Linearity Through Positive and Negative Parts
The integral of a signed function is defined by comparing the integrals of its nonnegative parts. Linearity is therefore not a consequence of simply distributing positive parts over sums: in general, those parts do not distribute. Instead, we use a valid decomposition into nonnegative integrable functions, together with additivity for nonnegative integrals.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; no finiteness, completeness, or sigma-finiteness assumption is imposed. Functions in this tutorial are real-valued and \(\mathcal{M}\)-measurable, and function identities are pointwise unless almost-everywhere equality is stated. A real-valued measurable \(f\) is integrable when \(\int |f|\,d\mu<\infty\). Its positive and negative parts are \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\), so \(f=f^+-f^-\) and \(|f|=f^++f^-\) everywhere.
We use the standard properties of the nonnegative Lebesgue integral: it is monotone, additive on nonnegative measurable functions, and homogeneous under multiplication by a nonnegative finite scalar. These properties hold even when an integral is infinite; whenever we subtract two such integrals below, we will first establish that both are finite.
A Difference Representation
For additivity, we need to handle representations of a signed function as a difference other than its own positive and negative parts. The following observation justifies that step.
Proof. Since \(|A-B|\le A+B\) pointwise, monotonicity and additivity of the nonnegative integral give \(\int |h|\,d\mu\le\int A\,d\mu+\int B\,d\mu<\infty\). Thus \(h\) is integrable. The pointwise identity \(h=A-B=h^+-h^-\) implies \(A+h^-=B+h^+\). All four functions are nonnegative and integrable: \(h^+,h^-\le |h|\), while \(A,B\) are integrable by assumption. Integrating the identity using nonnegative additivity gives \(\int A\,d\mu+\int h^-\,d\mu=\int B\,d\mu+\int h^+\,d\mu\). All quantities are finite, so rearrangement yields \(\int h^+\,d\mu-\int h^-\,d\mu=\int A\,d\mu-\int B\,d\mu\). The left side is \(\int h\,d\mu\) by definition. \(\square\)
Additivity
Proof. The sum \(f+g\) is measurable. Pointwise, \(|f+g|\le |f|+|g|\), so monotonicity and nonnegative additivity imply \(\int|f+g|\,d\mu\le\int|f|\,d\mu+\int|g|\,d\mu<\infty\). Hence \(f+g\) is integrable.
Now set \(A=f^++g^+\) and \(B=f^-+g^-\). These are nonnegative measurable functions. They are integrable because nonnegative additivity gives \(\int A\,d\mu=\int f^+\,d\mu+\int g^+\,d\mu<\infty\) and \(\int B\,d\mu=\int f^-\,d\mu+\int g^-\,d\mu<\infty\). The identities \(f=f^+-f^-\) and \(g=g^+-g^-\) give \(f+g=A-B\). Apply the difference-representation lemma to obtain \[ \int(f+g)\,d\mu =\int A\,d\mu-\int B\,d\mu =\left(\int f^+\,d\mu-\int f^-\,d\mu\right) +\left(\int g^+\,d\mu-\int g^-\,d\mu\right). \] By the definition of the signed integral, the last expression is \(\int f\,d\mu+\int g\,d\mu\). This proves the claim. \(\square\)
Worked Example: Adding Two Signed Functions on an Interval
Take \(X=[0,2)\), with the Lebesgue sigma-algebra and Lebesgue measure \(m\) restricted to \(X\). Define \(f=2\) on \([0,1)\) and \(f=-1\) on \([1,2)\); define \(g=-1\) on \([0,1)\) and \(g=3\) on \([1,2)\). Each interval has measure \(1\). Compute the integrals and verify additivity by positive and negative parts.
For \(f\), the positive part is \(2\) on the first interval and \(0\) on the second, while the negative part is \(0\) on the first and \(1\) on the second. Therefore \(\int f^+\,dm=2\cdot1=2\), \(\int f^-\,dm=1\cdot1=1\), and \(\int f\,dm=2-1=1\).
For \(g\), its positive part is \(0\) on the first interval and \(3\) on the second; its negative part is \(1\) on the first and \(0\) on the second. Thus \(\int g^+\,dm=3\cdot1=3\), \(\int g^-\,dm=1\cdot1=1\), and \(\int g\,dm=3-1=2\).
Pointwise, \(f+g=1\) on \([0,1)\) and \(f+g=2\) on \([1,2)\). It is nonnegative, so its positive-part integral is \(1\cdot1+2\cdot1=3\), and its negative-part integral is \(0\). Hence \(\int(f+g)\,dm=3\). Check: the separate integrals give \(\int f\,dm+\int g\,dm=1+2=3\), agreeing with the direct calculation.
Scalar Homogeneity
Proof. If \(c=0\), then \(cf=0\), which is integrable and has integral \(0=0\int f\,d\mu\). Suppose \(c>0\). Then \(|cf|=c|f|\), so \(\int|cf|\,d\mu=c\int|f|\,d\mu<\infty\). Pointwise, \((cf)^+=cf^+\) and \((cf)^-=cf^-\). The definition of the signed integral and nonnegative homogeneity give \(\int cf\,d\mu=c\int f^+\,d\mu-c\int f^-\,d\mu=c\int f\,d\mu\).
