What Does an Infinite Function Value Contribute?
A nonnegative measurable function may take the value \(+\infty\), but the integral does not respond to that value in isolation: the measure of the set where it occurs matters. A single point has Lebesgue measure zero, so an infinite value there need not affect the integral. If the function is infinite on a set of positive measure, by contrast, its integral must be infinite. These statements concern the integral of the whole function and also apply to its restriction to any measurable set.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; no finiteness, completeness, or \(\sigma\)-finiteness assumption is imposed. The function \(f:X\to[0,\infty]\) is measurable, and its integral is the nonnegative Lebesgue integral, possibly \(+\infty\). Recall its definition as a supremum over nonnegative, finite-valued simple functions below \(f\):
For a simple function \(\varphi=\sum_{j=1}^k a_j\chi_{E_j}\), where the \(E_j\in\mathcal{M}\) are disjoint and \(a_j\ge0\) are finite, its integral is \(\int_X\varphi\,d\mu=\sum_{j=1}^k a_j\mu(E_j)\). In particular, \(\int_X c\chi_E\,d\mu=c\mu(E)\) for finite \(c\ge0\) and measurable \(E\), including when \(\mu(E)=+\infty\). For a measurable \(A\subseteq X\), the restricted integral means \(\int_A f\,d\mu=\int_X f\chi_A\,d\mu\).
The Set Where the Function Is Infinite
Proof. First suppose \(\mu(S)>0\). For every positive integer \(n\), \(n\chi_S\le f\) pointwise: on \(S\) the value of \(f\) is \(+\infty\), and on \(S^c\) the left side is zero. Monotonicity of the nonnegative integral gives \(\int_X f\,d\mu\ge n\mu(S)\). If \(\mu(S)=+\infty\), the right side is \(+\infty\) for every \(n\). If \(0<\mu(S)<+\infty\), the quantities \(n\mu(S)\) are unbounded as \(n\) increases. In either case, \(\int_X f\,d\mu=+\infty\).
Now suppose \(\mu(S)=0\), and put \(g=f\chi_{S^c}\). Since \(g\le f\), monotonicity gives \(\int_Xg\,d\mu\le\int_Xf\,d\mu\). For the reverse inequality, take any nonnegative simple \(\varphi\le f\). The simple function \(\varphi\chi_{S^c}\) satisfies \(0\le\varphi\chi_{S^c}\le g\). Also, \(\int_X\varphi\,d\mu=\int_X\varphi\chi_{S^c}\,d\mu\): the part removed is supported on \(S\), and each simple-function level set intersected with \(S\) has measure zero. Consequently, \(\int_X\varphi\,d\mu\le\int_Xg\,d\mu\). Taking the supremum over all such \(\varphi\) yields \(\int_X f\,d\mu\le\int_Xg\,d\mu\). The two integrals are equal. \(\square\)
The theorem gives an exact conclusion in the null-set case: \(f\) and the function obtained by replacing its values on \(S\) by zero have the same integral. It does not say that a null set makes the integral finite. Away from \(S\), the function can still have arbitrarily large values over enough measure to give an infinite integral.
Worked Example: An Infinite Value at a Lebesgue-Null Point
On \(X=[0,1]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\), define \(f(0)=+\infty\), \(f(x)=1\) for \(0<x\le\tfrac12\), and \(f(x)=0\) for \(\tfrac12<x\le1\). Compute \(\int_X f\,dm\).
Here \(S=\{0\}\), and \(m(S)=0\). Replacing the value at \(0\) by zero gives \(g=\chi_{(0,1/2]}\), so the theorem gives \(\int_X f\,dm=\int_Xg\,dm\). Since \(m((0,\tfrac12])=\tfrac12-0=\tfrac12\), the indicator integral is
Check: the other nonzero level of \(f\) is just the value \(+\infty\) at \(\{0\}\); its contribution is zero because that set has measure zero. The resulting value \(\tfrac12\) agrees with the measure of the interval on which the modified indicator equals one.
Worked Example: Infinity on a Set of Positive Measure
On \(X=[0,1]\) with Lebesgue measure \(m\), let \(f=+\infty\) on \([0,\tfrac13]\) and \(f=2\) on \((\tfrac13,1]\). Determine \(\int_X f\,dm\).
