Positive and Negative Parts of an Integrable Function

Splitting a Function by Its Sign

A real-valued function can take positive, negative, and zero values on different measurable regions. Separating these regions gives two nonnegative functions that retain the full information in the original function. This separation is useful because the Lebesgue integral is first defined for nonnegative functions, while an integrable signed function is handled through its positive and negative contributions.

Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Unless stated otherwise, functions are real-valued and measurable, and identities between functions are pointwise everywhere. A function \(f\) is integrable here if \(\int_X |f|\,d\mu<\infty\). Its signed Lebesgue integral is expressed using its positive and negative parts.

Definition: For a real-valued function \(f\) on \(X\), define \(f^+(x)=\max\{f(x),0\}\) and \(f^-(x)=\max\{-f(x),0\}\). The functions \(f^+\) and \(f^-\) are called the positive part and negative part of \(f\), respectively. Both take values in \([0,\infty)\).

The negative part is nonnegative: it records the magnitude of the negative values, rather than keeping their negative sign. At a point where \(f\ge0\), the pair is \((f^+,f^-)=(f,0)\); where \(f<0\), it is \((0,-f)\). These two cases lead to the basic identities.

Proposition: For every real-valued function \(f\) on a set \(X\), \(f=f^+-f^-\) and \(|f|=f^++f^-\) at every point of \(X\). Also, \(f^+f^-=0\) at every point.

Proof. Fix \(x\in X\). If \(f(x)\ge0\), then \(f^+(x)=f(x)\) and \(f^-(x)=0\). Thus \(f^+(x)-f^-(x)=f(x)\), \(f^+(x)+f^-(x)=f(x)=|f(x)|\), and their product is zero. If \(f(x)<0\), then \(f^+(x)=0\) and \(f^-(x)=-f(x)>0\). Thus \(f^+(x)-f^-(x)=f(x)\), \(f^+(x)+f^-(x)=-f(x)=|f(x)|\), and again their product is zero. These cases include every real value of \(f(x)\), including zero, so all three identities hold everywhere. \(\square\)

Measurability of the Parts

For a real-valued function \(g\) on a measurable space \((X,\mathcal{M})\), measurability can be tested by asking whether \(\{x:g(x)>a\}\in\mathcal{M}\) for every real \(a\). This criterion gives a direct proof for the two parts and makes no assumption that \(\mu\) is complete.

Theorem: If \(f:X\to\mathbb{R}\) is measurable on a measurable space \((X,\mathcal{M})\), then \(f^+\) and \(f^-\) are measurable. Consequently, their pointwise sum and difference in the identities above are measurable.

Proof. Fix \(a\in\mathbb{R}\). If \(a\ge0\), then \(\{x:f^+(x)>a\}=\{x:f(x)>a\}\), because \(\max\{f(x),0\}>a\) for nonnegative \(a\) exactly when \(f(x)>a\). If \(a<0\), then \(f^+(x)\ge0>a\) for every \(x\), so \(\{x:f^+(x)>a\}=X\). Both sets are measurable. For \(f^-\), if \(a\ge0\), then \(\{x:f^-(x)>a\}=\{x:f(x)<-a\}\); this is measurable because \(\{f<-a\}\) is measurable for measurable \(f\). If \(a<0\), then \(f^-(x)\ge0>a\) everywhere, so the set is \(X\). The level-set criterion proves that both parts are measurable. Their sum and difference are measurable by the standard closure of real-valued measurable functions under addition and subtraction. \(\square\)

The last sentence does not assert that a difference of arbitrary extended-real-valued functions is defined: the functions here are real-valued. Moreover, \(f^+\) and \(f^-\) cannot both be positive at the same point, as the pointwise proposition established.

Worked Examples

Worked Example: A Linear Function That Changes Sign

On \(X=[0,3]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\), let \(f(x)=x-1\). Find \(f^+\) and \(f^-\), and compute the integrals of both parts.

The sign changes at \(x=1\). For \(0\le x<1\), \(f(x)<0\), so \(f^+(x)=0\) and \(f^-(x)=1-x\). For \(1\le x\le3\), \(f(x)\ge0\), so \(f^+(x)=x-1\) and \(f^-(x)=0\). Therefore, including the zero value at the boundary in the nonnegative region,

f^+(x)= \begin{cases} 0,&0\le x<1,\\ x-1,&1\le x\le3, \end{cases} \qquad f^-(x)= \begin{cases} 1-x,&0\le x<1,\\ 0,&1\le x\le3. \end{cases}

Integrating over the intervals on which the parts are nonzero gives \(\int_X f^+\,dm=\int_1^3(x-1)\,dx=[(x-1)^2/2]_1^3=2\), and \(\int_X f^-\,dm=\int_0^1(1-x)\,dx=[x-x^2/2]_0^1=1/2\). Thus \(\int_X |f|\,dm=2+1/2=5/2\), using \(|f|=f^++f^-\). The signed integral, defined by the difference of these finite integrals, is \(2-1/2=3/2\). Check: direct integration gives \(\int_0^3(x-1)\,dx=[x^2/2-x]_0^3=9/2-3=3/2\), agreeing with the signed integral from the parts.