Suppose instead that \(c<0\), and write \(d=-c>0\). Again, \(|cf|=d|f|\), so \(cf\) is integrable. Since multiplication by a negative number reverses signs, \((cf)^+=d f^-\) and \((cf)^-=d f^+\). Therefore \(\int cf\,d\mu=d\int f^-\,d\mu-d\int f^+\,d\mu=-d\int f\,d\mu=c\int f\,d\mu\). The cases \(c=0\), \(c>0\), and \(c<0\) exhaust all real scalars. \(\square\)
Worked Example: A Negative Scalar on Counting Measure
Let \(X=\{1,2,3\}\) with counting measure, so the integral of a function on \(X\) is the sum of its three values. Set \(f(1)=2\), \(f(2)=-1\), and \(f(3)=4\), and take \(c=-2\). Compute both sides of scalar homogeneity.
The positive part of \(f\) has values \(2,0,4\), so \(\int f^+\,d\mu=2+0+4=6\). Its negative part has values \(0,1,0\), so \(\int f^-\,d\mu=0+1+0=1\). Thus \(\int f\,d\mu=6-1=5\).
The function \(-2f\) has values \(-4,2,-8\). Its positive part has values \(0,2,0\), with integral \(2\); its negative part has values \(4,0,8\), with integral \(12\). Consequently \(\int(-2f)\,d\mu=2-12=-10\). Check: scalar homogeneity predicts \((-2)\int f\,d\mu=(-2)\cdot5=-10\). Direct summation also gives \(-4+2-8=-10\).
A Common Mistake: Positive Parts Do Not Distribute
In general, it is false that \((f+g)^+=f^++g^+\), or that \((f+g)^-=f^-+g^-\). For example, on \(\mathbb{R}\) with Lebesgue measure, let \(A=[0,1/3]\), \(f=3\chi_A\), and \(g=-3\chi_A\). Then \(f+g=0\) everywhere, so \((f+g)^+=0\). But \(f^+=3\chi_A\) and \(g^+=0\), giving \(f^++g^+=3\chi_A\), which is nonzero on \(A\). Indeed, at every \(x\in A\), the proposed identity would say \(0=3\).
The corresponding negative parts also show the failure: \((f+g)^-=0\), whereas \(f^-+g^-=3\chi_A\). The measures and integrals here are exact: \(m(A)=1/3\), so \(\int f\,dm=3(1/3)=1\), \(\int g\,dm=-3(1/3)=-1\), and \(\int(f+g)\,dm=0\). Additivity correctly gives \(1+(-1)=0\).
There is a valid difference representation in this example: \[ f+g=(f^++g^+)-(f^-+g^-). \] The two nonnegative terms on the right are both \(3\chi_A\), so their integral difference is \(1-1=0\). This representation is legitimate by the difference-representation lemma. What is not legitimate is calling \(f^++g^+\) and \(f^-+g^-\) the positive and negative parts of \(f+g\): the actual parts of the zero function are both zero.
What Linearity Does—and Does Not—Say
Additivity and scalar homogeneity combine to give linearity for any finite real linear combination of integrable real-valued functions. For instance, if \(f_1,\ldots,f_k\) are integrable and \(a_1,\ldots,a_k\in\mathbb{R}\), repeated application of the two theorems gives \(\int\sum_{j=1}^k a_jf_j\,d\mu=\sum_{j=1}^k a_j\int f_j\,d\mu\). Each intermediate sum remains integrable by the integrability argument in the additivity proof, and each scalar multiple is integrable by the homogeneity proof. This statement concerns finite sums; an infinite sum requires additional hypotheses before an integral may be interchanged with a limit.
Linearity is the central algebraic rule for the integral of integrable functions, but its proof relies on keeping signed quantities under control. The positive and negative parts guarantee finite quantities that can safely be subtracted. In applications, first check integrability, then use the exact pointwise decomposition; do not infer identities for positive or negative parts by distributing those operations across sums.
Check Your Understanding
For each question, keep separate the pointwise identities for positive and negative parts and the integral identities they justify.
- On a measure space, suppose \(f\) and \(g\) are integrable. Explain why \(|f+g|\le |f|+|g|\) implies that \(f+g\) is integrable.
- Give functions \(f,g\) for which \((f+g)^+\ne f^++g^+\), and compute all four functions on a set where the identity fails.
- If \(f\) is integrable, derive the formulas for \((cf)^+\) and \((cf)^-\) separately when \(c>0\) and when \(c<0\).
- For counting measure on \(\{1,2\}\), take \(f(1)=-3\), \(f(2)=2\), and \(c=-1/2\). Compute \(\int f\), \(\int cf\), and \(c\int f\).
- Why is \(A-B\) a valid representation for computing an integral when \(A,B\) are nonnegative and integrable, even if they are not the positive and negative parts of \(A-B\)?