The Comparison Test for Integrability

Recall from The Comparison Test for Lebesgue Integrability that if $f$ is a Lebesgue measurable function defined on a Lebesgue measurable set $E$ and if there exists a nonnegative Lebesgue measurable function $g$ on $E$ such that:

• 1) $|f(x)| \leq g(x)$ almost everywhere on $E$.

• 2) $g$ is Lebesgue integrable on $E$.

Then $f$ is Lebesgue integrable on $E$.

We now state a similar theorem for general complete measure spaces.

Theorem 1 (The Comparison Test for Integrability): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be a measurable function defined on a measurable set $E$ and suppose there exists a nonnegative measurable function $g$ on $E$ such that: 1) $|f(x)| \leq g(x)$ $\mu$-almost everywhere on $E$. 2) $g$ is integrable on $E$. Then $f$ is integrable on $E$ and furthermore, $\displaystyle{\biggr \lvert \int_E f(x) \: d \mu \biggr \rvert \leq \int_E |f(x)| \: d \mu \leq \int_E g(x) \: d \mu}$.

• Proof: Since $g$ is integrable on $E$ and $|f(x)| \leq g(x)$ we have that:

(1)

\begin{align} \quad \int_E |f(x)| \: d \mu \leq \int_E g(x) \: d \mu < \infty \end{align}

• Since $f$ is a measurable function on $E$ and the above equality holds, we have that $f$ is integrable on $E$.

• Furthermore:

(2)

\begin{align} \quad \biggr \lvert \int_E f(x) \: d \mu \biggr \rvert &= \biggr \lvert \int_E f^+(x) \: d \mu - \int_E f^-(x) \: d \mu \biggr \rvert \\ & \leq \int_E f^+(x) \: d \mu + \int_E f^-(x) \: d \mu \\ & \leq \int_E |f(x)| \: d \mu \\ & \overset{\mathrm{monotonicity}} \leq \int_E g(x) \: d \mu \quad \blacksquare \end{align}