Common Errors With Nonnegative Integrals

Four Checks Before Trusting an Integral Calculation

Many errors with nonnegative integrals do not come from difficult estimates. They come from changing what a representation means, treating equality almost everywhere as equality at every point, manipulating \(0\cdot\infty\) as an ordinary product, or assuming that a supremum is attained. These distinctions matter on arbitrary measure spaces: a set of measure zero may be irrelevant to an integral, but it need not be empty, and an infinite measure space can make constant-function formulas especially easy to misuse.

Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it is not assumed finite, sigma-finite, or complete. All functions in this tutorial are measurable and take values in \([0,\infty]\), except where a finite-valued simple function is specified. The integral is the nonnegative Lebesgue integral. We use the previously established facts that the integral is monotone and additive for nonnegative functions, that restricting an integral to a measurable set means integrating the function times its indicator, and that a nonnegative function with integral zero vanishes \(\mu\)-a.e.

Overlapping Indicators: A Sum Is Not a Disjoint Level-Set List

Definition: A finite-valued nonnegative simple function can be written in disjoint level-set form \(\varphi=\sum_{j=1}^k a_j\chi_{A_j}\), where the measurable sets \(A_j\) are pairwise disjoint and the coefficients \(a_j\) are nonnegative. In that form, \(\int\varphi\,d\mu=\sum_{j=1}^k a_j\mu(A_j)\). A sum \(\sum_i a_i\chi_{E_i}\) with overlapping \(E_i\) is also a valid expression for a function, but its terms need not be its disjoint level sets.

There are two correct ways to integrate a finite sum of nonnegative indicators. One can use additivity to integrate each summand, or one can first partition the space into disjoint sets on which the whole sum has a fixed value. The mistake is to use the disjoint-level-set formula while treating overlapping sets as if each carried just its own coefficient.

Worked Example: Disjointifying Two Overlapping Sets

On \(X=[0,3]\) with Lebesgue measure \(m\), let \(E=[0,2]\), \(F=[1,3]\), and \(f=\chi_E+2\chi_F\). Compute the integral in both ways.

The function equals \(1\) on \([0,1)\), \(3\) on \([1,2]\), \(2\) on \((2,3]\), and \(0\) nowhere else in \(X\). These three level sets are disjoint. Their measures are \(1\), \(1\), and \(1\), respectively, so

\int_X f\,dm=1\cdot 1+3\cdot 1+2\cdot 1=6.

Alternatively, additivity and positive homogeneity give \(\int_X f\,dm=\int_X\chi_E\,dm+\int_X2\chi_F\,dm=m(E)+2m(F)\). Here \(m(E)=2-0=2\) and \(m(F)=3-1=2\), so the result is \(2+2\cdot2=6\). Check: the disjoint-level-set computation gives \(1+3+2=6\), the same value as the additive computation \(2+4=6\). The set endpoints do not change these measures.

The additive computation is valid precisely because \(f\) is the sum of two functions. It would be incorrect to describe \(f\) as taking value \(1\) throughout \(E\) and value \(2\) throughout \(F\): on \(E\cap F=[1,2]\), its value is \(1+2=3\), not either of those two values alone.

Proposition: If \(E_1,\ldots,E_r\in\mathcal{M}\) and \(a_1,\ldots,a_r\ge0\) are finite, then the function \(f=\sum_{i=1}^r a_i\chi_{E_i}\) has integral \(\int f\,d\mu=\sum_{i=1}^r a_i\mu(E_i)\), even if the sets overlap. If one instead uses its disjoint level sets, the coefficient on each part is the sum of the \(a_i\) for the sets containing that part.

