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}

• The function $\max \{ u(x), v(x) \}$ is an upper function on $I$ as we saw on The Maximum and Minimum Functions as Upper Functions page, so $f^+$ can be expressed as the difference of two upper functions on $I$, so $f^+$ is Lebesgue integrable on $I$.

• 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 $I$. $\blacksquare$