The Value of a Constant Integral
A constant function has only one value, but its integral depends on the measure of the space on which it is integrated. On a finite-measure space, the integral is the constant multiplied by the total measure. On a space of infinite measure, every strictly positive finite constant has infinite integral. The zero constant is different: its integral remains zero even when the space has infinite measure.
Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space; it need not be complete, finite, or sigma-finite. Let \(c\) be a finite real number with \(c\ge0\), and define \(f:X\to[0,\infty)\) by \(f(x)=c\) for every \(x\in X\). The function is measurable: for any Borel set \(B\subseteq[0,\infty)\), its inverse image is either \(X\) or \(\varnothing\). Integrals of nonnegative functions are allowed to take the value \(+\infty\).
Proof. If \(c=0\), then \(f\) is the zero function. Its nonnegative integral is zero by the definition of the integral of a simple function (equivalently, its only positive level sets are empty). This gives the first case, including when \(\mu(X)=+\infty\). If \(c>0\), then \(f=c\chi_X\), a nonnegative simple function. The simple-function integral formula gives \(\int_X c\chi_X\,d\mu=c\,\mu(X)\). If \(\mu(X)<+\infty\), this is the ordinary finite product, including the value zero if \(\mu(X)=0\). If \(\mu(X)=+\infty\), the positive coefficient \(c\) multiplied by infinite measure gives \(+\infty\). Thus the stated cases hold. If \(X=\varnothing\), then \(\mu(X)=0\) and the function is identically zero, which is also covered. \(\square\)
The formula involves the measure of the whole space, not the number of points in it. A space can contain infinitely many points and still have finite total measure; conversely, an infinite total measure forces the integral of every strictly positive finite constant to be infinite.
Integrating over a Measurable Part of the Space
The same calculation applies when the constant is integrated only over a measurable set. The integral over \(E\in\mathcal{M}\) means \(\int_X f\chi_E\,d\mu\). The function \(f\chi_E\) is zero outside \(E\) and equals \(c\) on \(E\), so its integral is controlled by \(\mu(E)\), rather than by \(\mu(X)\).
Proof. By definition, \(\int_E c\,d\mu=\int_X c\chi_E\,d\mu\). If \(c=0\), this is the integral of the zero function and equals zero. If \(c>0\), \(c\chi_E\) is a nonnegative simple function, so its integral is \(c\mu(E)\). This remains true when \(\mu(E)=+\infty\), in which case the value is \(+\infty\). \(\square\)
One immediate consequence identifies exactly when a positive constant has finite integral. For \(c>0\), the theorem gives \(\int_X c\,d\mu<+\infty\) if and only if \(\mu(X)<+\infty\). The positivity assumption matters: for \(c=0\), the integral is finite regardless of whether the space has finite measure.
Worked Calculations
Worked Example: A Constant on a Finite Lebesgue Interval
Let \(X=[-2,3]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra restricted to \(X\), and let \(\mu\) be Lebesgue measure \(m\). Compute the integral of the constant function \(f(x)=7\).
The interval has length \(3-(-2)=5\), so \(m(X)=5\). Since \(7>0\), the theorem gives $$ \int_{[-2,3]}7\,dm=7\,m([-2,3])=7\cdot5=35. $$ This calculation uses only that \(7\chi_X\) is a simple function. As a second check, split the interval at \(0\): the two pieces have measures \(2\) and \(3\). The simple-function integral over the pieces is \(7\cdot2+7\cdot3=14+21=35\), agreeing with the whole-interval calculation.
Worked Example: A Constant under a Dirac Measure
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\), and \(\mu=\delta_0\), the Dirac measure at \(0\). Thus \(\delta_0(X)=1\). For the constant \(f(x)=\tfrac52\), compute its integral.
The total measure is \(1\), and the constant is strictly positive. Therefore $$ \int_{\mathbb{R}}\frac52\,d\delta_0 =\frac52\,\delta_0(\mathbb{R}) =\frac52\cdot1 =\frac52. $$ As a check using the defining point-mass property of the Dirac integral, the integral of a nonnegative measurable function is its value at \(0\); here \(f(0)=\tfrac52\). The result does not depend on the fact that \(\mathbb{R}\) has infinitely many points: the relevant quantity is \(\delta_0(\mathbb{R})\), which is finite.
Worked Example: A Positive Constant on an Infinite-Measure Space
On \(X=\mathbb{R}\) with Lebesgue measure \(m\), let \(f(x)=\tfrac34\). Show that its integral is infinite.