Proof. By additivity for nonnegative measurable functions, \(\int f\,d\mu=\sum_{i=1}^r\int a_i\chi_{E_i}\,d\mu\). Positive homogeneity and the indicator integral formula give \(\int a_i\chi_{E_i}\,d\mu=a_i\mu(E_i)\); if \(a_i=0\) and \(\mu(E_i)=\infty\), this term is the integral of the zero function and equals \(0\). For the disjoint description, for each subset \(J\subseteq\{1,\ldots,r\}\) form the measurable set \(A_J=(\bigcap_{i\in J}E_i)\cap(\bigcap_{i\notin J}E_i^c)\), with an empty intersection interpreted as \(X\). These sets partition \(X\). On \(A_J\), the function has value \(\sum_{i\in J}a_i\). Thus its disjoint level-set description assigns that summed value to each \(A_J\), combining sets with the same value if desired. This proves both claims. \(\square\)

Almost Everywhere Is Not Everywhere

Two measurable functions are equal \(\mu\)-a.e. when the set on which they differ has measure zero. For nonnegative measurable functions, equality \(\mu\)-a.e. implies equality of their integrals, including when the common integral is infinite. It does not imply that the functions agree at every point.

Proposition: Let \(f,g:X\to[0,\infty]\) be measurable on \((X,\mathcal{M},\mu)\). If \(f=g\) \(\mu\)-a.e., then \(\int f\,d\mu=\int g\,d\mu\). The conclusion is an equality of integrals; it need not be true that \(f=g\) everywhere.

Proof. Let \(N=\{x:f(x)\ne g(x)\}\), which is measurable and has \(\mu(N)=0\). On \(N^c\), the functions agree pointwise. For a nonnegative measurable function \(h\), its integral over \(N\) is zero: every nonnegative simple function bounded above by \(h\chi_N\) has integral zero, since its nonzero level sets are contained in \(N\). Hence \(\int_N f\,d\mu=\int_N g\,d\mu=0\). Decomposing each integral over \(N\) and \(N^c\), using additivity on disjoint measurable sets, gives \(\int f\,d\mu=\int_{N^c}f\,d\mu=\int_{N^c}g\,d\mu=\int g\,d\mu\). This argument permits the common value to be \(+\infty\). \(\square\)

Worked Example: A Pointwise Difference on a Null Set

On \(\mathbb{R}\) with Lebesgue measure \(m\), let \(f=\chi_{\{0\}}\) and \(g=0\). Both functions are measurable. They differ at \(0\), so they are not equal everywhere, but \(m(\{0\})=0\). Therefore

\int_{\mathbb{R}} f\,dm=1\cdot m(\{0\})=0, \qquad \int_{\mathbb{R}} g\,dm=0.

Check: the functions differ on exactly one point, a set of measure zero, and their integrals are both \(0\). For contrast, under counting measure on \(X=\{0\}\), the same functions have integrals \(1\) and \(0\): the singleton has counting measure \(1\), not \(0\). Thus the relevance of a point depends on the measure, not just on the fact that it is a single point.

Zero Times Infinity Requires Looking at the Function

The expression \(0\cdot\infty\) is not an ordinary real product. In integral calculations it can appear if one mechanically combines the coefficient of a constant function with the measure of its domain. The safe rule is to identify the function first. In particular, the function that is identically zero has integral zero on every measure space, including one with \(\mu(X)=\infty\). This is the convention \(0\cdot\mu(E)=0\), including when \(\mu(E)=\infty\), in the indicator formula for the zero function.

Worked Example: An Infinite Value on a Null Set

On \(\mathbb{R}\) with Lebesgue measure, define \(f(0)=\infty\) and \(f(x)=0\) for \(x\ne0\). This is a nonnegative measurable function. It equals the zero function \(m\)-a.e., since the exceptional set \(\{0\}\) has measure zero. The a.e.-invariance result therefore gives \(\int_{\mathbb{R}}f\,dm=0\).

The same conclusion follows directly from the definition of the nonnegative integral: every nonnegative simple function \(\varphi\le f\) must vanish off \(\{0\}\), and every measurable subset of \(\{0\}\) has measure zero. Thus each such \(\varphi\) has integral zero, and the supremum of these integrals is zero. Check: \(f\) is not the zero function pointwise, but it is zero outside a null set and its integral is \(0\). Writing \(\infty\cdot m(\{0\})\) does not compute this integral; the function-based argument does.

