Square Lebesgue Integrable Functions

We will now discuss another class of functions known as square Lebesgue integrable functions which we define below.

Definition: A function $f$ is said to be Square Lebesgue Integrable on an interval $I$ if $f$ is a measurable function on $I$ and $f^2$ is Lebesgue integrable on $I$. The set of all square Lebesgue integrable functions on $I$ is denoted $L^2(I)$.

Note that in the definition above that $f$ itself is not required to be Lebesgue integrable - only its square is!

For example, let $f$ be any continuous function on a closed and bounded interval $I$. Then $f$ is Riemann integrable on $I$, so $f$ is Lebesgue integrable on $I$ and furthermore, $f$ is measurable on $I$.

The product of two continuous functions is still continuous, so $f^2$ is also Riemann integrable on $I$ and hence $f^2$ is Lebesgue integrable on $I$.

Since $f$ is measurable on $I$ and $f^2$ is Lebesgue integrable on $I$ we see that any continuous function on a closed and bounded interval $I$ is square Lebesgue integrable.

Let's look at an example of a function that is not square Lebesgue integrable. Consider the function $\displaystyle{f(x) = \frac{1}{\sqrt{x}}}$ on the interval $(0, 1]$. Then $f$ is measurable on $I$ since the following sequence of step functions converges to $f$ on $(0, 1]$:

(1)

\begin{align} \quad f_n(x) = \left\{\begin{matrix} 0 & \mathrm{if} \: x \in \left (0, \frac{1}{n} \right )\\ \inf \left \{ f(x) : \frac{1}{n} \leq x < \frac{2}{n} \right \} & \mathrm{if} \: x \in \left [ \frac{1}{n}, \frac{2}{n} \right ) \\\inf \left \{ f(x) : \frac{2}{n} \leq x < \frac{3}{n} \right \} & \mathrm{if} \: x \in \left [ \frac{2}{n}, \frac{3}{n} \right ) \\ \vdots & \vdots \\ \inf \left \{ f(x) : \frac{(n-1)}{n} \leq x < 1\right \} & \mathrm{if} \: x \in \left [ \frac{n-1}{n}, 1 \right ] \end{matrix}\right. \end{align}

It should not be too hard to see that $(f_n(x))_{n=1}^{\infty}$ converges to $f$. The graphs of $f_2$, $f_3$, and $f_4$ are given below:

However, the function $\displaystyle{f^2(x) = \frac{1}{x}}$ is not Lebesgue integrable on $(0, 1]$. In fact, the Lebesgue integral of $f^2$ on $(0, 1]$ is the Riemann integral of $f^2$ on $(0, 1]$ since $f^2$ is continuous and nonnegative, so:

(2)

\begin{align} \quad \int_0^1 \frac{1}{x} \: dx = \lim_{b \to 0} \int_b^1 \frac{1}{x} \: dx = [\lim_{b \to 0} \ln x]_b^1 = \lim_{b \to 0} [\ln(1) - \ln(b)] = \lim_{b \to 0} \ln \left ( \frac{1}{b} \right ) = \infty \end{align}

So $f^2$ is not Lebesgue integrable on $(0, 1]$ which shows that $f$ is not a square Lebesgue integrable function on $(0, 1]$.