What Does It Mean to Change a Function on a Null Set?
Lebesgue integration often identifies functions that agree except on a set of measure zero. This identification is useful, but it does not mean that every property of a function is unchanged: an integral ignores a null set, while a pointwise value does not. Measurability also requires care, because a subset of a null set need not be measurable unless the measure space is complete.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, with no assumption that \(\mu\) is finite or sigma-finite unless stated. A set \(N\in\mathcal{M}\) is a null set if \(\mu(N)=0\). We say that two functions \(f,g\) agree \(\mu\)-almost everywhere, written \(f=g\) \(\mu\)-a.e., if there is a measurable null set \(N\) such that \(f(x)=g(x)\) for every \(x\in N^c\). A measure space is complete if every subset of every measurable null set belongs to \(\mathcal{M}\).
We will distinguish three questions: whether a modification remains measurable; whether a measurable function’s integral or norm changes; and whether a pointwise statement remains true. They have different answers.
Measurability and Completeness
Proof. It suffices to treat the extended real-valued case. Let \(N\in\mathcal{M}\) be null and suppose \(f=g\) on \(N^c\). For any real \(a\), the sets \(\{g>a\}\) and \(\{f>a\}\) agree outside \(N\). Their symmetric difference is therefore a subset of \(N\). Since \(f\) is measurable, \(\{f>a\}\in\mathcal{M}\); completeness implies that every subset of \(N\) is measurable. Thus \(\{g>a\}\in\mathcal{M}\). The sets \((a,\infty]\), \(a\in\mathbb{R}\), generate the Borel sigma-algebra of \([-\infty,\infty]\), so measurability of all these inverse images proves that \(g\) is measurable. For complex-valued functions, apply the same argument to the real and imaginary parts, or use that their inverse images of open sets differ from those of \(f\) only on \(N\). \(\square\)
Completeness is a sufficient condition, not a detail to omit. On an incomplete space, a change on a measurable null set can produce a nonmeasurable function. If both \(f\) and \(g\) are already known to be measurable, integral invariance below does not require completeness.
Worked Example: A Null-Set Change Can Destroy Measurability
Let \(X=[0,1]\), let \(\mathcal{M}\) be the Borel subsets of \([0,1]\), and let \(\mu\) be Lebesgue measure restricted to \(\mathcal{M}\). This is a finite but incomplete measure space. Let \(C\) be the Cantor set, so \(C\) is Borel and \(\mu(C)=0\). Choose a subset \(A\subset C\) that is not Borel; such a subset exists because \(C\) has strictly more subsets than there are Borel sets. Define \(f=0\) everywhere and \(g=\chi_A\).
The function \(f\) is Borel measurable. Also \(f=g\) outside \(C\), a measurable null set. However, \(g^{-1}(\{1\})=A\notin\mathcal{M}\), so \(g\) is not measurable. Check: the failure is exactly the missing completeness condition: \(A\) is a subset of the measurable null set \(C\), but is not in \(\mathcal{M}\). In the completed Lebesgue sigma-algebra, \(A\) would be measurable and this particular modification would be measurable.
Integrals and Norms Ignore Null-Set Values
Proof. Choose a measurable null set \(N\) outside which \(f=g\). For any nonnegative measurable function \(h\) that vanishes on \(N^c\), its integral is zero: every nonnegative simple function \(\varphi\le h\) vanishes on \(N^c\), and each nonzero term in its simple-function integral is supported on a measurable subset of \(N\), hence contributes zero. Taking the supremum over such \(\varphi\) gives \(\int h\,d\mu=0\).
Now write \(f=f\chi_{N^c}+f\chi_N\) and \(g=g\chi_{N^c}+g\chi_N\). The first terms are equal everywhere, since \(f=g\) on \(N^c\), and the second terms have integral zero by the preceding argument. Additivity of the nonnegative integral gives \(\int f\,d\mu=\int g\,d\mu\). For real-valued integrable functions, apply the nonnegative result to the positive and negative parts \(f^+,f^-\) and \(g^+,g^-\), which agree a.e.; the signed integrals are the differences of the finite integrals of these parts. For complex-valued integrable functions, apply this real-valued argument to the real and imaginary parts. \(\square\)
This result concerns integrals, not the separate property of being measurable. Its hypotheses explicitly require both functions to be measurable. The theorem also explains why integrability is invariant for measurable functions: if two such functions agree a.e., one has finite absolute integral exactly when the other does.
Proof. For \(1\le p<\infty\), the nonnegative measurable functions \(|f|^p\) and \(|g|^p\) agree a.e., so the theorem gives \(\int |f|^p\,d\mu=\int |g|^p\,d\mu\). Taking the \(p\)-th root proves the assertion, including whether this value is finite. For the essential supremum, define it as the infimum of those \(K\ge0\) for which \(\mu(\{|f|>K\})=0\), with value \(+\infty\) if no such finite \(K\) exists. Since the level sets for \(f\) and \(g\) agree outside a null set, their measures are equal for each fixed \(K\), so the same \(K\)'s qualify for both functions. Their essential suprema are equal. \(\square\)
Worked Example: The Integral and Norms Can Stay Fixed
On \([0,1]\) with Lebesgue measure \(m\), let \(f(x)=1\) for every \(x\). Define \(g(x)=2\) for rational \(x\in[0,1]\), and \(g(x)=1\) for irrational \(x\in[0,1]\). Both functions are measurable, and they differ only on the countable null set \(\mathbb{Q}\cap[0,1]\).