A related error is to infer an infinite integral merely because a function takes the value \(+\infty\) somewhere. The example shows that an infinite value on a null set contributes nothing. Conversely, an infinite value on a set of positive measure forces an infinite integral: for every finite \(M>0\), the function is at least \(M\) on that set, so monotonicity bounds its integral below by \(M\) times the measure of the set. One must check the measure of the set, not just the value at a point.

A Supremum Need Not Be a Maximum

For a nonnegative measurable function, the integral can be defined as the supremum of the integrals of nonnegative simple functions lying below it. A supremum is the least upper bound; it need not be the value of any member of the family. Approximating values from below can establish the supremum without producing a simple function that attains it.

Worked Example: Approximating the Integral of \(x\) Without Attaining It

Take \(X=[0,1]\) with Lebesgue measure and \(f(x)=x\). For \(n\ge1\), define \(\varphi_n(x)=2^{-n}\lfloor2^nx\rfloor\) for \(0\le x<1\), and set \(\varphi_n(1)=1\). Each \(\varphi_n\) is a finite-valued measurable simple function and \(0\le\varphi_n\le f\). For \(k=0,\ldots,2^n-1\), it equals \(k/2^n\) on an interval of measure \(2^{-n}\); its value at the endpoint \(1\) does not affect the integral. Hence

\int_0^1\varphi_n\,dm =\sum_{k=0}^{2^n-1}\frac{k}{2^n}\,2^{-n} =\frac{(2^n-1)2^n/2}{4^n} =\frac{2^n-1}{2^{n+1}}.

These values approach \(1/2\). Since \(\varphi_n\le x\), monotonicity bounds them above by \(\int_0^1x\,dm=1/2\), the usual Lebesgue integral of \(x\) (which agrees with its elementary Riemann integral). Thus the supremum of the integrals of simple functions below \(x\) is \(1/2\). Check: substituting \(n=1\) gives \((2-1)/2^2=1/4\); substituting \(n=2\) gives \((4-1)/2^3=3/8\). Both are below \(1/2\), and the displayed formula tends to \(1/2\).

No finite-valued simple function below \(x\) attains that supremum. If such a \(\varphi\) had integral \(1/2\), then additivity applied to \(x=\varphi+(x-\varphi)\) would give \(\int_0^1(x-\varphi)\,dm=0\). The zero-integral criterion implies \(x=\varphi\) \(m\)-a.e. But a finite-valued simple function takes only finitely many values, so \(\{x\in[0,1]:x=\varphi(x)\}\) is contained in a finite set. A finite set has Lebesgue measure zero, contradicting equality almost everywhere on an interval of measure \(1\). The supremum is therefore not a maximum.

A Practical Error Check

Before accepting a calculation of a nonnegative integral, ask what each expression represents. For overlapping indicators, decide whether the formula is a sum of functions or a list of disjoint level sets. For an equality claim, distinguish equality everywhere from equality \(\mu\)-a.e. For a product involving zero and infinity, return to the function and its values on the relevant sets. For an approximation argument, show both that the candidate is an upper bound and that admissible values approach it; do not claim attainment without producing a member that attains it.

Check Your Understanding

For each question, state the measure space and identify which distinction prevents the tempting shortcut.

On \([0,2]\) with Lebesgue measure, let \(h=\chi_{[0,1]}+\chi_{(1/2,\,3/2]}\). Find its disjoint level sets and compute its integral.
On \(\mathbb{R}\) with Lebesgue measure, compare the integrals of \(2\chi_{\{5\}}\) and the zero function. Are the functions equal everywhere? Are they equal almost everywhere?
On a measure space with \(\mu(X)=\infty\), what is the integral of the identically zero function? Explain why the expression \(0\cdot\infty\) is not an ordinary multiplication step.
For \(f(x)=x\) on \([0,1]\), why does a sequence of simple functions whose integrals approach \(1/2\) not by itself show that any one of them has integral \(1/2\)?
Under counting measure on a singleton, does changing a function at that point preserve its integral? Explain why the Lebesgue-measure example with a singleton does not give the same answer.