Pointwise Convergence Can Escape to Infinity
On a finite-measure space, Egorov’s theorem says that almost-everywhere convergence becomes uniform after removing a measurable set of arbitrarily small measure. The finiteness assumption is essential: on an infinite-measure space, there are pointwise-convergent sequences for which every set outside of which convergence is uniform has infinite measure.
We use Lebesgue measure \(m\) on \(\mathbb{R}\), with measurable sets in the Lebesgue sigma-algebra \(\mathcal{L}\). All functions in the examples below are real-valued and Borel measurable, hence Lebesgue measurable. Convergence is pointwise everywhere unless explicitly stated otherwise. A sequence \(f_n\) converges uniformly to \(f\) on a set \(E\subseteq\mathbb{R}\) if, for every \(\varepsilon>0\), there is an \(N\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(n\ge N\) and every \(x\in E\).
Worked Example: Indicators Moving Along the Real Line
For \(n\ge1\), define \(f_n=\chi_{[n,n+1]}\) on \(\mathbb{R}\). We first check pointwise convergence. Fix \(x\in\mathbb{R}\). If \(n>x\), then every \(y\in[n,n+1]\) satisfies \(y\ge n>x\), so \(x\notin[n,n+1]\) and \(f_n(x)=0\). Thus \(f_n(x)\) is eventually zero for each fixed \(x\), and \(f_n(x)\to0\) everywhere.
Now let \(A\in\mathcal{L}\) be a deleted set and put \(E=\mathbb{R}\setminus A\). Suppose \(f_n\to0\) uniformly on \(E\). Apply the definition of uniform convergence with \(\varepsilon=\tfrac12\). There must be an \(N\) such that \(f_n(x)<\tfrac12\) for every \(x\in E\) and every \(n\ge N\). Since \(f_n(x)=1\) at every \(x\in[n,n+1]\), this requires \(E\cap[n,n+1]=\varnothing\) for every \(n\ge N\). The union of these intervals is \([N,\infty)\), so \([N,\infty)\subseteq A\). Consequently \(m(A)=\infty\).
Check: the intervals \([n,n+1]\) each have measure \(1\), and their union for \(n\ge N\) is the whole tail, not just a collection of isolated pieces. Its measure is infinite. Therefore no deletion of finite measure—not even one of large but finite measure—makes this sequence converge uniformly on the remainder.
A Criterion for These Indicator Sequences
Proof. Suppose first that \(g_n\to0\) uniformly on \(E\). With \(\varepsilon=\tfrac12\), there is an \(N\) such that \(g_n(x)<\tfrac12\) for all \(x\in E\) and \(n\ge N\). If \(x\in E\cap S_n\), then \(g_n(x)=1\), contradicting that inequality. Thus \(E\cap S_n=\varnothing\) for every \(n\ge N\). Conversely, suppose such an \(N\) exists. Then \(g_n(x)=0\) for every \(x\in E\) and every \(n\ge N\). For any \(\varepsilon>0\), this gives \(|g_n(x)-0|=0<\varepsilon\) for all those \(n\) and \(x\), which is uniform convergence on \(E\). \(\square\)
In the first example, the proposition says that uniform convergence on \(E\) requires \(E\) to avoid every interval \([n,n+1]\) from some index onward. Since those intervals cover a tail of \(\mathbb{R}\), their removal necessarily costs infinite measure. The obstruction is not that the functions fail to become small at any fixed point. It is that points where the functions equal \(1\) keep appearing farther out, and uniform convergence must control all such points at once.
Worked Example: Sparse Supports Still Force Infinite Deletion
Define \(h_n=\chi_{[n^2,n^2+1]}\) on \((\mathbb{R},\mathcal{L},m)\). For each fixed \(x\in\mathbb{R}\), choose an integer \(N\) with \(N^2>x\). For every \(n\ge N\), \(n^2\ge N^2>x\), so \(x\notin[n^2,n^2+1]\) and \(h_n(x)=0\). Hence \(h_n\to0\) everywhere.
Suppose \(h_n\to0\) uniformly on \(E=\mathbb{R}\setminus A\), where \(A\in\mathcal{L}\). By the proposition, for some \(N\), \(E\cap[n^2,n^2+1]=\varnothing\) for all \(n\ge N\). Therefore \(\bigcup_{n\ge N}[n^2,n^2+1]\subseteq A\). These intervals are pairwise disjoint: for \(n\ge1\), \(n^2+1<(n+1)^2\), since \((n+1)^2-(n^2+1)=2n>0\). Each interval has measure \(1\). Countable additivity gives $$ m\left(\bigcup_{n\ge N}[n^2,n^2+1]\right) =\sum_{n=N}^{\infty}m([n^2,n^2+1]) =\sum_{n=N}^{\infty}1 =\infty. $$ Thus \(m(A)=\infty\). Check: although the intervals become increasingly far apart, there are infinitely many disjoint intervals of length \(1\); their total measure remains infinite.
