. $\blacksquare$