Worked Example: Parts of a Quadratic Function

On \(X=[-2,2]\) with Lebesgue measure, let \(f(x)=x^2-1\). Determine the parts and compute their integrals. The function is nonnegative when \(|x|\ge1\) and negative when \(|x|<1\). In particular, it is zero at \(x=-1\) and \(x=1\).

It follows that \(f^+(x)=x^2-1\) for \(x\in[-2,-1]\cup[1,2]\), and is zero for \(-1<x<1\). The negative part is \(f^-(x)=1-x^2\) for \(-1<x<1\), and is zero on the remaining points. The endpoint choices at \(\pm1\) do not affect the formulas there because both expressions vanish.

Using symmetry and evaluating the antiderivative,

\int_{-2}^{2}f^+(x)\,dx =2\int_1^2(x^2-1)\,dx =2\left[\frac{x^3}{3}-x\right]_1^2 =2\left(\frac{8}{3}-2-\frac{1}{3}+1\right) =\frac{8}{3}.

For the negative part,

\int_{-2}^{2}f^-(x)\,dx =\int_{-1}^{1}(1-x^2)\,dx =\left[x-\frac{x^3}{3}\right]_{-1}^{1} =\left(1-\frac13\right)-\left(-1+\frac13\right) =\frac43.

Hence \(\int_{-2}^2|f(x)|\,dx=8/3+4/3=4\), and the signed integral from the parts is \(8/3-4/3=4/3\). Check: direct integration of \(x^2-1\) gives \(\left[x^3/3-x\right]_{-2}^{2}=(8/3-2)-(-8/3+2)=4/3\), as required.

Worked Example: A Function Defined on Measurable Pieces

On \(X=[0,4]\) with Lebesgue measure, define \(f(x)=3\) on \([0,1)\), \(f(x)=-2\) on \([1,3)\), and \(f(x)=0\) on \([3,4]\). The three pieces are measurable and partition \(X\). Find the parts and their integrals, keeping track of the half-open endpoints.

The positive values occur exactly on \([0,1)\), a set of measure \(1\), and the negative values occur exactly on \([1,3)\), a set of measure \(2\). Thus \(f^+=3\chi_{[0,1)}\) and \(f^-=2\chi_{[1,3)}\). The endpoint \(x=1\) belongs to the negative piece, while \(x=3\) belongs to the zero piece. Consequently, \(\int_X f^+\,dm=3\cdot1=3\) and \(\int_X f^-\,dm=2\cdot2=4\). The integral of the absolute value is \(3+4=7\), and the signed integral is \(3-4=-1\).

Check: Integrating the original function piece by piece gives \(3\cdot m([0,1))-2\cdot m([1,3))+0\cdot m([3,4])=3\cdot1-2\cdot2+0\cdot1=-1\). The part calculation gives the same signed value.

Parts and the Integral of an Integrable Function

Now let \((X,\mathcal{M},\mu)\) be any measure space and let \(f:X\to\mathbb{R}\) be measurable and integrable, meaning \(\int_X|f|\,d\mu<\infty\). By the pointwise identity, \(0\le f^+\le|f|\) and \(0\le f^-\le|f|\). Monotonicity of the nonnegative Lebesgue integral therefore gives \(0\le\int_X f^+\,d\mu\le\int_X|f|\,d\mu<\infty\) and \(0\le\int_X f^-\,d\mu\le\int_X|f|\,d\mu<\infty\). Both parts have finite integrals. The signed integral is consequently well-defined by

\int_X f\,d\mu = \int_X f^+\,d\mu-\int_X f^-\,d\mu.

In particular, the sign of the signed integral records which contribution is larger; it is not the same as integrating \(|f|\), which adds the magnitudes. The pointwise identity \(|f|=f^++f^-\) also expresses the absolute value as the sum of those magnitudes. These pointwise identities require no integrability assumption; integrability is used here to ensure the two part integrals are finite.

A Common Sign Error

A frequent mistake is to define the negative part as \(\min\{f,0\}\). That quantity is nonpositive, not the nonnegative negative part used here. For example, at a point where \(f=-3\), the correct values are \(f^+=\max\{-3,0\}=0\) and \(f^-=\max\{3,0\}=3\). The correct reconstruction is \(f=f^+-f^-=0-3=-3\), while \(|f|=f^++f^-=0+3=3\).

If instead one sets a purported \(f^-\) equal to \(\min\{f,0\}\), then at \(f=-3\) that value is \(-3\), and the formula \(f=f^+-f^-\) would give \(0-(-3)=3\), the wrong sign. One can use a signed negative truncation with a different convention, but its identities must also be changed. With the present convention, both parts are nonnegative and the minus sign appears explicitly in \(f=f^+-f^-\).

Check Your Understanding

Use the definitions with the nonnegative negative part \(f^-=\max\{-f,0\}\). In each question, state whether the identities are pointwise or only almost everywhere.

For \(f(x)=2-x\) on \([0,3]\) with Lebesgue measure, write \(f^+\) and \(f^-\) on each sign region.
Prove directly from the definition that \(f^+f^-=0\) for every real-valued function \(f\).
For \(f(x)=1-x^2\) on \([-2,2]\), identify the sets where each part is nonzero.
If \(f\) is integrable on a measure space, explain why both \(\int f^+\,d\mu\) and \(\int f^-\,d\mu\) are finite.
At a point where \(f=-5\), calculate \(f^+\), \(f^-\), \(f^+-f^-\), and \(f^++f^-\).