The set \(S=\{f=+\infty\}=[0,\tfrac13]\) has measure \(m(S)=\tfrac13-0=\tfrac13>0\). For each positive integer \(n\), the simple function \(n\chi_S\) lies below \(f\), so
As \(n\) increases, \(n/3\) is unbounded. Therefore \(\int_X f\,dm=+\infty\). Check: even the particular minorant \(3\chi_S\) has integral \(3\cdot\tfrac13=1\); the available lower bounds continue to grow without limit as the coefficient increases.
A Null Infinite Set Does Not Guarantee a Finite Integral
When \(\mu(S)=0\), the theorem reduces the integral to the part on \(S^c\), where \(f\) is finite-valued. But a finite value at every point does not imply a finite integral. The following example shows this directly from the supremum definition, without assuming a theorem about improper integrals.
Worked Example: A Finite-Valued Function with Infinite Integral
On \(X=[0,1]\) with Lebesgue measure, define \(f(0)=0\) and \(f(x)=1/x\) for \(0<x\le1\). The function is finite at every point, so \(S=\varnothing\). Show that its integral is nevertheless infinite.
For each positive integer \(N\), define the simple function
Each interval in the sum lies in \((0,1]\), and the intervals are pairwise disjoint. If \(x\in[1/(k+1),1/k)\), then \(x<1/k\), so \(1/x>k\); hence \(\varphi_N(x)\le f(x)\). Outside these intervals, \(\varphi_N(x)=0\le f(x)\). The length of the \(k\)-th interval is \(1/k-1/(k+1)=1/(k(k+1))\). Therefore
These partial sums are unbounded. Indeed, for each integer \(j\ge1\), the \(2^j\) terms with \(2^j+1\le k+1\le2^{j+1}\) are each at least \(1/2^{j+1}\), so their sum is at least \(2^j/2^{j+1}=1/2\). Including more such disjoint blocks makes the partial sums arbitrarily large. Since every \(\varphi_N\) is a simple minorant of \(f\), the supremum defining \(\int_X f\,dm\) is \(+\infty\).
Check: for \(N=2\), the computed minorant integral is \(1/2+1/3=5/6\); for \(N=3\), it is \(1/2+1/3+1/4=13/12\). These values increase, and the block estimate proves they are unbounded. Thus the integral is infinite even though \(f(x)\) is finite for every \(x\).
Restricted Integrals and a Common Mistake
The same reasoning applies to a measurable set \(A\subseteq X\). The function \(f\chi_A\) is measurable and takes the value \(+\infty\) exactly on \(S\cap A\). Applying the theorem to \(f\chi_A\) shows that \(\int_A f\,d\mu=+\infty\) if \(\mu(S\cap A)>0\); if \(\mu(S\cap A)=0\), the values \(+\infty\) on that restricted null set contribute nothing. This distinction is about measure, not the number of points: an uncountable set can be null, and a single point can have positive measure under some measures.
Common mistake: treating an infinite function value as if it automatically contributed \(+\infty\) to the integral. For Lebesgue measure on \([0,1]\), the function \(h(0)=+\infty\), \(h(x)=0\) for \(x>0\), has \(\{h=+\infty\}=\{0\}\) with measure zero. The theorem gives \(\int h\,dm=0\), not \(+\infty\). The correction is to check the measure of the entire set where the function is infinite. Conversely, measure zero only removes that set's contribution; the example \(f(x)=1/x\) above has no infinite values at all and still has infinite integral.
Check Your Understanding
In each question, specify whether the integral is finite or infinite and identify the role of the measure of the infinite-value set.
- On \([0,1]\) with Lebesgue measure, let \(f=+\infty\) on \(\{\tfrac12\}\) and \(f=3\) elsewhere. What is \(\int f\,dm\)?
- On \([0,1]\) with Lebesgue measure, let \(g=+\infty\) on \([\tfrac14,\tfrac12]\) and \(g=0\) elsewhere. Prove that \(\int g\,dm=+\infty\) using simple minorants.
- Can a function finite at every point have infinite integral? Explain using the minorant construction for \(x\mapsto1/x\) on \((0,1]\), with value zero at zero.
- For a measurable \(A\subseteq X\), what condition on \(A\cap\{f=+\infty\}\) forces \(\int_A f\,d\mu=+\infty\)?
- Explain why the expression “infinity times zero” is not a valid way to determine the integral over a null set where \(f=+\infty\).