Lebesgue Integrability of the Positive and Negative Parts of a Function

Lebesgue Integrability of the Positive and Negative Parts of a Function

Recall from The Positive and Negative Parts of a Function page that if $f$ is a function defined on an interval $I$ then the positive part of $f$ denoted $f^+$ is the function defined for all $x \in I$ by:

(1)
\begin{align} \quad f^+(x) = \max \{ f(x), 0 \} \end{align}

Similarly, the negative part of $f$ denoted $f^-$ is the function defined for all $x \in I$ by:

(2)
\begin{align} \quad f^-(x) = \max \{ -f(x), 0 \} \end{align}

We also proved the following two identities:

(3)
\begin{align} \quad f(x) = f^+(x) - f^-(x) \end{align}
(4)
\begin{align} \quad \mid f(x) \mid = f^+(x) + f^-(x) \end{align}

We will now see that if $f$ is Lebesgue integrable on $I$ then both the positive and negative parts of $I$ are Lebesgue integrable on $I$.

 Theorem 1: Let $f$ be Lebesgue integrable on $I$. Then $f^+$ and $f^-$ are Lebesgue integrable on $I$.
• Proof: Let $f$ be Lebesgue integrable on $I$. Then there exists upper functions $u$ and $v$ on $I$ such that:
(5)
\begin{align} \quad f = u - v \end{align}
• Therefore we have that $f^+$ is defined for all $x \in I$ by:
(6)
\begin{align} \quad f^+(x) = \max \{ f(x), 0 \} = \max \{ u(x) - v(x), 0 \} = \max \{ u(x), v(x) \} - v(x) \end{align}
• Similarly, we have that $f^-$ is defined for all $x \in I$ by:
(7)
\begin{align} \quad f^-(x) = \max \{ -f(x), 0 \} = \max \{ v(x) - u(x), 0 \} = \max \{ u(x), v(x) \} - u(x) \end{align}
• So $f^-$ can be expressed as the difference of two upper functions on $I$, so $f^-$ is Lebesgue integrable on $I$. $\blacksquare$