Products of Square Leb. Integrable Functions are Leb. Integrable

# Products of Square Lebesgue Integrable Functions are Lebesgue Integrable

Recall from the Square Lebesgue Integrable Functions page that a function $f$ is said to be square Lebesgue integrable on an interval $I$ if $f$ is measurable on $I$ and $f^2$ is Lebesgue integrable on $I$.

We noted that a function $f$ may be square Lebesgue integrable but need not be Lebesgue integrable itself.

We will now look at a nice result which says that if we have two square Lebesgue integrable functions on an interval $I$ then their product is a Lebesgue integrable function on $I$.

Theorem 1: Let $f$ and $g$ be square Lebesgue integrable functions on a interval $I$. Then $fg$ is Lebesgue integrable on $I$. |

**Proof:**Let $f$ and $g$ be square Lebesgue integrable functions on $I$. Since $f, g \in L^2(I)$ we have that $f, g \in M(I)$. So there exists sequences of step functions $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ such that

\begin{align} \quad \lim_{n \to \infty} f_n(x) = f(x) \quad \mathrm{and} \quad \lim_{n \to \infty} g_n(x) = g(x) \end{align}

- Consider the product sequence of functions $(f_ng_n(x))_{n=1}^{\infty}$ which is also a sequence of step functions. Then this sequence of step functions converges to $fg$ everywhere except at the points in the union of the measure zero sets for which $(f_n(x))_{n=1}^{\infty}$ or $(g_n(x))_{n=1}^{\infty}$ does not converge to $f$ or $g$, i.e., almost everywhere on $I$. So $fg \in M(I)$.

- Now consider the inequality $(f(x) - g(x))^2 \geq 0$. Expanding this yields:

\begin{align} \quad [f(x)]^2 - 2f(x)g(x) + [g(x)]^2 & \geq 0 \\ \quad \frac{1}{2} \left ( [f(x)]^2 + [g(x)]^2 \right ) & \geq f(x)g(x) \\ \quad \frac{1}{2} \left ( [f(x)]^2 + [g(x)]^2 \right ) & \geq \mid f(x)g(x) \mid \\ \end{align}

- Since $f, g \in L^2(I)$ we have that $f^2, g^2 \in L(I)$. From the Linearity of Lebesgue Integrals page, we have that then $\displaystyle{\frac{1}{2} \left ( [f(x)]^2 + [g(x)]^2 \right )}$ is Lebesgue integrable.

- From the Criterion for a Measurable Function to be Lebesgue Integrable page we see that since $fg$ is a measurable function such that $\displaystyle{\mid fg \mid \leq \frac{1}{2} [f^2 + g^2]}$ almost everywhere on $I$ (where $\displaystyle{\frac{1}{2}[f^2 + g^2]}$ is Lebesgue integrable on $I$), that then $fg$ is Lebesgue integrable on $I$. $\blacksquare$