Proof of Egorov's Theorem by Exceptional Sets

The Exceptional-Set Strategy

Egorov's theorem converts almost-everywhere convergence into uniform convergence after removing a set of small measure. The key is to organize the possible failures of a fixed error bound into measurable sets that shrink as the starting index increases. Almost-everywhere convergence makes their intersection null; finiteness of the measure then lets us make one of these sets as small as desired.

Throughout, $(X,\mathcal{M},\mu)$ is a measure space with $\mu(X)<\infty$. The functions $f_n:X\to\mathbb{R}$ and $f:X\to\mathbb{R}$ are $\mathcal{M}$-measurable and finite-valued. The statement $f_n\to f$ $\mu$-a.e. means that there is a measurable set $Z$ with $\mu(Z)=0$ such that $f_n(x)\to f(x)$ for every $x\in X\setminus Z$. Uniform convergence on a set $E\subseteq X$ means that for every $\eta>0$, there is an $N$ such that $|f_n(x)-f(x)|<\eta$ for every $x\in E$ and every $n\ge N$.

Definition: For a threshold $1/k$, where $k$ is a positive integer, and a starting index $n$, define the tail exceptional set $$ A_{n,k}=\bigcup_{j=n}^{\infty}\{x\in X:|f_j(x)-f(x)|\ge 1/k\}. $$ It consists of points where the error is at least $1/k$ at some index $j\ge n$. The union over all later indices is essential: a point is excluded if the error threshold fails even once in the tail.

Each set $A_{n,k}$ is measurable. Indeed, $f_j-f$ is measurable, so $\{|f_j-f|\ge 1/k\}$ is measurable, and $A_{n,k}$ is a countable union of such sets. For fixed $k$, the sets decrease as $n$ increases: $A_{n+1,k}\subseteq A_{n,k}$.

Why the Tail Sets Shrink to a Null Set

For each fixed $k$, a point $x$ belongs to every $A_{n,k}$ exactly when the error is at least $1/k$ at arbitrarily late indices. At every point where $f_n(x)\to f(x)$, that cannot happen: eventually $|f_j(x)-f(x)|<1/k$ for all later $j$. Consequently, $\bigcap_{n=1}^{\infty}A_{n,k}\subseteq Z$, so this intersection has measure zero.

Lemma (Continuity from above for finite measure): If $B_1\supseteq B_2\supseteq\cdots$ are measurable sets in $(X,\mathcal{M},\mu)$ and $\mu(B_1)<\infty$, then $$ \lim_{n\to\infty}\mu(B_n)=\mu\left(\bigcap_{n=1}^{\infty}B_n\right). $$

Proof. Set $C_n=B_1\setminus B_n$. Then $C_n$ increases and $\bigcup_n C_n=B_1\setminus\bigcap_n B_n$. By continuity from below, $\mu(C_n)\to\mu(\bigcup_nC_n)$. Since $B_n$ and $C_n$ partition $B_1$, and $\mu(B_1)<\infty$, we have $\mu(B_n)=\mu(B_1)-\mu(C_n)$. Taking limits gives the stated equality. $\square$

Apply the lemma to $B_n=A_{n,k}$. Its finite-measure hypothesis holds because $\mu(A_{1,k})\le\mu(X)<\infty$, and its intersection is null. Thus, for each fixed $k$, $\mu(A_{n,k})\to0$ as $n\to\infty$. The index at which this measure becomes small may depend on $k$; the proof chooses a separate index for each threshold.

Egorov's Theorem, Proved by Exceptional Sets

Theorem (Egorov): Let $(X,\mathcal{M},\mu)$ be a finite measure space, and let $f_n,f:X\to\mathbb{R}$ be measurable. If $f_n\to f$ $\mu$-a.e., then for every $\varepsilon>0$ there is a measurable $E\subseteq X$ such that $\mu(X\setminus E)<\varepsilon$ and $f_n\to f$ uniformly on $E$.

