Why Values on Null Sets Can Be Ignored
In Lebesgue integration, two functions may differ at some points and still have the same integral. The relevant condition is not that the exceptional set has few points, but that it has measure zero. This distinction matters: a singleton has Lebesgue measure zero on \(\mathbb{R}\), while it has counting measure one. We will prove the invariance statement directly from the definition of the nonnegative integral as a supremum of simple minorants.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space. No assumption that \(\mu\) is finite, sigma-finite, or complete is required. A null set means a set \(N\in\mathcal{M}\) with \(\mu(N)=0\). Functions in this tutorial are measurable and take values in \([0,\infty]\). A finite-valued nonnegative simple function is a measurable function with finite range contained in \([0,\infty)\). Its integral is computed by writing it as a finite sum \(\varphi=\sum_{j=1}^k a_j\chi_{E_j}\) on disjoint measurable sets, where \(a_j\geq0\), and setting \(\int\varphi\,d\mu=\sum_{j=1}^k a_j\mu(E_j)\).
The proof relies on a basic fact about simple functions: restricting a simple function to the complement of a null set does not change its integral. We show this explicitly, since it is the step that permits us to compare minorants of two functions that need not agree on the exceptional set.
Proof. Write \(\varphi=\sum_{j=1}^k a_j\chi_{E_j}\), where the \(E_j\) are disjoint members of \(\mathcal{M}\) and \(a_j\geq0\). Then \(\varphi\chi_{N^c}=\sum_{j=1}^k a_j\chi_{E_j\setminus N}\), also a finite-valued nonnegative simple function. For every \(j\), \(E_j\cap N\) is measurable and has measure zero, since \(E_j\cap N\subseteq N\). Countable additivity, applied to the disjoint union \(E_j=(E_j\setminus N)\cup(E_j\cap N)\), gives \(\mu(E_j\setminus N)=\mu(E_j)\), including when this value is infinite. Therefore \(\int\varphi\chi_{N^c}\,d\mu=\sum_{j=1}^k a_j\mu(E_j\setminus N)=\sum_{j=1}^k a_j\mu(E_j)=\int\varphi\,d\mu\). \(\square\)
Invariance Under Changes on Null Sets
Proof. Choose a measurable null set \(N\) such that \(f=g\) on \(N^c\). We first prove \(\int f\,d\mu\leq\int g\,d\mu\). Let \(\varphi\) be any finite-valued nonnegative simple function with \(0\leq\varphi\leq f\). Define \(\psi=\varphi\chi_{N^c}\). This is simple and measurable. On \(N^c\), \(\psi=\varphi\leq f=g\); on \(N\), \(\psi=0\leq g\), because \(g\) is nonnegative. Thus \(0\leq\psi\leq g\), so the definition of the integral gives \(\int\psi\,d\mu\leq\int g\,d\mu\). The lemma gives \(\int\varphi\,d\mu=\int\psi\,d\mu\). Hence every simple minorant \(\varphi\) of \(f\) has integral at most \(\int g\,d\mu\). Taking the supremum over all such \(\varphi\) yields \(\int f\,d\mu\leq\int g\,d\mu\). The same argument with \(f\) and \(g\) exchanged gives \(\int g\,d\mu\leq\int f\,d\mu\). The two inequalities imply equality, also when either side is \(+\infty\). \(\square\)
The proof does not discard the exceptional values by assuming the functions are bounded or that their integrals are finite. Instead, it takes each simple minorant of one function, sets it to zero on the null set, and obtains a minorant of the other function with exactly the same integral. This works even if the original function takes the value \(+\infty\) on the exceptional set.
Worked Example: Changing an Endpoint Under Lebesgue Measure
On \((\mathbb{R},\mathcal{L},m)\), let \(f=\chi_{[0,1]}\) and \(g=\chi_{(0,1]}\). These functions differ only at \(0\): \(f(0)=1\), whereas \(g(0)=0\). The exceptional set \(\{0\}\) is measurable and has Lebesgue measure zero, so \(f=g\) \(m\)-almost everywhere.
Both functions are simple, so their integrals are computed directly. The interval \([0,1]\) has Lebesgue measure \(1\), and the interval \((0,1]\) also has Lebesgue measure \(1\). Therefore $$ \int f\,dm=1\cdot m([0,1])=1\cdot1=1,\qquad \int g\,dm=1\cdot m((0,1])=1\cdot1=1. $$ Check: the theorem predicts equal integrals because the only differing point lies in a null set; the direct simple-function calculations give \(1=1\).
Worked Example: Changing a Finite Value to Infinity at One Point
On \((\mathbb{R},\mathcal{L},m)\), define \(f=\chi_{[0,1]}\), and define \(g\) to equal \(f\) except that \(g(1/2)=+\infty\). These functions are measurable and take values in \([0,\infty]\). They differ only on the measurable null set \(\{1/2\}\), so the theorem gives \(\int g\,dm=\int f\,dm\).