Since \(m([0,1])=1\) and \(m(\mathbb{Q}\cap[0,1])=0\),
For every \(1\le p<\infty\), \(\|f\|_p=(\int_0^1 1^p\,dm)^{1/p}=1\), and \(\|g\|_p=(2^p\cdot0+1^p\cdot1)^{1/p}=1\). Both essential supremum norms equal \(1\), because \(\{|g|>1\}=\mathbb{Q}\cap[0,1]\) has measure zero, while \(\{|g|>K\}=[0,1]\) for \(0\le K<1\). Yet the pointwise supremum of \(g\) is \(2\), whereas the pointwise supremum of \(f\) is \(1\). Check: the integral and essential norms ignore the rational values, but the pointwise supremum includes them.
What Is Not Preserved Pointwise?
Almost-everywhere equality preserves statements formulated up to null sets, but it does not preserve statements about every point. The following distinctions are useful:
| Property or statement | Preserved under measurable a.e. equality? | Reason |
|---|---|---|
| Integral and \(L^p\) norm | Yes, when defined as above | The integrands agree outside a null set. |
| Essential supremum | Yes | For each fixed threshold \(K\), the strict tail sets \(\{|f|>K\}\) and \(\{|g|>K\}\) have the same measure. |
| Pointwise values or pointwise supremum | No | A value on the exceptional set can change. |
| Continuity at every point | No | A null-set modification can create a discontinuity. |
| Boundedness everywhere | No | Values on a countable null set can be made unbounded. |
Worked Example: An Integral Does Not Detect Pointwise Unboundedness
On \([0,1]\) with Lebesgue measure \(m\), let \(f(x)=0\). Define \(g(1/n)=n\) for each positive integer \(n\), and \(g(x)=0\) at all other points. The set \(N=\{1/n:n\ge1\}\) is countable, hence measurable with \(m(N)=0\). The function \(g\) is measurable: for any real \(a\), its superlevel set \(\{g>a\}\) is a subset of the countable set \(N\), hence is Borel.
Both functions are nonnegative and agree outside \(N\), so the invariance theorem gives \(\int_0^1g\,dm=\int_0^1f\,dm=0\). Directly, \(g=0\) off \(N\), and the integral over \(N\) is zero because \(N\) has measure zero. Nevertheless \(g(1/n)=n\), so \(g\) is unbounded, while \(f\) is bounded. Check: for every finite \(K\ge0\), choose an integer \(n>K\); then \(1/n\in\{g>K\}\), so \(g\) is not bounded everywhere. The set \(\{g>K\}\) is still countable and null, which is why \(\|g\|_\infty=0=\|f\|_\infty\) despite pointwise unboundedness.
The distinction also applies to sequences. If \(f_n=g_n\) outside a null set \(N_n\) for each \(n\), and each \(N_n\) is measurable and null, then \(N=\bigcup_{n=1}^{\infty}N_n\) is measurable and null by countable subadditivity. Off \(N\), the two sequences agree term by term. Consequently, any pointwise convergence statement for one sequence holds for the other at every point outside \(N\); in particular, almost-everywhere convergence is unchanged by making separate null-set modifications to countably many terms. Pointwise convergence at every point and uniform convergence need not be preserved.
Worked Example: Almost-Everywhere Convergence Need Not Be Pointwise Everywhere
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=0\) for every \(n\) and every \(x\). Define \(g_n(x)=0\) when \(x\ne0\), and \(g_n(0)=(-1)^n\). Each \(g_n\) is measurable and agrees with \(f_n\) off the null set \(\{0\}\).
For every \(x\ne0\), \(g_n(x)=0\) for all \(n\), so \(g_n(x)\to0\). Thus \(g_n\to0\) almost everywhere. At \(x=0\), the sequence alternates between \(1\) and \(-1\), so it does not converge there. Also \(\sup_{x\in[0,1]}|g_n(x)|=1\) for every \(n\), so \(g_n\) does not converge uniformly to zero. In contrast, \(f_n\to0\) pointwise and uniformly. Check: the exceptional set is exactly \(\{0\}\), and \(m(\{0\})=0\); the a.e. convergence conclusion holds, while the stronger pointwise-everywhere and uniform conclusions fail.
A Common Mistake
A tempting but incorrect argument is: “The functions differ only on a null set, so the modified function is measurable and all its properties are unchanged.” The integral conclusion requires measurability of both functions; the measurability conclusion requires completeness if values are changed arbitrarily on a measurable null set; and pointwise properties can change even on a complete space.
Wrong calculation: in the Borel example above, one might write \(g^{-1}(\{1\})=A\subset C\), then infer that \(g^{-1}(\{1\})\) is Borel because \(C\) is a Borel null set. This inference is invalid: Borel sigma-algebras are closed under Borel subsets of Borel sets, not under arbitrary subsets of them. Here \(A\) was chosen non-Borel. Correction: either work on a complete measure space, where every subset of \(C\) is measurable, or verify the inverse images of Borel sets directly before calling the modified function measurable. Once both functions are measurable and equal a.e., integral invariance applies; it does not supply measurability that was not established.
Check Your Understanding
For each question, distinguish assertions about all points from assertions that hold almost everywhere.
- On a complete measure space, prove that changing a measurable real-valued function on a measurable null set preserves measurability.
- Give an incomplete measure space and a function modification on a measurable null set that is not measurable.
- If measurable nonnegative functions \(f\) and \(g\) agree a.e., explain why their integrals are equal even when the common value is \(+\infty\).
- Can two measurable functions that agree a.e. have different pointwise suprema but the same essential supremum? Give an example.
- If each term of a sequence is modified on its own measurable null set, why is almost-everywhere convergence preserved, and why need not pointwise-everywhere convergence be preserved?