From Almost-Everywhere to Uniform Convergence
Almost-everywhere convergence allows a sequence of measurable functions to fail to converge on a null set. Even where convergence holds, however, the rate can depend on the point: no single index may work simultaneously across the whole space. Egorov's theorem gives a useful remedy on a finite-measure space. By removing a measurable set of arbitrarily small measure, one can make the convergence uniform on what remains.
Let \((X,\mathcal{M},\mu)\) be a measure space with \(\mu(X)<\infty\); the measure need not be complete. All functions in this tutorial are real-valued and \(\mathcal{M}\)-measurable. The statement \(f_n\to f\) \(\mu\)-a.e. means that there is a measurable null set \(N\) such that \(f_n(x)\to f(x)\) for every \(x\in X\setminus N\). Uniform convergence on a subset \(A\subseteq X\) means that for every \(\eta>0\), there is an \(N_0\) such that \(|f_n(x)-f(x)|<\eta\) for every \(x\in A\) and every \(n\ge N_0\). If \(A\) is empty, this condition holds vacuously.
The conclusion is uniform convergence on the complement of a small exceptional set, not necessarily on all of \(X\). The smallness is measured by \(\mu(E)\), and the bound is strict: the theorem guarantees \(\mu(E)<\varepsilon\). It does not claim that a single exceptional set works for every \(\varepsilon\).
Understanding the Hypotheses
For real-valued measurable functions, the set where \((f_n(x))\) converges is measurable. This ensures that the exceptional set where convergence fails can be treated as a measurable null set when using the definition of almost-everywhere convergence.
Proof. A real sequence converges if and only if it is Cauchy, because \(\mathbb{R}\) is complete. Thus \(x\in C\) precisely when, for every positive integer \(r\), there is a positive integer \(N\) such that \(|f_p(x)-f_q(x)|<1/r\) for all \(p,q\ge N\). Therefore
Each difference \(f_p-f_q\) is measurable, so each set inside the innermost intersection is measurable. The displayed expression uses only countable intersections and unions, which preserve membership in \(\mathcal{M}\). Hence \(C\in\mathcal{M}\). \(\square\)
Finite measure is a condition on the whole space: \(\mu(X)<\infty\), not merely that \(X\) is covered by countably many sets of finite measure. That latter property is sigma-finiteness, which is weaker. Egorov's conclusion can fail on a sigma-finite space of infinite measure. The theorem also requires almost-everywhere convergence; measurability by itself is not enough.
Proof. Fix \(x\in A\) and \(\eta>0\). By uniform convergence, there is an \(N_0\) such that \(|f_n(y)-f(y)|<\eta\) for every \(y\in A\) and every \(n\ge N_0\). In particular, the inequality holds at \(y=x\), so \(|f_n(x)-f(x)|<\eta\) for all \(n\ge N_0\). This is the definition of \(f_n(x)\to f(x)\). If \(A\) is empty, there are no points for which pointwise convergence must be checked, so the implication remains valid. \(\square\)
Worked Examples
Worked Example: Powers on the Unit Interval
On \(X=[0,1]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\), let \(f_n(x)=x^n\). Determine the pointwise limit and exhibit a small exceptional set outside which convergence is uniform.
If \(0\le x<1\), then \(x^n\to0\), while \(f_n(1)=1\) for every \(n\). Thus the pointwise limit is \(f(x)=0\) for \(x<1\), with \(f(1)=1\); in fact convergence holds everywhere. It is not uniform on all of \([0,1]\): for \(x<1\), \(|f_n(x)-f(x)|=x^n\), and the supremum over \(0\le x<1\) is \(1\) for every \(n\). The supremum equals \(1\), although no \(x<1\) attains it.
For a specific choice, take \(E=[3/4,1]\). Its measure is \(m(E)=1-3/4=1/4\). On \(X\setminus E=[0,3/4)\), the pointwise limit is zero and
Hence the supremum of the errors on \(X\setminus E\) is at most \((3/4)^n\), which tends to zero. For example, at \(n=1\) the bound is \(3/4\); at \(n=2\) it is \(9/16\); and at \(n=3\) it is \(27/64\). More generally, given any \(\varepsilon>0\), choose \(0<\delta<\min(\varepsilon,1)\) and set \(E=[1-\delta,1]\). Then \(m(E)=\delta<\varepsilon\) and the error on the complement is bounded by \((1-\delta)^n\to0\). Check: with \(\delta=1/4\), the exact bound at \(n=2\) is \((3/4)^2=9/16\), as above, and the bound tends to zero since \(0<3/4<1\).
Worked Example: Shrinking Spikes
On \([0,1]\) with Lebesgue measure, define \(f_n=n\chi_{(0,1/n)}\). Show that the functions converge almost everywhere but not uniformly on the whole interval, and construct an exceptional set of measure less than a prescribed \(\varepsilon>0\).