Since \(m(\mathbb{R})=+\infty\) and \(\tfrac34>0\), the simple-function formula gives $$ \int_{\mathbb{R}}\frac34\,dm =\frac34\,m(\mathbb{R}) =+\infty. $$ The conclusion can also be checked by restricting to bounded intervals. For each positive integer \(N\), the interval \([-N,N]\) has measure \(2N\), and hence $$ \int_{[-N,N]}\frac34\,dm =\frac34\cdot 2N =\frac{3N}{2}. $$ Given any finite threshold \(M\ge0\), choose a positive integer \(N>\tfrac{2M}{3}\). Then \(\tfrac{3N}{2}>M\). The integral over the whole line is at least the integral over this interval, so it exceeds every finite threshold and must be \(+\infty\). This agrees with the direct simple-function calculation.
Worked Example: The Zero Constant on an Infinite-Measure Space
Again take \(\mathbb{R}\) with Lebesgue measure, but now let \(f(x)=0\) for every \(x\). What is its integral?
The function is the zero function, so its nonnegative integral is $$ \int_{\mathbb{R}}0\,dm=0. $$ This conclusion follows directly from the zero-function case in the theorem; no finite-measure assumption is needed. It can also be checked on every bounded interval: for each positive integer \(N\), the integral over \([-N,N]\) is \(0\cdot 2N=0\). The whole-space integral remains zero, not \(+\infty\).
Worked Example: A Positive Constant on a Null Space
Let \(X=\{a,b\}\), let \(\mathcal{M}\) be the power set of \(X\), and define \(\mu(A)=0\) for every \(A\in\mathcal{M}\). This is a measure space with \(\mu(X)=0\). Compute the integral of the constant \(f=5\).
The function is the simple function \(5\chi_X\), so $$ \int_X5\,d\mu=5\,\mu(X)=5\cdot0=0. $$ For a direct check, its only nonzero level has coefficient \(5\) on a set of measure \(0\); the simple-function integral is therefore \(5\cdot0=0\). Thus a strictly positive constant need not have a positive integral when the whole space has measure zero.
Common Pitfalls and the Role of the Assumptions
Pitfall: treating \(0\cdot(+\infty)\) as \(+\infty\). A mistaken calculation might say $$ \int_{\mathbb{R}}0\,dm=0\cdot m(\mathbb{R})=0\cdot(+\infty)=+\infty. $$ That conclusion is wrong. The zero function has integral zero by its definition, so the correct convention in the compact product formula is \(0\cdot(+\infty)=0\). The separate cases in the theorem prevent this ambiguity.
Pitfall: assuming an infinite domain alone makes the integral infinite. The set \(\mathbb{R}\) is infinite, but with Dirac measure its total measure is \(1\), so the constant \(\tfrac52\) has integral \(\tfrac52\), as computed above. What matters is total measure. Lebesgue measure on \(\mathbb{R}\) is infinite, while Dirac measure on \(\mathbb{R}\) is finite.
Pitfall: overlooking the strict positivity condition. It is correct that a positive constant on an infinite-measure space has infinite integral; it is not correct to apply that statement to \(c=0\). For example, the zero constant on \(\mathbb{R}\) has integral \(0\), whereas the constant \(\tfrac34\) has integral \(+\infty\).
Finally, finite measure and sigma-finiteness are different conditions. Finite measure means \(\mu(X)<+\infty\). Sigma-finiteness means that \(X\) is a countable union of measurable sets of finite measure; it does not require \(\mu(X)\) itself to be finite. Lebesgue measure on \(\mathbb{R}\) is sigma-finite but has infinite total measure, and therefore every strictly positive finite constant on it has infinite integral. The constant-integral formula requires neither finiteness nor sigma-finiteness.
Check Your Understanding
For each question, identify the total measure and check whether the constant is zero or strictly positive before applying the formula.
- On a measure space with \(\mu(X)=\tfrac{7}{3}\), compute the integral of the constant \(c=6\).
- On a space with \(\mu(X)=+\infty\), what is the integral of the constant \(c=0\)? Explain why the product convention matters.
- Suppose \(E\in\mathcal{M}\), \(\mu(E)=0\), and \(c=12\). Compute \(\int_E c\,d\mu\).
- For a finite constant \(c>0\), prove that \(\int_X c\,d\mu<+\infty\) if and only if \(\mu(X)<+\infty\).
- Can an infinite underlying set have finite integral for a positive constant? Give a measure-space example and justify it using the total measure.