Why Additivity Requires a Proof
For nonnegative simple functions, additivity follows by refining their measurable level sets into a common partition. For arbitrary nonnegative functions, the integral is defined through simple functions lying below them, so additivity does not follow just by writing down the same finite sum. The main issue is the reverse inequality: a simple function below \(f+g\) need not split immediately into a simple function below \(f\) and another below \(g\). A measurable partition and a small controlled loss make that split possible.
Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, sigma-finite, or complete. Functions \(f,g:X\to[0,\infty]\) are measurable, and equalities and inequalities between functions in the theorem are pointwise everywhere. The integral is the nonnegative Lebesgue integral, with values in \([0,\infty]\). A nonnegative simple function is a measurable function with finitely many values in \([0,\infty)\); its integral may nevertheless be infinite if it is positive on a set of infinite measure.
The pointwise sum \(f+g\), with the convention \(\infty+a=\infty\) for \(a\in[0,\infty]\), is measurable and takes values in \([0,\infty]\). We will show that its integral has the same extended-real addition convention.
The Additivity Theorem
Proof. Write \(I(h)=\int h\,d\mu\). We first prove \(I(f+g)\ge I(f)+I(g)\). If \(\varphi\le f\) and \(\psi\le g\) are nonnegative simple functions, then \(\varphi+\psi\) is simple and satisfies \(\varphi+\psi\le f+g\). The definition of the integral and additivity for simple functions give $$ I(f+g)\ge\int(\varphi+\psi)\,d\mu=\int\varphi\,d\mu+\int\psi\,d\mu. $$ Taking suprema over the two minorants gives the claimed lower bound. More explicitly, for two nonempty sets of nonnegative extended numbers, the supremum of all pairwise sums is the sum of the two suprema: if both suprema are finite, choose elements within any prescribed positive error of each; if either supremum is infinite, elements in that set are arbitrarily large (or one is already infinite), while the other set contains a nonnegative element. Thus the argument also covers infinite integrals.
For the reverse bound, fix any nonnegative simple \(\varphi\le f+g\). Write its positive levels on disjoint measurable sets as \(\varphi=\sum_{j=1}^{r}a_j\chi_{E_j}\), where each \(a_j>0\). Any zero-level part can be omitted. On each \(E_j\), \(f+g\ge a_j\). Fix \(0<\varepsilon<1\), and put \(b_j=(1-\varepsilon)a_j\) and \(\delta_j=\varepsilon a_j\). We construct simple functions \(\varphi_1,\varphi_2\) with \(\varphi_1+\varphi_2=(1-\varepsilon)\varphi\), \(\varphi_1\le f\), and \(\varphi_2\le g\).
On \(E_j\cap\{f\ge b_j\}\), set \(\varphi_1=b_j\) and \(\varphi_2=0\). On \(E_j\cap\{f<b_j\}\), set $$ \varphi_1=\delta_j\left\lfloor\frac{f}{\delta_j}\right\rfloor,\qquad \varphi_2=b_j-\varphi_1. $$ Here \(f\) is finite on the second set. The floor takes only finitely many values there, since \(f<b_j\), so both functions are simple after combining the finitely many level sets for all \(j\). The sets used in the construction are measurable, and the floor of a nonnegative measurable function is measurable. On the first set, \(\varphi_1\le f\) and \(\varphi_2=0\le g\). On the second, \(\varphi_1\le f\) and \(\varphi_1>f-\delta_j\). Hence $$ \varphi_2=b_j-\varphi_1<b_j-f+\delta_j=a_j-f\le g, $$ where the last inequality follows from \(f+g\ge a_j\). Thus the constructed functions are nonnegative simple minorants of \(f\) and \(g\), and their sum is exactly \((1-\varepsilon)\varphi\).
By simple-function additivity, positive homogeneity, and the definition of the integral, $$ (1-\varepsilon)\int\varphi\,d\mu =\int\varphi_1\,d\mu+\int\varphi_2\,d\mu \le I(f)+I(g). $$ If \(\int\varphi\,d\mu=\infty\), this inequality for any \(0<\varepsilon<1\) implies \(I(f)+I(g)=\infty\). If the simple integral is finite, letting \(\varepsilon\) decrease to zero gives \(\int\varphi\,d\mu\le I(f)+I(g)\). Consequently every simple minorant of \(f+g\) has integral at most \(I(f)+I(g)\). Taking the supremum over all such minorants yields \(I(f+g)\le I(f)+I(g)\). The two inequalities prove the theorem. \(\square\)
Worked Applications
Worked Example: Additivity on a Finite Counting Space
Let \(X=\{1,2,3\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure, so every point has measure \(1\). Define \(f(1)=1,\ f(2)=0,\ f(3)=2\), and \(g(1)=2,\ g(2)=3,\ g(3)=1\). Every function on this finite space is measurable, and its integral is the sum of its values.