The Finite-Measure Contrast
The conclusion is not that uniform convergence can never be obtained by deleting a small set. Rather, the examples show that on an infinite-measure space pointwise convergence alone does not guarantee this. On a finite-measure space, Egorov’s theorem supplies the guarantee: if \((X,\mathcal{M},\mu)\) has \(\mu(X)<\infty\), and measurable real-valued functions converge \(\mu\)-a.e., then for every \(\delta>0\) there is a measurable exceptional set \(A\) with \(\mu(A)<\delta\) such that convergence is uniform on \(X\setminus A\). The following example exhibits this behavior directly on a finite interval.
Worked Example: Removing a Small Interval on a Finite Space
Take \(X=[0,1]\), with the restriction of Lebesgue measure, and set \(q_n=\chi_{(0,1/n)}\) on \(X\). At \(x=0\), every \(q_n(x)=0\). At any fixed \(x>0\), choose \(N\) with \(N>1/x\). For \(n\ge N\), \(1/n\le1/N<x\), so \(x\notin(0,1/n)\) and \(q_n(x)=0\). Thus \(q_n\to0\) at every point of \(X\).
Let \(\delta>0\). Choose an integer \(N\ge1\) with \(N>1/\delta\), and take \(A=(0,1/N)\). Its measure is \(m(A)=1/N<\delta\). For every \(n\ge N\), \((0,1/n)\subseteq(0,1/N)=A\). Hence \(q_n(x)=0\) for every \(x\in X\setminus A\) and every \(n\ge N\), so convergence on \(X\setminus A\) is uniform. Check: the endpoints are handled by the open intervals: \(0\notin A\) and \(q_n(0)=0\), while \(1/N\notin A\) and is outside every \((0,1/n)\) for \(n\ge N\). The exceptional measure is exactly \(1/N\), which is strictly less than \(\delta\) by the choice of \(N\).
What Fails Without Finite Measure
The real-line example pinpoints the role of the hypothesis \(\mu(X)<\infty\) in Egorov’s theorem. For the sequence \(f_n=\chi_{[n,n+1]}\), we proved more than the failure of a particular choice of exceptional set: every measurable \(A\) for which convergence is uniform on \(\mathbb{R}\setminus A\) must contain \([N,\infty)\) for some \(N\), and therefore must have infinite measure. In particular, for any prescribed finite \(\delta>0\), no set \(A\) with \(m(A)<\delta\) can suffice.
A common mistake is to reason: “At each fixed \(x\), \(f_n(x)\) is eventually zero, so after removing a set of small measure, the convergence must be uniform.” This replaces the uniform quantifier—one index \(N\) must work for every point remaining—with pointwise quantifiers, where the index may depend on \(x\). For example, choosing \(x=n+\tfrac12\) gives \(f_n(x)=1\) at that point for each \(n\). These points move with \(n\), so they do not contradict pointwise convergence at any fixed \(x\); but they do obstruct a single uniform index on the whole remainder.
For the first example, the tempting calculation “\(\lim_{n\to\infty}f_n(x)=0\) for each \(x\), therefore \(\sup_{x\in\mathbb{R}}f_n(x)\to0\)” is wrong. In fact, for every \(n\), the interval \([n,n+1]\) is nonempty and \(f_n=1\) there, so \(\sup_{x\in\mathbb{R}}f_n(x)=1\). After deleting a finite-measure set, the same obstruction persists on every sufficiently distant interval, since uniform convergence would require the deleted set to contain the entire tail. Pointwise convergence controls each fixed location; uniform convergence controls all locations at the same index. On an infinite-measure space, the locations where convergence is slow can escape to infinity, and removing them may require infinite measure.
Check Your Understanding
Use the uniform-convergence quantifiers explicitly in your reasoning.
- For \(f_n=\chi_{[n,n+1]}\) on \(\mathbb{R}\), why does pointwise convergence hold even though \(\sup_{x\in\mathbb{R}}f_n(x)=1\) for every \(n\)?
- Suppose \(f_n=\chi_{S_n}\). Prove that if \(f_n\to0\) uniformly on \(E\), then there is an \(N\) such that \(E\cap S_n=\varnothing\) for every \(n\ge N\).
- For \(h_n=\chi_{[n^2,n^2+1]}\), compute the measure of the union of the supports for \(n\ge N\), and explain why spacing the intervals farther apart does not make that measure finite.
- In the finite-interval example, verify that \(A=(0,1/N)\) works for all indices \(n\ge N\), not just for \(n=N\).
- State the finite-measure assumption in Egorov’s theorem and explain which step of the real-line examples shows that a finite-measure exceptional set cannot always exist.