Proof. Fix $\varepsilon>0$. For each positive integer $k$, let $\delta_k=\varepsilon/2^{k+1}$. Since $\mu(A_{n,k})\to0$, choose $N_k$ so that $\mu(A_{N_k,k})<\delta_k$. Define $$ E=X\setminus\left(Z\cup\bigcup_{k=1}^{\infty}A_{N_k,k}\right). $$ The set $E$ is measurable. Countable subadditivity and $\mu(Z)=0$ give $$ \mu(X\setminus E)\le\sum_{k=1}^{\infty}\mu(A_{N_k,k}) <\sum_{k=1}^{\infty}\frac{\varepsilon}{2^{k+1}} =\frac{\varepsilon}{2}<\varepsilon. $$ The series equals $\varepsilon/2$ because $\sum_{k=1}^{\infty}2^{-(k+1)}=1/2$.

It remains to verify uniform convergence on $E$. Fix any $\eta>0$. Choose a positive integer $k$ with $1/k<\eta$. If $x\in E$, then $x\notin A_{N_k,k}$. By the definition of that tail set, for every $n\ge N_k$ we have $|f_n(x)-f(x)|<1/k<\eta$. The same $N_k$ works for every $x\in E$, which is precisely uniform convergence. Points in $E$ also lie outside $Z$, though the tail-set bounds already give the uniform error estimate there. This proves the theorem. $\square$

Worked Constructions

Worked Example: Moving Spikes on the Unit Interval

On $[0,1]$ with Lebesgue measure $m$, set $f_n=\chi_{(0,1/n)}$ and $f=0$. Every point has limit zero: at $x=0$ every value is zero, and for $x>0$, $f_n(x)=0$ once $n\ge 1/x$. For every $k\ge1$, the condition $|f_j-f|\ge1/k$ is equivalent to $f_j=1$, since the only values are zero and one. Therefore $$ A_{N,k}=\bigcup_{j\ge N}(0,1/j)=(0,1/N),\qquad m(A_{N,k})=1/N. $$ The equality of sets holds because these intervals decrease as $j$ increases, so their union over $j\ge N$ is the first interval.

Take $\varepsilon=1/4$ and choose $N_k=2^{k+3}+1$. Then $$ m(A_{N_k,k})=\frac{1}{2^{k+3}+1}<\frac{1}{2^{k+3}}=\frac{\varepsilon}{2^{k+1}}. $$ The sets $(0,1/N_k)$ decrease as $k$ increases, so their union is $(0,1/17)$, since $N_1=17$. The exceptional-set construction gives $E=[0,1]\setminus(0,1/17)$, with $m([0,1]\setminus E)=1/17<1/4$. For every $n\ge17$, $f_n=0$ everywhere on $E$, so convergence there is uniform. Check: $1/17<1/4$ because $4<17$; also, $(0,1/n)$ does not contain $1/17$ when $n\ge17$, so the boundary point of $E$ causes no exception.

Worked Example: Powers Approaching a Discontinuous Limit

On $[0,1]$ with Lebesgue measure, let $f_n(x)=x^n$. The pointwise limit is $f(x)=0$ for $0\le x<1$ and $f(1)=1$. Fix $k\ge2$. For $x<1$, the condition $|x^j-f(x)|\ge1/k$ is $x^j\ge1/k$, equivalent to $x\ge k^{-1/j}$; at $x=1$, the error is zero. As $j$ increases, $k^{-1/j}$ increases, so the intervals $[k^{-1/j},1)$ decrease. Hence $$ A_{N,k}=[k^{-1/N},1),\qquad m(A_{N,k})=1-k^{-1/N}. $$ For $k=1$, the tail exceptional set is empty, since the error is always less than $1$.

For each fixed $k\ge2$, $k^{-1/N}\to1$, so $m(A_{N,k})\to0$. Thus one can choose $N_k$ with $1-k^{-1/N_k}<\varepsilon/2^{k+1}$, as required in the proof. For example, at threshold $k=2$ and index $N=32$, the exceptional set has measure $1-2^{-1/32}$. Writing this as $1-e^{-(\ln 2)/32}$, the inequality $1-e^{-u}<u$ for $u>0$ gives $$ 1-2^{-1/32}<\frac{\ln 2}{32}<\frac{1}{32}, $$ where $\ln 2<1$. Thus this particular tail set has measure less than $1/32$. Check: its exact measure is also the length of $[2^{-1/32},1)$, namely $1-2^{-1/32}$. The convergence is not uniform on all of $[0,1]$: for each $n$, $\sup_{0\le x<1}|x^n-f(x)|=1$, since $x^n$ approaches $1$ as $x$ approaches $1$ from below.

