The Meaning of “Almost Everywhere”
A null set may contain points even though its measure is zero. The phrase “almost everywhere” uses this fact to describe a property that may fail, but only on a negligible set. In the previous tutorial, we saw that countable unions of measurable null sets are null. That result is the key reason countably many almost-everywhere statements can be made to hold at once.
Let \((X,\mathcal{F},\mu)\) be a measure space, and let \(P(x)\) be a property whose truth or failure is defined for each \(x\in X\). The points where it fails are its exceptional points. The exceptional points need not themselves form a measurable set. Accordingly, the definition is phrased using a measurable null set that contains them.
The definition does not require the exceptional set itself to be measurable. It only requires that the exceptional set be covered by a measurable set of measure zero. If the measure space is complete, the previous tutorial’s Subsets of Null Sets in a Complete Measure Space theorem ensures that every subset of a null set is measurable and null. In a space that is not complete, the distinction can matter.
A statement holds almost everywhere on a measurable set \(A\) if its failures within \(A\) are contained in a measurable null set. Equivalently, the property holds for every point of \(A\setminus N\) for some null set \(N\). This relative formulation is useful when a function or property is being considered only on part of the space.
Examples of Almost-Everywhere Statements
Worked Example: Changing a Function at One Point
Use Lebesgue measure on the Borel subsets of \([0,1]\). Define \(f:[0,1]\to\mathbb{R}\) by $$ f(x)= \begin{cases} 17,&x=\frac{1}{3},\\ x,&x\ne\frac{1}{3}. \end{cases} $$ Let \(g(x)=x\) for every \(x\in[0,1]\). The equality \(f(x)=g(x)\) fails exactly at \(x=1/3\). The singleton \(\{1/3\}\) is measurable and null, so \(f=g\) almost everywhere.
The functions are not equal at every point: \(f(1/3)=17\), whereas \(g(1/3)=1/3\). Their almost-everywhere equality means precisely that the discrepancy is confined to a null set; it does not mean that the functions are pointwise identical.
Worked Example: Almost Everywhere for Counting Measure
Let \(X=\mathbb{N}\), let \(\mathcal{F}=\mathcal{P}(\mathbb{N})\), and let \(\mu\) be counting measure. A set has counting measure zero only if it is empty: any nonempty set contains some \(k\), and its counting measure is at least \(1\). Thus the only null set is \(\varnothing\).
Consequently, a property on this space holds almost everywhere exactly when it holds at every natural number. For example, the property \(n^2\geq 1\) holds almost everywhere because it holds for every \(n\in\mathbb{N}\). There is no nonempty exceptional set that can be discarded. The meaning of “almost everywhere” therefore depends on the measure, not just on the underlying set of points.
Worked Example: A Nonmeasurable Exceptional Set Inside a Null Set
Take the Borel subsets of \(\mathbb{R}\) with Lebesgue measure restricted to that sigma-algebra. Let \(C\) be the Cantor set. It is a Borel null set. There exists a subset \(E\subseteq C\) that is not Borel: the family of Borel subsets of \(\mathbb{R}\) has cardinality at most that of \(\mathbb{R}\), whereas the power set of \(C\) has strictly larger cardinality.
Consider the property \(P(x)\) that \(x\notin E\). It fails precisely for \(x\in E\), and this failure set is not measurable in the chosen measure space. Nevertheless, \(P\) holds almost everywhere: all its failures lie inside the measurable null set \(C\). This example explains why the definition uses containment in a null set rather than requiring the failure set itself to be measurable.
Countably Many Properties Hold Simultaneously
For each individual statement that holds almost everywhere, there may be a different null set of exceptions. The countable-union result from Countable Subadditivity shows how to combine those exceptional sets into one null set. This yields a common exceptional set outside which all the statements hold.
Proof. For each positive integer \(n\), choose a measurable null set \(N_n\) such that \(P_n(x)\) holds for every \(x\notin N_n\). Define $$ N=\bigcup_{n=1}^{\infty}N_n. $$ The set \(N\) is measurable, and it is null by the previously established Countable Union of Null Sets theorem. If \(x\notin N\), then \(x\notin N_n\) for every \(n\). Therefore \(P_n(x)\) holds for every positive integer \(n\). This gives one null exceptional set for all the properties, as required. \(\square\)
Worked Example: Avoiding a Countable List of Points
On \([0,1]\) with Lebesgue measure, let \(x_n=1/(n+2)\) and define \(P_n(x)\) to mean \(x\ne x_n\). Each \(P_n\) fails only at the singleton \(\{x_n\}\), which is null. The theorem shows that all the statements \(P_n\) hold simultaneously outside $$ N=\bigcup_{n=1}^{\infty}\left\{\frac{1}{n+2}\right\}. $$ This union is countable and hence null by the Every Countable Subset of the Real Line Is Lebesgue Null theorem. Thus every point of \([0,1]\setminus N\) avoids every point in the listed sequence. The common exceptional set matters: knowing each \(P_n\) separately holds almost everywhere would not, by itself, specify where all of them hold together.
