Representatives and the Meaning of Almost Everywhere
Lebesgue integration and many convergence statements ignore changes on sets of measure zero. Pointwise convergence at every point does not: altering even one value can change whether a sequence converges at that point. The distinction is important when functions are treated as equivalence classes, as they are in spaces such as \(L^p(\mu)\). The central fact here is that almost-everywhere convergence depends only on the almost-everywhere equivalence classes of the sequence and its proposed limit.
Let \((X,\mathcal{M},\mu)\) be a measure space; it need not be finite, sigma-finite, or complete. All functions in this tutorial are measurable and real-valued. A set \(N\in\mathcal{M}\) is a null set if \(\mu(N)=0\). Two functions \(f,g:X\to\mathbb{R}\) are equal \(\mu\)-almost everywhere if \(\{x:f(x)\ne g(x)\}\) is contained in a null set. Since the functions are measurable, this disagreement set itself belongs to \(\mathcal{M}\), so it is also null.
This relation is an equivalence relation. Each function equals itself everywhere. If \(f=g\) \(\mu\)-a.e., then \(g=f\) \(\mu\)-a.e. For transitivity, if \(f=g\) outside a null set \(N_1\) and \(g=h\) outside a null set \(N_2\), then \(f=h\) outside \(N_1\cup N_2\), which is null by countable subadditivity. Thus it makes sense to speak of a function’s almost-everywhere equivalence class.
For a sequence \((f_n)\) and a measurable real-valued function \(f\), pointwise convergence to \(f\) means \(f_n(x)\to f(x)\) for every \(x\in X\). Almost-everywhere convergence means that there is a null set \(N\) such that \(f_n(x)\to f(x)\) for every \(x\in X\setminus N\). The exceptional null set may depend on the sequence and the proposed limit. This is a statement about convergence outside a null set, not a claim that convergence holds at the exceptional points.
A Common Exceptional Set
When comparing sequences, it is essential to collect all their exceptional sets. A different null set may be involved for each index \(n\), and the appropriate common set is their countable union. Countable subadditivity ensures this union is still null; no completeness or finiteness assumption on \(\mu\) is required.
Proof. For each \(n\), let \(D_n=\{x:f_n(x)\ne g_n(x)\}\), and let \(D=\{x:f(x)\ne g(x)\}\). These sets are measurable and null by the hypotheses. The countable union \(D\cup\bigcup_{n=1}^{\infty}D_n\) is measurable and null, because its measure is at most \(\mu(D)+\sum_{n=1}^{\infty}\mu(D_n)=0\). Suppose first that \(f_n\to f\) \(\mu\)-a.e.; then there is a null set \(N_0\) such that convergence holds at each point outside \(N_0\). The set \(N=N_0\cup D\cup\bigcup_{n=1}^{\infty}D_n\) is still null. At every \(x\notin N\), we have \(g_n(x)=f_n(x)\) for every \(n\), and \(g(x)=f(x)\). Therefore \(g_n(x)\to g(x)\), proving \(g_n\to g\) \(\mu\)-a.e. The reverse implication follows by the same argument with the roles of \(f_n,f\) and \(g_n,g\) interchanged. \(\square\)
The proposition implies that the sequences and their limits agree pointwise outside a common null set, but it does not imply that they agree at every point; nor does it imply that pointwise convergence at every point is preserved. The common set is null because there are only countably many functions in the sequence, together with the proposed limit. Taking an uncountable union of null sets would not in general give a null set.
Worked Example: A Single Point Can Destroy Pointwise Convergence
Work on \(X=[0,1]\) with its Lebesgue sigma-algebra and Lebesgue measure \(m\). Define \(f_n(x)=0\) for every \(x\in[0,1]\) and every \(n\). Define \(g_n(x)=0\) for \(x\ne0\), and \(g_n(0)=(-1)^n\). Both functions are measurable: \(g_n\) differs from the constant zero function only on the measurable singleton \(\{0\}\).
For every \(n\), the disagreement set is either \(\{0\}\) or empty, and in fact \(g_n\ne f_n\) at \(0\); hence \(\{x:f_n(x)\ne g_n(x)\}=\{0\}\). Its measure is \(m(\{0\})=0\), so \(f_n=g_n\) \(m\)-a.e. Take the proposed limit \(f=g=0\), which agrees with itself everywhere. The sequence \(f_n\) converges to zero at every \(x\). For \(x\ne0\), \(g_n(x)=0\) for all \(n\), so \(g_n(x)\to0\). At \(x=0\), the even-indexed terms are \(g_{2k}(0)=(-1)^{2k}=1\), while the odd-indexed terms are \(g_{2k+1}(0)=(-1)^{2k+1}=-1\). These subsequences have different limits, so \(g_n(0)\) does not converge.
Check: the two sequences agree at every point outside \(\{0\}\), and \([0,1]\setminus\{0\}\) has full measure \(1\). Thus both converge to zero almost everywhere, even though only the first sequence converges at every point.
The Proposed Limit Also Has a Representative
It is not enough to change the sequence terms; the proposed limit may also be replaced by an almost-everywhere equal function. The proposition allows this change because it includes the condition \(f=g\) \(\mu\)-a.e. Without that condition, convergence to the chosen representative can fail at points where the representatives differ, even if the sequence itself is unchanged.
