From Almost-Everywhere to Uniform Convergence with a Measure Budget
Egorov's theorem says that almost-everywhere convergence on a finite-measure space becomes uniform after removing a set of arbitrarily small measure. In applications, “arbitrarily small” often needs to mean less than a particular tolerance \(\delta\). The exceptional-set proof gives a direct way to meet that budget: assign part of \(\delta\) to each accuracy threshold, then choose a tail index that makes the corresponding exceptional set small enough.
Throughout, let \((X,\mathcal{M},\mu)\) be a measure space with \(\mu(X)<\infty\). Let \(f_n:X\to\mathbb{R}\) and \(f:X\to\mathbb{R}\) be measurable functions, and suppose \(f_n\to f\) \(\mu\)-a.e. No completeness assumption is needed: the sets used below are measurable because the functions are measurable. For a measurable set \(G\), convergence is uniform on \(G\) if for every \(\varepsilon>0\) there is an integer \(N\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in G\) and every \(n\ge N\).
The strict inequality in this definition matters. Outside \(E_{N,k}\), every \(n\ge N\) satisfies \(|f_n-f|\le 1/k\). To obtain the strict bound required in the definition of uniform convergence, choose \(k\) so that \(1/k<\varepsilon\). The resulting error is then at most \(1/k\), and hence strictly less than \(\varepsilon\).
The Prescribed-Error Construction
Proof. Fix \(k\ge1\). The sets \(E_{N,k}\) decrease with \(N\), and their intersection is $$ \bigcap_{N\ge1}E_{N,k} =\{x:|f_n(x)-f(x)|>1/k\text{ for infinitely many }n\}. $$ If \(f_n(x)\to f(x)\), then for this fixed \(k\) the error is eventually at most \(1/k\), so \(x\) is not in the intersection. Consequently, \(\bigcap_NE_{N,k}\) is contained in the null set where convergence fails. Its measure is zero. Since \(\mu(E_{1,k})\le\mu(X)<\infty\), continuity from above for measures gives \(\mu(E_{N,k})\to0\) as \(N\to\infty\). We may therefore choose \(N_k\) with \(\mu(E_{N_k,k})<\delta/2^{k+1}\).
The set \(G\) is measurable, as it is the complement of a countable union of measurable sets. Countable subadditivity gives $$ \mu(X\setminus G) =\mu\left(\bigcup_{k\ge1}E_{N_k,k}\right) \le\sum_{k\ge1}\mu(E_{N_k,k}) \le\sum_{k\ge1}\frac{\delta}{2^{k+1}} =\frac{\delta}{2}<\delta. $$ For uniform convergence, let \(\varepsilon>0\) and choose \(k\) with \(1/k<\varepsilon\). If \(x\in G\), then \(x\notin E_{N_k,k}\), so \(|f_n(x)-f(x)|\le1/k<\varepsilon\) for every \(n\ge N_k\). The index depends on \(\varepsilon\) through \(k\), but not on \(x\). This is uniform convergence on \(G\). \(\square\)
The allocation \(\delta/2^{k+1}\) is convenient rather than unique. Its sum is \(\delta/2\), leaving a margin below the requested tolerance. The chosen indices \(N_k\) need not be increasing: each threshold is used separately, and the set \(G\) excludes all the corresponding tail sets.
Worked Example: An Explicit Good Set for Indicator Functions
On \([0,1]\) with Lebesgue measure \(m\), let \(f_n=\chi_{[0,1/n]}\) and \(f=\chi_{\{0\}}\). The sequence converges to \(f\) at every point: at \(0\) all values equal \(1\), while at each \(x>0\), \(f_n(x)=0\) as soon as \(n>1/x\). Choose the measure-loss tolerance \(\delta=1/4\).
For \(k\ge2\), the error is either zero or one, so the condition \(|f_n-f|>1/k\) holds exactly when \(0<x\le1/n\). Thus, for \(N\ge1\), $$ E_{N,k}=\bigcup_{n\ge N}(0,1/n]=(0,1/N], \qquad m(E_{N,k})=\frac1N. $$ For \(k=1\), the condition is error \(>1\), which never occurs, so \(E_{N,1}=\varnothing\). The prescribed allowance for \(k\ge2\) is $$ \frac{\delta}{2^{k+1}}=\frac{1}{2^{k+3}}. $$ Taking \(N_k=2^{k+3}+1\) gives \(m(E_{N_k,k})=1/(2^{k+3}+1)<1/2^{k+3}\), as required.
