The Comparison Test for Lebesgue Integrability

# The Comparison Test for Lebesgue Integrability

 Theorem 1 (The Comparison Test for Lebesgue Integrability): Let $f$ be a Lebesgue measurable function and let $g$ be a nonnegative Lebesgue measurable function defined on a Lebesgue measurable set $E$. Suppose that: 1) $|f(x)| \leq g(x)$ almost everywhere on $E$. 2) $g$ is Lebesgue integrable on $E$. Then $f$ is a Lebesgue integrable on $E$.
• Proof: For all $x \in E$ we can write:
(1)
\begin{align} \quad |f(x)| = f^+(x) + f^-(x) \end{align}
• From (1) we then have that $f^+(x) \leq g(x)$ and $f^-(x) \leq g(x)$ for all $x \in E$. Now from (2), since $g$ is Lebesgue integrable on $E$ we have that $\displaystyle{\int_E g < \infty}$ and so:
(2)
\begin{align} \quad \int_E f^+ \leq \int_E g < \infty \quad \mathrm{and} \quad \int_E f^- \leq \int_E g < \infty \end{align}
• Therefore, since $f$ is a Lebesgue measurable function with $\displaystyle{\int_E f^+ < \infty}$ and $\displaystyle{\int_E f^- < \infty}$ we have by definition that $f$ is Lebesgue integrable on $E$.

## Example 1

Use the comparison test to prove that the function $\displaystyle{f(x) = \left\{\begin{matrix} 1 & \mathrm{if} \: x \in \mathbb{Q} \cap [0, 1]\\ -1 & \mathrm{if} \: x \in [0, 1] \setminus \mathbb{Q} \end{matrix}\right.}$ is Lebesgue integrable on $[0, 1]$.

We have that $f$ is a Lebesgue measurable function on $[0, 1]$ which is a Lebesgue measurable set. Let $g(x) = 1$ on $[0, 1]$. Then $g$ is a nonnegative Lebesgue measurable function on $[0, 1]$. Furthermore $|f(x)| \leq |g(x)|$ everywhere on $[0, 1]$ and $g$ is Lebesgue integrable on $E$. Therefore $f$ is Lebesgue integrable on $E$.