Worked Example: Changing Only the Limit Function
Again take \(([0,1],\mathcal{L},m)\). For every \(n\), set \(f_n(x)=g_n(x)=0\) on \([0,1]\). Let \(f(x)=0\) for all \(x\), and let \(g=\chi_{\{0\}}\). The function \(g\) is measurable, and the disagreement set between \(f\) and \(g\) is exactly \(\{0\}\), of measure zero. Thus \(f=g\) \(m\)-a.e.
For every \(x\in[0,1]\), \(f_n(x)=0=f(x)\), so \(f_n\to f\) everywhere. For \(x\ne0\), \(g_n(x)=0=g(x)\), so \(g_n(x)\to g(x)\) there. At \(x=0\), however, \(g_n(0)=0\) for every \(n\), whereas \(g(0)=1\). The constant sequence with value zero does not converge to one. Therefore \(g_n\to g\) almost everywhere but not everywhere.
Check: the convergence set for \(g_n\to g\) is \((0,1]\), whose complement is \(\{0\}\) and has measure zero. The proposition applies because both the sequence representatives agree everywhere and the proposed limits agree almost everywhere.
Worked Example: A Countable Set of Nonconvergence Points
Let \(q_1,q_2,\ldots\) be an enumeration without repetitions of \(\mathbb{Q}\cap[0,1]\), and use \(([0,1],\mathcal{L},m)\). Define \(f_n(x)=x\). Define \(g_n(q_j)=(-1)^n\) for every rational \(q_j\), and define \(g_n(x)=x\) for irrational \(x\). Each \(g_n\) is measurable: it agrees with the continuous function \(x\mapsto x\) outside the countable measurable set \(\mathbb{Q}\cap[0,1]\), and modifying a measurable function on a countable set by specified values preserves Lebesgue measurability.
For each \(n\), the disagreement set \(\{x:f_n(x)\ne g_n(x)\}\) is contained in \(\mathbb{Q}\cap[0,1]\). This set is countable, so it has measure zero: if its points are listed as \(q_j\), countable subadditivity gives \(m(\mathbb{Q}\cap[0,1])\le\sum_{j=1}^{\infty}m(\{q_j\})=0\). Thus \(f_n=g_n\) \(m\)-a.e. for every \(n\). Take \(f(x)=x\). Then \(f_n\to f\) everywhere, and \(g_n=f\) at every irrational, so \(g_n\to f\) there. For each rational \(q_j\), the subsequence values \(g_{2k}(q_j)=1\) and \(g_{2k+1}(q_j)=-1\) do not converge to each other or to a common limit. Hence \(g_n\) fails to converge at every rational.
Check: the set of failure points is exactly \(\mathbb{Q}\cap[0,1]\), a null set, so \(g_n\to f\) \(m\)-a.e. This example changes the sequence on the same countable null set at every index, while the preceding example changed values on a single point.
What the Invariance Does—and Does Not—Say
Almost-everywhere convergence is a property of equivalence classes in the following precise sense: if each term is replaced by an almost-everywhere equal measurable function, and the proposed limit is also replaced by an almost-everywhere equal measurable function, the truth of almost-everywhere convergence does not change. The reason is not that exceptional values are irrelevant at each individual point. Rather, all possible disagreements are confined to one null set formed by a countable union.
A common incorrect argument is: “Since \(f_n=g_n\) almost everywhere for each \(n\), convergence at every point is preserved.” The first clause gives only a potentially different null exceptional set for each \(n\). In fact, their union is null, which proves agreement of the sequences at every point outside that common null set—not at points inside it. In the first worked example, every term differs at \(0\), and the sequence of changed values oscillates there. Thus the claim of everywhere preservation is false, while almost-everywhere preservation remains valid.
When checking a specific problem, keep the two conclusions separate:
- For convergence almost everywhere, identify a common null set containing all termwise disagreement sets and any disagreement set between the proposed limits.
- For pointwise convergence everywhere, check every point, including points in those exceptional sets; changing values there can create or remove convergence.
No assumption that \(\mu(X)\) is finite is involved in this invariance result. It uses only that each exceptional set has measure zero and that a countable union of null sets is null. It also does not assert convergence in measure, \(L^p\) convergence, or uniform convergence; those are different modes of convergence with their own definitions and hypotheses.
Check Your Understanding
For each question, distinguish carefully between convergence at every point and convergence outside a null set.
- On \([0,1]\) with Lebesgue measure, let \(f_n=0\) everywhere and let \(g_n\) equal \((-1)^n\) at \(1/2\) and zero elsewhere. To which functions do the two sequences converge almost everywhere, and does \(g_n\) converge at \(1/2\)?
- Suppose \(f_n=g_n\) \(m\)-a.e. for every \(n\), but no condition is given on proposed limits \(f\) and \(g\). Which part of the invariance proposition is missing, and why can convergence to both proposed limits not be concluded?
- For measurable real-valued functions on an arbitrary measure space, explain why the union of the termwise disagreement sets remains null when there are countably many terms.
- In the rational-set example, identify the exact set where \(g_n\) fails to converge and compute its Lebesgue measure.
- Give a reason the proposition does not claim that the two sequences agree at every point, even though each pair of terms agrees almost everywhere.