Why Partitions Make Linearity Precise
A nonnegative simple function has only finitely many values, so its integral is a finite weighted sum of measures. The difficulty in adding two such functions is that their values may be specified on different measurable sets. The sets can overlap without either partition refining the other. Taking all pairwise intersections gives a common partition on which both functions, and therefore their sum, are constant. This finite refinement is the mechanism behind additivity.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space: \(\mu\) need not be finite, \(\sigma\)-finite, or complete. A nonnegative simple function means a measurable function \(f:X\to[0,\infty)\) with finite range. Functions and identities in this tutorial are pointwise, everywhere on \(X\). The integral of such a function may equal \(+\infty\).
A partition may include empty sets or sets on which the function is zero; neither affects the value of the sum. This formulation also covers \(X=\varnothing\): the integral is the empty sum, equal to zero. If a function is initially described on finitely many disjoint measurable sets that do not cover \(X\), include their complement as an additional cell and assign it value zero.
Refining a Partition
Suppose \(A_1,\ldots,A_r\) and \(B_1,\ldots,B_s\) are finite measurable partitions of \(X\). Their common refinement consists of the \(rs\) sets \(A_i\cap B_j\). These intersections are measurable and pairwise disjoint, and their union is \(X\): every \(x\in X\) lies in exactly one \(A_i\) and exactly one \(B_j\).
If \(f=a_i\) on \(A_i\), then \(f=a_i\) on every \(A_i\cap B_j\). Finite additivity of a measure gives \(\mu(A_i)=\sum_{j=1}^s\mu(A_i\cap B_j)\). Consequently, refining a partition does not change the integral's finite sum. For \(a_i>0\), multiplying this identity by \(a_i\) is valid even if one or more measures are infinite; for \(a_i=0\), all corresponding contributions are defined as zero.
This also verifies that the formula for \(\int f\,d\mu\) is independent of the partition used to describe \(f\). Indeed, given two such descriptions, refine their partitions together. On each intersection, the two assigned values must agree because both descriptions give the value of the same pointwise function there. The resulting finite sums therefore agree.
Additivity and Positive Homogeneity
Proof. Choose finite measurable partitions \(A_1,\ldots,A_r\) and \(B_1,\ldots,B_s\) on which \(f\) and \(g\) are constant, respectively. Write \(f=a_i\) on \(A_i\) and \(g=b_j\) on \(B_j\), with \(a_i,b_j\ge0\). The common refinement \(A_i\cap B_j\) is a finite measurable partition, and \(f+g=a_i+b_j\) on each of its cells. This proves that \(f+g\) is simple and gives $$ \int(f+g)\,d\mu =\sum_{i=1}^{r}\sum_{j=1}^{s}(a_i+b_j)\mu(A_i\cap B_j). $$ For nonnegative extended-real terms, finite sums can be regrouped and distributed without subtraction. A zero coefficient contributes zero by the convention in the definition; a positive coefficient times an infinite measure contributes \(+\infty\). Thus the last sum equals $$ \sum_{i=1}^{r}a_i\sum_{j=1}^{s}\mu(A_i\cap B_j) +\sum_{j=1}^{s}b_j\sum_{i=1}^{r}\mu(A_i\cap B_j). $$ Finite additivity gives \(\sum_j\mu(A_i\cap B_j)=\mu(A_i)\) and \(\sum_i\mu(A_i\cap B_j)=\mu(B_j)\). The displayed expression is therefore \(\sum_i a_i\mu(A_i)+\sum_j b_j\mu(B_j)=\int f\,d\mu+\int g\,d\mu\). This proves additivity, including when either side is infinite.
Now let \(c>0\). On each \(A_i\), the function \(cf\) has the finite nonnegative value \(ca_i\), so it is simple and \(\int(cf)\,d\mu=\sum_i(ca_i)\mu(A_i)=c\sum_i a_i\mu(A_i)=c\int f\,d\mu\). If the sum is infinite, its product by the positive finite number \(c\) is infinite, as is the sum with coefficients \(ca_i\). If \(c=0\), then \(0f\) is the zero function everywhere, whose integral is zero by its partition formula. This proves both assertions. \(\square\)
Worked Calculations
Worked Example: A Common Refinement for Lebesgue Measure
Take \(X=[0,3)\) with its Borel sigma-algebra and Lebesgue measure \(m\). Define \(f=2\) on \([0,1)\), \(f=5\) on \([1,3)\), and \(g=3\) on \([0,2)\), \(g=1\) on \([2,3)\). The common refinement is \([0,1),[1,2),[2,3)\); each cell has measure \(1\). On these cells the values of \(f\) are \(2,5,5\), the values of \(g\) are \(3,3,1\), and the values of \(f+g\) are \(5,8,6\).
Using the partition formula, $$ \int f\,dm=2\cdot1+5\cdot1+5\cdot1=12,\qquad \int g\,dm=3\cdot1+3\cdot1+1\cdot1=7. $$ The refined sum gives $$ \int(f+g)\,dm=5\cdot1+8\cdot1+6\cdot1=19. $$ The right side of additivity is \(12+7=19\), so the two calculations agree. As a homogeneity check, \(\int(\tfrac12 f)\,dm=1\cdot1+\tfrac52\cdot1+\tfrac52\cdot1=6\), while \(\tfrac12\int f\,dm=\tfrac12\cdot12=6\).
Check: Adding the integrals of the three values of \(f+g\) gives \(5+8+6=19\), the same as \(12+7=19\).
Worked Example: Overlapping Supports Under Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure. Define \(f(1)=2,\ f(2)=1,\ f(3)=0,\ f(4)=3\), and \(f(n)=0\) for \(n\ge5\). Define \(g(1)=1,\ g(2)=4,\ g(3)=2,\ g(4)=0\), and \(g(n)=0\) for \(n\ge5\). Counting measure assigns measure \(1\) to each singleton, so each integral is the sum of the function's values at the points.
