When Positive and Negative Mass Are Both Infinite
For a signed function, cancellation is legitimate only when the Lebesgue integral is defined. In particular, an expression that informally resembles “infinity minus infinity” does not specify a number. The issue is not merely a technical restriction on notation: finite truncations of the same function can produce different answers, and product-space theorems do not repair the ambiguity by themselves.
Let \((X,\mathcal{M},\mu)\) be any measure space; no assumption of finiteness or sigma-finiteness is needed for the definitions in this paragraph. Let \(f:X\to\mathbb{R}\) be measurable and real-valued. Its positive and negative parts are the nonnegative measurable functions \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Pointwise, \(f=f^+-f^-\) and \(|f|=f^++f^-\). The nonnegative integrals \(\int f^+\,d\mu\) and \(\int f^-\,d\mu\) are each well-defined in \([0,\infty]\). The extended Lebesgue integral of \(f\) is defined as their difference only when they are not both infinite.
Thus, if \(\int f^+=\infty\) and \(\int f^-<\infty\), the integral is \(+\infty\); if \(\int f^+<\infty\) and \(\int f^-=\infty\), it is \(-\infty\). If both are finite, the integral is finite. Only the remaining case, in which both are infinite, has no value under this definition.
Integrability and Absolute Integrability
Proof. Pointwise, \(|f|=f^++f^-\). Additivity of the nonnegative Lebesgue integral gives \(\int |f|\,d\mu=\int f^+\,d\mu+\int f^-\,d\mu\), with extended nonnegative values allowed. The sum is finite exactly when both summands are finite. If both are finite, their difference defining \(\int f\,d\mu\) is a finite real number. Conversely, finiteness of \(\int |f|\,d\mu\) forces each nonnegative summand to be finite, since each is bounded above by \(|f|\) pointwise and the integral is monotone. \(\square\)
This criterion matters for limits as well as for the definition of an integral. If \(f\) is integrable and \(E_n\in\mathcal{M}\) increase to \(X\), then truncating to \(E_n\) recovers the integral independently of the exhaustion. Indeed, \(f\chi_{E_n}\to f\) pointwise, and \(|f\chi_{E_n}|\le |f|\), whose integral is finite. The Dominated Convergence Theorem therefore gives \(\int_{E_n}f\,d\mu\to\int_X f\,d\mu\). If both parts have infinite integral, this argument is unavailable: there is no integrable dominating function \(|f|\).
Worked Example: Two Exhaustions, Two Limits
On \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure, define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=1\) on \([2k,2k+1)\), \(f(x)=-1\) on \([2k+1,2k+2)\), for each integer \(k\ge0\), and \(f(x)=0\) for \(x<0\). Each positive interval has length \(1\), and each negative interval has length \(1\). Consequently, \(\int f^+\,dm=\sum_{k=0}^\infty 1=\infty\) and \(\int f^-\,dm=\sum_{k=0}^\infty 1=\infty\); the signed integral is undefined.
Now take the increasing sets \(E_N=[0,2N)\), for integers \(N\ge1\). They exhaust \([0,\infty)\), and each contains exactly \(N\) positive intervals and \(N\) negative intervals. Hence \(\int_{E_N}f\,dm=N-N=0\). Instead take \(F_N=[0,2N+1)\). These also increase to \([0,\infty)\), but contain \(N+1\) positive intervals and \(N\) negative intervals, so \(\int_{F_N}f\,dm=(N+1)-N=1\).
Check: \(E_N\uparrow[0,\infty)\) and \(F_N\uparrow[0,\infty)\), but the truncated integrals converge to \(0\) and \(1\), respectively. Both sequences consist of finite-measure sets, and the finite calculations use only intervals of length \(1\). The discrepancy is possible because \(f\) is not integrable, not because either finite truncation was miscomputed.
What Tonelli Can—and Cannot—Say
On a product space, the same distinction must be maintained. Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and let \(f:X\times Y\to\mathbb{R}\) be measurable with respect to \(\mathcal{M}\otimes\mathcal{N}\). Under these hypotheses, Tonelli's theorem applies separately to the nonnegative functions \(f^+\) and \(f^-\), using the product measure \(\mu\times\nu\). Each application may yield \(+\infty\). It does not follow that the difference of the resulting two values exists. Sigma-finiteness is stated here because nonnegativity alone does not justify invoking the standard product-measure form of Tonelli on arbitrary factor spaces.
Proof. The functions \(f^+\) and \(f^-\) are nonnegative and measurable because they are obtained from the measurable real-valued function \(f\) by continuous operations. The factors are sigma-finite, so Tonelli applies to each function. It identifies its product integral with either iterated integral, allowing infinite values. If both resulting values are \(+\infty\), their formal difference is \(+\infty-(+\infty)\), which is undefined in the extended real numbers and is excluded by the definition of the signed integral. Thus the separate Tonelli conclusions give no value for \(\int f\,d(\mu\times\nu)\). \(\square\)
Worked Example: Tonelli Separately on the Positive and Negative Parts
On \((\mathbb{R}^2,\mathcal{L}\otimes\mathcal{L},m\times m)\), define \(f(x,y)=1\) when \(x>0\) and \(0\le y\le1\), define \(f(x,y)=-1\) when \(x<0\) and \(0\le y\le1\), and set \(f(x,y)=0\) otherwise. Both factor spaces are sigma-finite, since \(\mathbb{R}=\bigcup_{n\ge1}[-n,n]\) and each interval in this union has finite measure. The function is measurable because its defining regions are measurable rectangles or differences of such sets.