For \(f\), the simple-function formula gives \(\int f\,dm=1\cdot m([0,1])=1\). We can also verify the value for \(g\) directly from the supremum definition. The simple function \(\chi_{[0,1]\setminus\{1/2\}}\) is at most \(g\) everywhere and has integral \(m([0,1]\setminus\{1/2\})=1\), so \(\int g\,dm\geq1\). Conversely, let \(\varphi\) be any finite-valued nonnegative simple function with \(\varphi\leq g\). Outside \([0,1]\), \(g=0\), so \(\varphi=0\) there. On \([0,1]\setminus\{1/2\}\), \(g=1\), so \(\varphi\leq1\). Its value at \(1/2\) contributes no integral: for any simple representation, the part supported on \(\{1/2\}\) has integral zero because \(m(\{1/2\})=0\). Consequently \(\int\varphi\,dm\leq m([0,1])=1\). Taking the supremum shows \(\int g\,dm\leq1\), hence \(\int g\,dm=1\).
Check: the lower bound is \(1\) and the upper bound is \(1\), so the direct calculation agrees with the theorem. This example also shows why the extended value \(+\infty\) at a null point is permitted.
Worked Example: A Dirac Measure Ignores Changes Away from Its Atom
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)\), let \(f=\chi_{[0,1]}\) and \(g=\chi_{[-2,3]}\), where \(\delta_0(E)=1\) if \(0\in E\) and \(\delta_0(E)=0\) if \(0\notin E\). The functions differ on \([-2,0)\cup(1,3]\), which excludes \(0\). That set has \(\delta_0\)-measure zero. Both functions equal \(1\) at \(0\), so they agree \(\delta_0\)-almost everywhere.
For an indicator, the simple-function integral is \(\int\chi_E\,d\delta_0=\delta_0(E)\). Since \(0\in[0,1]\), \(\int f\,d\delta_0=\delta_0([0,1])=1\). Since \(0\in[-2,3]\), \(\int g\,d\delta_0=\delta_0([-2,3])=1\). Check: each set contains the atom \(0\), so both values are \(1\), as the theorem requires.
When a Point Change Does Matter
A common mistake is to reason that changing a function at one point never changes its integral. That statement is false without specifying the measure. Under counting measure on \((\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)\), where \(\mathbb{N}=\{1,2,\ldots\}\) and \(\mu(E)\) is the number of elements of \(E\), the singleton \(\{1\}\) has measure \(1\), not \(0\).
Worked Example: A Singleton Is Not Null for Counting Measure
Let \(f=\chi_{\{1\}}\) and \(g=0\) on \(\mathbb{N}\), with counting measure \(\mu\). The functions differ exactly on \(\{1\}\), and this is not a null set because \(\mu(\{1\})=1\). Their integrals are $$ \int f\,d\mu=1\cdot\mu(\{1\})=1\cdot1=1,\qquad \int g\,d\mu=0. $$ Incorrect calculation: “They differ at only one point, so their integrals are equal.” This treats a singleton as though it had measure zero. Correction: the invariance theorem applies only when the difference is confined to a measurable null set. Here the exceptional set has measure \(1\), and the direct integrals are \(1\) and \(0\).
More generally, the relevant set is the set where the functions actually differ, not merely a set one informally regards as small. For measurable extended nonnegative functions, if they are equal \(\mu\)-almost everywhere, their discrepancy is contained in a measurable null set and the theorem applies. In a measure space with atoms, a point carrying positive measure cannot be ignored on that basis. In a space with Lebesgue measure, individual points are null, but not every small-looking set is null.
The simple-minorant argument is useful because it establishes invariance before any finiteness or integrability assumptions are imposed. It concerns nonnegative measurable functions and their extended nonnegative integrals; it does not assert that the functions are equal everywhere, nor does it depend on the measure space being complete. The equality of integrals follows precisely from the nullity of the exceptional set.
Check Your Understanding
For each question, identify the measure space and verify whether the exceptional set is null before applying the theorem.
- On \(\mathbb{R}\) with Lebesgue measure, compare \(\chi_{[2,5]}\) and \(\chi_{(2,5]}\). What are their integrals, and which set contains their discrepancy?
- State the simple-minorant modification used to prove \(\int f\,d\mu\leq\int g\,d\mu\) when \(f=g\) outside a measurable null set \(N\).
- On \(\mathbb{N}\) with counting measure, can two functions differing only at \(n=4\) be declared equal almost everywhere? Explain using the measure of \(\{4\}\).
- On \(\mathbb{R}\) with Lebesgue measure, let \(f=0\) everywhere and let \(g(0)=+\infty\), \(g(x)=0\) for \(x\neq0\). What does the theorem imply about their integrals?
- Does the invariance theorem require either integral to be finite, or the measure space to be complete? Give the relevant hypotheses precisely.