First compute the two integrals: $$ \int f\,d\mu=1+0+2=3,\qquad \int g\,d\mu=2+3+1=6. $$ Pointwise, \((f+g)(1)=1+2=3\), \((f+g)(2)=0+3=3\), and \((f+g)(3)=2+1=3\). Therefore $$ \int(f+g)\,d\mu=3+3+3=9. $$ Check: \(\int f\,d\mu+\int g\,d\mu=3+6=9\), agreeing with the direct sum. The calculation also checks the point where one function vanishes: at \(2\), the sum is \(0+3=3\).
Worked Example: Infinite Integrals on Counting Measure
Take \(X=\mathbb{N}=\{1,2,\ldots\}\) with its power-set sigma-algebra and counting measure. Let \(f(n)=1\) for even \(n\) and \(f(n)=0\) for odd \(n\); let \(g(n)=1\) for odd \(n\) and \(g(n)=0\) for even \(n\). The functions are measurable, nonnegative, and their supports are disjoint.
Among \(1,\ldots,2N\), there are exactly \(N\) even and \(N\) odd integers. Thus the partial sums of \(f\) and \(g\) over these points are each \(N\). As \(N\) increases without bound, both counting-measure integrals are \(+\infty\). At every \(n\), exactly one of \(f(n)\) and \(g(n)\) equals \(1\), so \((f+g)(n)=1\). Its partial sum over \(1,\ldots,2N\) is \(2N\), and its integral is also \(+\infty\).
Check: the right side is \(+\infty+(+\infty)=+\infty\), matching the left side. The partial-sum calculation verifies each divergence directly; no subtraction of infinite quantities is involved.
Worked Example: Overlapping Supports in Lebesgue Measure
On \(X=[0,1]\) with the Borel sigma-algebra and Lebesgue measure restricted to it, set \(f=\chi_{[0,1/2)}\) and \(g=\chi_{[1/4,3/4]}\). These are measurable nonnegative functions. The first interval has measure \(1/2\), and the second has measure \(3/4-1/4=1/2\), so \(\int f\,dm=1/2\) and \(\int g\,dm=1/2\).
To compute the sum directly, partition the interval at \(1/4\), \(1/2\), and \(3/4\). On \([0,1/4)\), only \(f\) is \(1\), so \(f+g=1\); on \([1/4,1/2)\), both are \(1\), so \(f+g=2\); on \([1/2,3/4]\), only \(g\) is \(1\), so \(f+g=1\). Outside these pieces both vanish. The endpoint \(1/2\) belongs to the third piece, not the second, because the first interval is half-open; the endpoint choices do not change these Lebesgue measures. Therefore $$ \int(f+g)\,dm =1\cdot\frac14+2\cdot\frac14+1\cdot\frac14 =\frac14+\frac12+\frac14=1. $$ Check: \(\int f\,dm+\int g\,dm=1/2+1/2=1\). The direct partition also shows why the overlap contributes twice.
A Common Error: Counting the Overlap Only Once
In the preceding example, a tempting but incorrect calculation is $$ \int(f+g)\,dm=m([0,1/2)\cup[1/4,3/4])=\frac34. $$ This treats \(f+g\) as the indicator of the union. But on \([1/4,1/2)\), which has measure \(1/2-1/4=1/4\), both indicators equal \(1\), so their sum equals \(2\), not \(1\). The correct integral is the measure of the two non-overlap pieces, \(1/4+1/4\), plus twice the measure of the overlap, \(2(1/4)\), giving \(1\). An indicator of a union represents the pointwise maximum of these indicators, not their sum.
The theorem does not require disjoint supports, finite integrals, or a finite measure space. Its proof instead relies on the definition by simple minorants. The lower bound combines minorants of \(f\) and \(g\); the upper bound carefully partitions the level sets of a minorant of \(f+g\). The small factor \(1-\varepsilon\) allows the partition to use finitely many values while keeping each part below the corresponding function.
Check Your Understanding
For each question, pay attention to whether the functions are nonnegative and whether the integral may be infinite.
- In the theorem’s upper-bound proof, why is it useful to work with \((1-\varepsilon)\varphi\) rather than trying to split \(\varphi\) exactly?
- For counting measure on \(\mathbb{N}\), let \(f(n)=1\) for \(n\) divisible by \(3\) and \(0\) otherwise, and let \(g=1-f\). Compute both integrals and the integral of \(f+g\).
- On \([0,1]\) with Lebesgue measure, take \(f=\chi_{[0,2/3)}\) and \(g=\chi_{[1/3,1]}\). Compute the integral of their sum by splitting at \(1/3\) and \(2/3\).
- Explain why \(\int(f+g)\,d\mu\) is not generally the measure of the union of the supports of \(f\) and \(g\).
- Which assumptions in the additivity theorem ensure that the expression \(\int f\,d\mu+\int g\,d\mu\) has no undefined subtraction of infinities?