The intervals \((0,1/N_k]\) shrink as \(k\) increases, and the largest one occurs at \(k=2\), where \(N_2=33\). Therefore their union is \((0,1/33]\), and the construction gives $$ G=[0,1]\setminus(0,1/33]=\{0\}\cup(1/33,1], \qquad m([0,1]\setminus G)=\frac1{33}<\frac14. $$ For every \(x\in G\) and \(n\ge33\), either \(x=0\), where \(f_n(x)=f(x)=1\), or \(x>1/33\ge1/n\), where \(f_n(x)=f(x)=0\). Hence the functions agree on \(G\) for all \(n\ge33\). In particular, \(N=33\) is a uniform-convergence index for \(\varepsilon=1/2\). It is the least such index: \(x=1/32\) belongs to \(G\), and \(f_{32}(1/32)=1\) whereas \(f(1/32)=0\), so an index at most \(32\) fails for \(\varepsilon=1/2\). Check: the removed interval has length \(1/33\), and \(1/33<1/4\) because \(4<33\).
Worked Example: Tail Bounds for Powers on the Unit Interval
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\) and let \(f=\chi_{\{1\}}\). For \(0\le x<1\), \(x^n\to0\), while \(f_n(1)=1\) for all \(n\); hence convergence holds everywhere. Take \(\delta=1/2\).
For \(k\ge2\) and \(x<1\), the sequence \(x^n\) decreases with \(n\). Therefore some \(n\ge N\) has \(x^n>1/k\) exactly when \(x^N>1/k\). The point \(x=1\) has zero error because \(f(1)=1\). It follows that $$ E_{N,k}=(k^{-1/N},1), \qquad m(E_{N,k})=1-k^{-1/N}. $$ For \(k=1\), the exceptional set is empty. For \(k\ge2\), use \(1-e^{-u}\le u\) for \(u\ge0\), with \(u=(\ln k)/N\), to obtain $$ m(E_{N,k})=1-e^{-(\ln k)/N}\le\frac{\ln k}{N}. $$ Choose \(N_k=\lceil 2^{k+2}\ln k\rceil+1\) for \(k\ge2\), and \(N_1=1\). Then \(N_k>2^{k+2}\ln k\), so \(m(E_{N_k,k})<1/2^{k+2}=\delta/2^{k+1}\). The proposition gives a set \(G\) with \(m([0,1]\setminus G)\le\sum_{k\ge2}2^{-(k+2)}=1/8<1/2\).
For a concrete uniform index, choose \(\varepsilon=1/10\) and \(k=11\); then \(1/11<1/10\). Here $$ N_{11}=\lceil8192\ln 11\rceil+1=19645, $$ since \(8192\ln11\) is approximately \(19643.558\). By the definition of \(G\), every \(x\in G\) satisfies \(|x^n-f(x)|\le1/11<1/10\) for every \(n\ge19645\). Thus \(19645\) is a uniform-convergence index on this particular \(G\) for the chosen tolerance. Check: \(8192=2^{13}\), and the allowance at \(k=11\) is \(\delta/2^{12}=1/8192\); the bound \(\ln11/N_{11}<1/8192\) verifies the required strict measure estimate.
Common Pitfall: Meeting the Budget Strictly
A frequent error is to choose an index that makes the exceptional-set measure equal to its allowance, then claim the strict inequality required by the construction. In the indicator example, at \(k=2\) the allowance is \(1/32\). Choosing \(N_2=32\) gives $$ m(E_{32,2})=m((0,1/32])=\frac1{32}, $$ which is not less than \(1/32\). The corrected choice \(N_2=33\) gives \(m(E_{33,2})=1/33<1/32\). Exact endpoints do not change the length here, but they do not turn equality of measures into a strict bound either.
Another point to check is which way the tail condition is used. Outside \(E_{N,k}\), the error is at most \(1/k\), not necessarily strictly less than \(1/k\). This is still enough: for a requested \(\varepsilon>0\), choose \(k\) with \(1/k<\varepsilon\). The inequality \(|f_n-f|\le1/k\) then implies the required strict inequality \(|f_n-f|<\varepsilon\).
The method is quantitative once the tail measures can be calculated or bounded. It separates the task into two steps: first spend a summable measure budget across accuracy levels, then find an index for each level. The resulting good set may depend on \(\delta\) and on the chosen indices; the index that proves uniform convergence depends on the requested accuracy but is valid for every point of that good set.
Check Your Understanding
For each question, pay attention to whether the exceptional condition uses \(>\) or \(\ge\), and verify that the measure bound is strict when required.
- Why does \(\bigcap_{N\ge1}E_{N,k}\) have measure zero for each fixed \(k\)? Which convergence fact is used?
- If the prescribed allocations are \(\delta/2^{k+1}\), what is their sum, and why does this guarantee a loss strictly below \(\delta\)?
- For \(f_n=\chi_{[0,1/n]}\) on \([0,1]\), find the least \(N\) for which \(m(E_{N,3})<1/64\).
- For \(f_n(x)=x^n\) and \(k\ge2\), derive the formula for \(m(E_{N,k})\) and explain why \(x=1\) is not in that tail exceptional set.
- Explain why the bound \(|f_n-f|\le1/k\) on the good set suffices for uniform convergence at tolerance \(\varepsilon\), provided \(1/k<\varepsilon\).