Worked Example: A Weighted Countable Space

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{M}$ be all subsets of $\mathbb{N}$, and define $\mu(A)=\sum_{j\in A}2^{-j}$. This is a finite measure, since $\mu(X)=\sum_{j=1}^{\infty}2^{-j}=1$. Set $f_n(j)=1$ if $j\ge n$, and $f_n(j)=0$ if $j<n$. For each fixed $j$, $f_n(j)=0$ whenever $n>j$, so $f_n(j)\to0$.

For every $k\ge1$, the error is either zero or one, and $1\ge1/k$. Thus $$ A_{N,k}=\{j\in\mathbb{N}:j\ge N\},\qquad \mu(A_{N,k})=\sum_{j=N}^{\infty}2^{-j}=2^{1-N}. $$ Choose $\varepsilon=1/8$ and $N_k=k+6$. Direct substitution gives $\mu(A_{N_k,k})=2^{-k-5}<2^{-k-4}=\varepsilon/2^{k+1}$. The union of these tails is $\{j\ge7\}$, because the largest tail occurs at $k=1$, where $N_1=7$. Therefore $E=\{1,2,3,4,5,6\}$, and $$ \mu(X\setminus E)=\sum_{j=7}^{\infty}2^{-j}=2^{-6}=\frac{1}{64}<\frac18. $$ For $n\ge7$, every $j\in E$ satisfies $j<n$, so $f_n(j)=0$ on all of $E$. Check: the total mass is $\mu(E)+\mu(X\setminus E)=\sum_{j=1}^{6}2^{-j}+2^{-6}=(1-2^{-6})+2^{-6}=1$, as required.

A Common Error: Testing Only the Current Index

To control uniform convergence, it is not enough to bound the measure of $\{|f_n-f|\ge1/k\}$ at one selected index $n$. Uniform convergence requires an error bound at every later index, so the exceptional set must include the whole tail. For a concrete example on $[0,1]$, take $f_1=0$, $f_2=1$, and $f_n=0$ for every $n\ge3$, with $f=0$. This sequence converges everywhere to zero. At threshold $1/2$, the current-index bad set $\{|f_1-f|\ge1/2\}$ is empty. But the tail exceptional set $A_{1,1}=\bigcup_{j\ge1}\{|f_j-f|\ge1\}$ is all of $[0,1]$, because $f_2=1$. The correction is to use $A_{N,k}$, which tests every $j\ge N$, and only then use $X\setminus A_{N,k}$ to guarantee that all later errors are strictly less than $1/k$.

The proof's two measure-theoretic ingredients have distinct roles. Almost-everywhere convergence makes each fixed-threshold intersection $\bigcap_n A_{n,k}$ null. Finite total measure permits continuity from above, which turns that null intersection into small measures for sufficiently late tail sets. Choosing those indices separately for thresholds $1/k$, then taking a countable union of the controlled exceptional sets, is what yields one large set with uniform convergence.

Check Your Understanding

Use the tail-set definition and keep track of whether the error condition is strict or non-strict.

Why does $x\in X\setminus A_{N,k}$ imply $|f_n(x)-f(x)|<1/k$ for every $n\ge N$, rather than only at $n=N$?
For $f_n=\chi_{(0,1/n)}$ on $[0,1]$, compute $A_{N,k}$ and its Lebesgue measure for a fixed positive integer $k$.
In the powers example, explain why $A_{N,k}=[k^{-1/N},1)$ for $k\ge2$, including the treatment of the endpoint $x=1$.
Where in the proof is the assumption $\mu(X)<\infty$ used to control the measures of the tail exceptional sets?
Give a sequence that converges everywhere but for which a bad set at a single index fails to control the whole tail.