Products of Lebesgue Measurable Functions

# Products of Lebesgue Measurable Functions

Recall from Linearity Properties of Lebesgue Measurable Functions page that if $f$ and $g$ are Lebesgue measurable functions with common domain then $f + g$ is a Lebesgue measurable function. Furthermore, if $c \in \mathbb{R}$ then $cf$ is also a Lebesgue measurable function.

We now show that the product of two Lebesgue measurable functions with common domain is also a Lebesgue measurable function.

Lemma 1: 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 2: 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:

\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}

- From the previous lemma and from the fact that the sum / multiples of Lebesgue measurable functions we see that $fg$ is a Lebesgue measurable function.