The values of \(f+g\) at \(1,2,3,4\) are, respectively, \(2+1=3,\ 1+4=5,\ 0+2=2,\ 3+0=3\). Hence $$ \int f\,d\mu=2+1+0+3=6,\qquad \int g\,d\mu=1+4+2+0=7, $$ and $$ \int(f+g)\,d\mu=3+5+2+3=13=6+7. $$ For positive homogeneity with \(c=3\), the values of \(3f\) are \(6,3,0,9\), so \(\int3f\,d\mu=6+3+0+9=18\). The other side is \(3\int f\,d\mu=3\cdot6=18\).
Check: Pointwise addition gave total \(13\), and the separate totals give \(6+7=13\). Scaling each nonzero value of \(f\) by \(3\) gave \(18\), equal to \(3\cdot6\).
Worked Example: The Same Values Under a Dirac Measure
Let \(X=\mathbb{R}\) with its Borel sigma-algebra and the Dirac measure \(\delta_2\), defined by \(\delta_2(E)=1\) if \(2\in E\) and \(\delta_2(E)=0\) otherwise. Set \(f=4\chi_{[0,3)}+\chi_{[3,5)}\) and \(g=7\chi_{[1,4)}+2\chi_{[4,6)}\). At the point \(2\), which is the only point with positive Dirac mass, \(f(2)=4\) and \(g(2)=7\).
The partition formula gives \(\int f\,d\delta_2=4\cdot1+1\cdot0=4\), since \(2\in[0,3)\) and \(2\notin[3,5)\). It gives \(\int g\,d\delta_2=7\cdot1+2\cdot0=7\), since \(2\in[1,4)\) and \(2\notin[4,6)\). At \(2\), the sum has value \(4+7=11\); its integral is \(11\cdot1=11\). Thus \(\int(f+g)\,d\delta_2=11=4+7\). Scaling \(f\) by \(2\) gives value \(2\cdot4=8\) at \(2\), so \(\int2f\,d\delta_2=8=2\cdot4\).
Check: Each integral is determined by the value at \(2\): the values \(4,7,11,8\) give the stated integrals directly.
Worked Example: Infinite Integral and Zero Scaling
On \(\mathbb{R}\) with Lebesgue measure \(m\), let \(f=\chi_{\mathbb{R}}\) and \(g=2\chi_{(0,1)}\). Both are nonnegative simple functions. Since \(m(\mathbb{R})=\infty\), \(\int f\,dm=1\cdot\infty=\infty\). Since \(m((0,1))=1\), \(\int g\,dm=2\cdot1=2\).
The function \(f+g\) has value \(3\) on \((0,1)\) and value \(1\) on its complement. Its integral is $$ \int(f+g)\,dm=3\cdot m((0,1))+1\cdot m(\mathbb{R}\setminus(0,1)) =3\cdot1+1\cdot\infty=\infty. $$ The sum of the separate integrals is \(\infty+2=\infty\), in agreement. But \(0f\) is the zero function everywhere, so \(\int(0f)\,dm=0\). This case is why the zero-scalar identity is stated separately rather than written as \(0\int f\,dm\).
Check: The refined partition produces an infinite contribution from the complement of \((0,1)\); the integral is therefore \(\infty\), matching \(\infty+2=\infty\).
A Common Error: Adding Coefficients on Overlapping Sets
Suppose, again on \(\mathbb{R}\) with Lebesgue measure, that \(f=2\chi_{[0,2)}\) and \(g=3\chi_{[1,3)}\). A tempting but incorrect calculation treats \(f+g\) as having value \(2+3=5\) throughout the union \([0,3)\), and concludes that its integral is \(5\cdot3=15\). The coefficients should be added only where both functions are nonzero, not throughout the union.
The correct common refinement is \([0,1),[1,2),[2,3)\). On these intervals the values of \(f+g\) are \(2,5,3\), respectively. Therefore \(\int(f+g)\,dm=2\cdot1+5\cdot1+3\cdot1=10\). Separately, \(\int f\,dm=2\cdot2=4\) and \(\int g\,dm=3\cdot2=6\), so their sum is also \(10\). This illustrates the general procedure: refine first, then add the values on each cell.
Additivity and positive homogeneity here require only finite partitions and the measure axioms; no convergence theorem is involved. The conclusions are specifically about nonnegative simple functions. Extending additivity from this finite-valued setting to arbitrary nonnegative measurable functions requires additional reasoning.
Check Your Understanding
Use finite measurable partitions and the integral formula to justify each answer.
- If \(f=a\chi_A\) and \(g=b\chi_B\) for measurable sets \(A,B\) and finite \(a,b\ge0\), what partition should be used to compute \(\int(f+g)\,d\mu\)?
- On counting measure on \(\mathbb{N}\), let \(f(1)=3,\ f(2)=2\) and let \(f(n)=0\) for \(n\ge3\). Compute \(\int 4f\,d\mu\) and compare it with \(4\int f\,d\mu\).
- Explain why the formula \(0\int f\,d\mu\) is not used when \(\int f\,d\mu=\infty\), and compute \(\int(0f)\,d\mu\) directly.
- Under \(\delta_0\), suppose \(f(0)=2\) and \(g(0)=5\). What are \(\int f\,d\delta_0\), \(\int g\,d\delta_0\), and \(\int(f+g)\,d\delta_0\)?
- For \(f=2\chi_{[0,2)}\) and \(g=3\chi_{[1,3)}\), identify the value of \(f+g\) on each cell of the common refinement and compute its Lebesgue integral.