At \(x=0\), every \(f_n(x)=0\). At each fixed \(x>0\), choose an integer \(N>1/x\). For every \(n\ge N\), \(1/n\le1/N<x\), so \(x\notin(0,1/n)\) and \(f_n(x)=0\). Thus \(f_n(x)\to0\) for every \(x\in[0,1]\). On the other hand, \(\sup_{x\in[0,1]}|f_n(x)|=n\): the value \(n\) occurs at every \(x\in(0,1/n)\), a nonempty set. These suprema do not tend to zero, so convergence is not uniform on the whole interval.
Choose an integer \(N>1/\varepsilon\) and let \(E=(0,1/N)\). Then \(m(E)=1/N<\varepsilon\). For every \(n\ge N+1\), \(1/n\le1/(N+1)<1/N\), so the support \((0,1/n)\) lies inside \(E\). Consequently \(f_n=0\) on \([0,1]\setminus E\) for every \(n\ge N+1\). The convergence on that complement is eventually exact, hence uniform. Check: at \(n=N+1\), the support endpoint is \(1/(N+1)\), strictly less than \(1/N\), so the open support is indeed contained in \(E\).
Worked Example: A Finite Counting Measure Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be its power set, and assign masses \(\mu(\{a\})=1\), \(\mu(\{b\})=2\), and \(\mu(\{c\})=1\). Define \(f_n(a)=1/n\), \(f_n(b)=2/n\), and \(f_n(c)=(-1)^n/n\). Verify uniform convergence to zero.
The total measure is \(\mu(X)=1+2+1=4\), so the space has finite measure. Each point has positive measure, so a null subset cannot contain any of \(a,b,c\). The pointwise limits are all zero: \(1/n\to0\), \(2/n\to0\), and \(|(-1)^n/n|=1/n\to0\). At every point, \(|f_n(x)|\le2/n\), and equality occurs at \(b\). Therefore
This is uniform convergence on \(X\) itself, so one may choose \(E=\varnothing\), whose measure is \(0<\varepsilon\) for every \(\varepsilon>0\). Check: at \(n=4\), the three absolute values are \(1/4\), \(1/2\), and \(1/4\), and their maximum is \(1/2=2/4\), as the formula gives.
Why the Scope of the Theorem Matters
Worked Example: Infinite Measure Can Prevent Uniformization
On \(\mathbb{R}\) with Lebesgue measure \(m\), set \(f_n=\chi_{[n,\infty)}\) and \(f=0\). Show that \(f_n\to f\) pointwise but that no finite-measure exceptional set makes the convergence uniform.
For every fixed \(x\in\mathbb{R}\), all integers \(n>x\) satisfy \(x<n\), so \(x\notin[n,\infty)\) and \(f_n(x)=0\). Thus convergence to zero holds everywhere. Now let \(E\) be any measurable set with \(m(E)<\infty\). For every integer \(n\), the tail \([n,\infty)\) has infinite measure. It cannot be contained in \(E\), because then monotonicity of measure would give \(m([n,\infty))\le m(E)<\infty\). Hence there is an \(x\in[n,\infty)\setminus E\), and at that point \(|f_n(x)-f(x)|=1\). In particular, for every \(n\), there is a point outside \(E\) with error greater than \(1/2\). No index can make all errors on \(\mathbb{R}\setminus E\) less than \(1/2\), so convergence there is not uniform. The space has infinite measure, and the finite-measure hypothesis of Egorov's theorem is absent.
A related pitfall is to treat an exceptional set of small measure as though the sequence must converge uniformly on the original space. In the powers example, the incorrect calculation is to say that \(\sup_{x\in[0,1]}x^n=(3/4)^n\). That bound only applies when \(x<3/4\); on the full interval the supremum of \(|x^n-f(x)|\) is \(1\) for every \(n\). The correction is to specify the domain of the bound: on \([0,3/4)\), the error is at most \((3/4)^n\), while on the full interval it is not uniformly small.
The zero-measure case also deserves care. If \(\mu(X)=0\), take \(E=X\), not \(E=\varnothing\). Then \(\mu(E)=0<\varepsilon\), and \(X\setminus E=\varnothing\), where uniform convergence holds vacuously. Choosing \(E=\varnothing\) would leave all of \(X\) in the complement, and uniform convergence there need not hold merely because \(X\) has measure zero. The theorem's conclusion concerns the complement of the chosen exceptional set, so that complement must be identified correctly.
Check Your Understanding
For each question, identify the measure space and distinguish convergence on the whole space from convergence after removing an exceptional set.
- For \(f_n(x)=x^n\) on \([0,1]\), what is the pointwise limit at \(x=1\), and why does removing \([3/4,1]\) give uniform convergence on the complement?
- For \(f_n=n\chi_{(0,1/n)}\), verify that the sequence converges pointwise at \(x=0\) and at a fixed \(x>0\). Why is its convergence not uniform on \([0,1]\)?
- State the measure hypothesis in Egorov's theorem. Is sigma-finiteness alone enough to replace it?
- If \(\mu(X)=0\), which set should be chosen as the exceptional set to make the complement empty? Verify the strict inequality required for any \(\varepsilon>0\).
- For \(f_n=\chi_{[n,\infty)}\) on \(\mathbb{R}\), explain why every finite-measure exceptional set leaves a point with error greater than \(1/2\) for each \(n\).