Tonelli applied to \(f^+\) gives \(\int_{\mathbb{R}^2}f^+\,d(m\times m)=\int_0^\infty\int_0^1 1\,dy\,dx=\int_0^\infty 1\,dx=\infty\). For \(f^-\), it gives \(\int_{\mathbb{R}^2}f^-\,d(m\times m)=\int_{-\infty}^0\int_0^1 1\,dy\,dx=\int_{-\infty}^0 1\,dx=\infty\). The signed product integral is therefore undefined.
Integrating first in \(y\), the section integral is \(1\) for \(x>0\), \(-1\) for \(x<0\), and \(0\) at \(x=0\). Its integral over \(x\in\mathbb{R}\) is itself undefined, since its positive and negative parts each have integral \(\infty\). Integrating first in \(x\), for each \(y\in[0,1]\) the section has positive and negative integrals both equal to \(\infty\), so its signed section integral is undefined there. These computations do not produce a valid iterated integral of \(f\).
Check: On the finite rectangle \([-a,b]\times[0,1]\), with \(a,b>0\), the integral is \(b-a\): the positive region has area \(b\), and the negative region has area \(a\). Taking \(a=b\) gives \(0\); taking \(a=b+1\) gives \(-1\). All these rectangles can expand to \(\mathbb{R}\times[0,1]\), while their differences need not converge to the same value.
A Discrete Exhaustion with Arbitrary Cancellation
Worked Example: Rectangular Sums on a Counting Measure Space
Let \(X=Y=\mathbb{N}=\{1,2,\ldots\}\), each with its power-set sigma-algebra and counting measure. These spaces are sigma-finite because each is a countable union of singleton sets of finite measure. Define \(f:\mathbb{N}^2\to\mathbb{R}\) by \(f(n,1)=1\) for \(n\ge2\), \(f(1,n)=-1\) for \(n\ge2\), and \(f(m,n)=0\) at all other pairs. These rules have disjoint supports, so they assign exactly one value at every pair.
For the product counting measure, the positive-part integral is the sum of one over the points \((2,1),(3,1),\ldots\), hence \(\int f^+=\sum_{n=2}^\infty1=\infty\). The negative-part integral is also \(\sum_{n=2}^\infty1=\infty\), from the points \((1,2),(1,3),\ldots\). Thus the signed integral is undefined.
For integers \(M,N\ge1\), sum over the finite rectangle \(R_{M,N}=\{1,\ldots,M\}\times\{1,\ldots,N\}\). There are \(M-1\) positive entries in its first column and \(N-1\) negative entries in its first row; \((1,1)\) contributes zero. Therefore \(\sum_{(m,n)\in R_{M,N}}f(m,n)=(M-1)-(N-1)=M-N\). The rectangles \(R_{N,N}\) exhaust \(\mathbb{N}^2\) and give sums \(0\). The rectangles \(R_{N+1,N}\) also exhaust \(\mathbb{N}^2\) and give sums \(1\). The rectangles \(R_{2N,N}\) exhaust \(\mathbb{N}^2\) and give sums \(N\), which diverge to \(+\infty\).
Check: Substituting \(M=N\) gives \(M-N=0\); substituting \(M=N+1\) gives \(M-N=1\); substituting \(M=2N\) gives \(M-N=N\). Each family has both side lengths tending to infinity. The finite sums are exact, but they do not define an integral of the full signed function.
A Common Cancellation Error
A tempting calculation for the first example is to group each adjacent pair of intervals and write \(\int f\,dm=\sum_{k=0}^\infty(1-1)=0\). The correction is that the displayed series is not a definition of the Lebesgue integral of \(f\). By definition, the positive contribution is \(\sum_{k=0}^\infty1=\infty\) and the negative contribution is \(\sum_{k=0}^\infty1=\infty\), so their difference is undefined. The grouped finite sums describe one chosen truncation procedure; the odd-endpoint truncations in that same example give \(1\) instead.
Likewise, symmetry cannot assign zero to an undefined integral merely because a function is odd. On a symmetric finite interval, oddness may give a zero integral; on the whole line, the Lebesgue integral still requires the positive and negative parts not both to have infinite integral. The symmetric truncations in the product example give zero, but rectangular truncations with unequal side lengths give other values.
The practical test is direct: before subtracting, exchanging signed integrals, or using cancellation, compute or establish finiteness of the positive and negative parts. Tonelli is available for each nonnegative part when its product-space hypotheses hold. Fubini's theorem for a signed function requires integrability, in particular \(\int |f|<\infty\); it cannot be applied to turn two infinite nonnegative integrals into a signed value.
Check Your Understanding
For each question, distinguish the two nonnegative integrals from any proposed signed value.
- If \(\int f^+\,d\mu=7\) and \(\int f^-\,d\mu=\infty\), is \(\int f\,d\mu\) defined, and what is its value?
- For the alternating-interval function, compute \(\int_{[0,2N+2)}f\,dm\) for \(N\ge0\). What does this sequence show?
- On sigma-finite factor spaces, what does Tonelli's theorem assert when applied separately to \(f^+\) and \(f^-\)? Why might that still leave \(\int f\) undefined?
- In the counting-measure example, compute the rectangular sum for \(M=5\) and \(N=3\), and compare it with the sum for \(M=N=3\).
- State the condition on \(\int|f|\,d\mu\) that guarantees that finite-measure truncations \(f\chi_{E_n}\), for \(E_n\uparrow X\), recover \(\int f\,d\mu\).