Countability is essential to this argument. It does not say that an arbitrary, uncountable collection of properties can be made simultaneous by discarding a null set. For each \(a\in[0,1]\), the property \(x\ne a\) holds almost everywhere, since its failure set \(\{a\}\) is null. But the combined demand that \(x\ne a\) for every \(a\in[0,1]\) cannot hold at any \(x\in[0,1]\): taking \(a=x\) makes it fail. The union of these uncountably many singleton exceptional sets is the whole interval, which is not null.
Almost-Everywhere Equality and Pointwise Limits
For functions \(f,g:X\to\mathbb{R}\), the notation \(f=g\) almost everywhere means that the property \(f(x)=g(x)\) holds almost everywhere. No measurability of the functions is needed for that statement, provided the equality is defined pointwise and its failure set is contained in a measurable null set. In particular, it is not necessary that the set \(\{x:f(x)\ne g(x)\}\) itself be measurable.
Almost-everywhere equality is an equivalence relation. Its transitivity follows by combining two exceptional sets, and it is important when working with sequences: if two sequences of functions agree almost everywhere at each index, then they agree at every index outside one common null set. At those points, their pointwise convergence and limits can be compared directly.
Proof. For each \(n\), choose a measurable null set \(N_n\) outside which \(f_n(x)=g_n(x)\). Let \(N=\bigcup_{n=1}^{\infty}N_n\). By the Countable Union of Null Sets theorem, \(N\) is null. Fix \(x\notin N\). Then \(x\notin N_n\) for every \(n\), so \(f_n(x)=g_n(x)\) for every \(n\). If both real sequences converge, say \(f_n(x)\to L\) and \(g_n(x)\to M\), their terms are identical. The uniqueness of the limit of a real sequence gives \(L=M\). This proves both conclusions. \(\square\)
Worked Example: Two Sequences with the Same Almost-Everywhere Limit
On \([0,1]\) with Lebesgue measure, set \(f_n(x)=x^n\). Define \(g_n(x)=x^n\) when \(x\ne 1/2\), and \(g_n(1/2)=7\) for every \(n\). For each \(n\), the two functions agree outside \(\{1/2\}\). They therefore agree almost everywhere, and the theorem gives a common null set outside which their terms agree for every \(n\).
For \(0\leq x<1\), \(x^n\to 0\), while \(1^n=1\) for every \(n\). Hence \(f_n\) converges to \(0\) on \([0,1)\) and to \(1\) at \(x=1\). The sequence \(g_n\) has the same pointwise limits at every \(x\ne 1/2\). At \(x=1/2\), it is constantly \(7\), so its limit there is \(7\), whereas \(f_n(1/2)=(1/2)^n\to0\). The difference at that single point is consistent with equality of the sequences, and of their limits, almost everywhere.
What Almost Everywhere Does—and Does Not—Guarantee
Almost-everywhere statements disregard a null set, not necessarily an empty set. The first worked example had a genuine point of disagreement, and the third showed that the failure set can even be nonmeasurable in an incomplete space. What makes the property hold almost everywhere is the existence of a measurable null set containing every failure.
The countable-stability theorem also has a specific boundary: it combines a countable list of properties. The uncountable family of conditions \(x\ne a\), one for each \(a\in[0,1]\), demonstrates why the countable proof cannot simply be extended without additional hypotheses. Always identify whether the exceptional sets can be combined into a null set before making a simultaneous claim.
Finally, almost-everywhere equality is weaker than pointwise equality, but it is strong enough to preserve pointwise limits outside a common null set when the functions agree almost everywhere at each index. The word “common” is essential: countable stability supplies a single exceptional set that works for the whole sequence.
Check Your Understanding
Use the definition and results in this tutorial to answer the following questions.
- Why does the definition allow the exceptional set to be contained in a null set instead of requiring the exceptional set itself to be measurable?
- On a space with counting measure, what does it mean for a property to hold almost everywhere?
- How does the Countable Union of Null Sets theorem prove that countably many almost-everywhere properties hold simultaneously?
- If \(f_n=g_n\) almost everywhere for every \(n\), why can one choose a single null set outside which the sequences agree term by term?
- Why does the family of properties \(x\ne a\), for every \(a\in[0,1]\), show that uncountable simultaneous claims require care?