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 $\displaystyle{\biggr \lvert \int_I f(x) \: dx \biggr \rvert \leq \int_I \mid f(x) \mid \: dx}$. |

**Proof:**Let $f$ be Lebesgue integrable on $I$. Then from the theorem on the Lebesgue Integrability of the Positive and Negative Parts of a Function page we know that the positive and negative parts of $f$ are Lebesgue integrable on $I$, i.e., $f^+$ and $f^-$ are Lebesgue integrable on $I$. But $\mid f \mid = f^+ + f^-$, and so by the Linearity of Lebesgue Integrals we have that $\mid f \mid$ is Lebesgue integrable on $I$.

- Furthermore, for all $x \in I$ we know that:

\begin{align} \quad - \mid f(x) \mid \leq f(x) \leq \mid f(x) \mid \end{align}

- Since $- \mid f \mid$, $f$, and $\mid f \mid$ are all Lebesgue integrable on $I$, we have that by one of the theorems on the Comparison Theorems for Lebesgue Integrals page that:

\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 $\displaystyle{\biggr \lvert \int_I f(x) \: dx \biggr \rvert \leq \int_I \mid f(x) \mid \: dx}$. $\blacksquare$