Lebesgue Integrability of the Absolute Value of a Function

# Lebesgue Integrability of the Absolute Value of a Function

Recall from the Lebesgue Integrability of the Positive and Negative Parts of a Function page that if $f$ is a Lebesgue integrable function on $I$ then the positive and negative parts of $f$, $f^+$ and $f^-$, are both Lebesgue integrable on $I$.

We will now use these two results to prove a useful theorem which shows that $\mid f \mid$ is Lebesgue integrable as well and that the absolute value of the Lebesgue integral of $f$ on $I$ is always less than or equal to the Lebesgue integral of the absolute value of $f$ on $I$.

 Theorem 1: Let $f$ be Lebesgue integrable on $I$. Then $\mid f \mid$ is Lebesgue integrable on $I$ and $\displaystyle{\biggr \lvert \int_I f(x) \: dx \biggr \rvert \leq \int_I \mid f(x) \mid \: dx}$.
• Furthermore, for all $x \in I$ we know that:
(1)
\begin{align} \quad - \mid f(x) \mid \leq f(x) \leq \mid f(x) \mid \end{align}
(2)
\begin{align} \quad \int_I - \mid f(x) \mid \: dx \leq \int_I f(x) \: dx \leq \int_I \mid f(x) \mid \: dx \\ \quad - \int_I \mid f(x) \mid \: dx \leq \int_I f(x) \: dx \leq \int_I \mid f(x) \mid \: dx \end{align}
• Therefore $\displaystyle{\biggr \lvert \int_I f(x) \: dx \biggr \rvert \leq \int_I \mid f(x) \mid \: dx}$. $\blacksquare$