Uniform Convergence and Convergence in Measure on Finite Spaces

Two Different Ways to Control Error

Uniform convergence controls the size of the error at every point, once the index is large enough. Convergence in measure controls instead the size of the set on which the error exceeds a fixed threshold. These are different demands: a function can have a large error on a set whose measure is very small. In this tutorial, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu(X)<\infty\), unless a statement explicitly says otherwise. All functions \(f_n,f:X\to\mathbb{R}\) under discussion are \(\mathcal{M}\)-measurable, and convergence statements are everywhere-defined unless marked \(\mu\)-a.e.

Definition: The sequence \((f_n)\) converges uniformly to \(f\) on \(X\) if, for every \(\varepsilon>0\), there is an \(N\) such that \(n\ge N\) implies \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in X\). It converges in measure to \(f\) if, for every \(\varepsilon>0\), \(\mu(\{x\in X:|f_n(x)-f(x)|>\varepsilon\})\to0\) as \(n\to\infty\). The strict threshold \(>\varepsilon\) is part of this definition.

The uniform condition has the quantifier order “for every threshold, eventually for every point.” Convergence in measure asks that, at each fixed threshold, the measure of the points violating that threshold tends to zero. Measurability ensures that the error set whose measure is taken belongs to \(\mathcal{M}\). Neither definition requires the functions to be integrable.

Uniform Convergence Implies Convergence in Measure

Theorem: Let \((X,\mathcal{M},\mu)\) be a finite measure space, and let \(f_n,f:X\to\mathbb{R}\) be measurable. If \(f_n\to f\) uniformly on \(X\), then \(f_n\to f\) in measure.

Proof. Fix \(\varepsilon>0\), and define \(E_n(\varepsilon)=\{x\in X:|f_n(x)-f(x)|>\varepsilon\}\). By uniform convergence, there is an \(N\) such that for every \(n\ge N\) and every \(x\in X\), \(|f_n(x)-f(x)|<\varepsilon\). Thus no \(x\) belongs to \(E_n(\varepsilon)\) when \(n\ge N\), so \(E_n(\varepsilon)=\varnothing\) and \(\mu(E_n(\varepsilon))=0\). In particular, these measures tend to zero for every fixed \(\varepsilon>0\), which is convergence in measure. The proof does not use \(\mu(X)<\infty\); uniform convergence implies convergence in measure on an arbitrary measure space as well. \(\square\)

The finite-measure assumption is therefore unnecessary for this implication. It is important in other comparisons of convergence modes, but the direct reason here is stronger: beyond a suitable index, the error set is empty, not merely small. The converse is different. Convergence in measure allows error sets to remain nonempty, provided their measures tend to zero.

Worked Example: A Shrinking Spike

On \(X=[0,1]\) with the restricted Lebesgue measure \(m\), let \(f_n(x)=n\chi_{(0,1/n)}(x)\) and \(f(x)=0\). Each \(f_n\) is measurable and real-valued. For a fixed \(\varepsilon>0\), if \(n\le\varepsilon\), then \(f_n\) never exceeds \(\varepsilon\), so the error set is empty. If \(n>\varepsilon\), then the error exceeds \(\varepsilon\) exactly on \((0,1/n)\). Consequently,

m(\{|f_n-f|>\varepsilon\})= \begin{cases} 0,&n\le\varepsilon,\\[2pt] 1/n,&n>\varepsilon. \end{cases}

For each fixed \(\varepsilon>0\), eventually \(n>\varepsilon\), and then the error-set measure is \(1/n\to0\). Hence \(f_n\to0\) in measure. But uniform convergence fails: for every \(n\), at \(x=1/(2n)\) we have \(0<x<1/n\), so \(|f_n(x)-0|=n\). In particular, the uniform-convergence condition fails at the fixed threshold \(1/2\) for every \(n\). Check: the error set has measure \(1/n\) when \(n>\varepsilon\), while the error takes the value \(n\) at points in that set; its shrinking measure does not make its height small.

Two More Ways the Converse Can Fail

A shrinking spike has increasingly large values on increasingly small sets. The next example keeps the function values bounded, but moves their supports through the space. The third example uses a finite measure space with countably many points whose individual masses tend to zero.

Worked Example: Moving Dyadic Intervals

Take \(X=[0,1)\) with Lebesgue measure. For each integer \(k\ge0\), and each \(j\in\{0,\ldots,2^k-1\}\), assign the index \(n=2^k+j\) to the interval \(I_{k,j}=[j/2^k,(j+1)/2^k)\), and set \(f_n=\chi_{I_{k,j}}\). These blocks of indices are disjoint and cover the positive integers: the block for \(k\) runs from \(2^k\) through \(2^{k+1}-1\). Thus this defines one sequence.

Fix \(\varepsilon>0\). If \(0<\varepsilon<1\), the error set \(\{|f_n|>\varepsilon\}\) is exactly the interval assigned to \(n\), of measure \(2^{-k}\). As \(n\to\infty\), its block number \(k\to\infty\), and \(2^{-k}\to0\). If \(\varepsilon\ge1\), the error set is empty because \(f_n\) only takes the values zero and one, and the threshold is strict. Therefore \(f_n\to0\) in measure. On the other hand, every assigned interval is nonempty, and \(f_n=1\) throughout that interval. Hence \(\sup_{x\in X}|f_n(x)|=1\) for every \(n\), so the sequence does not converge uniformly to zero. Check: at level \(k=2\), each interval has measure \(2^{-2}=1/4\); at level \(k=3\), each has measure \(2^{-3}=1/8\). The error-set measures decrease across levels, but the supremum error stays equal to \(1\).

In fact, this sequence does not converge pointwise at any \(x\in[0,1)\): at every level \(k\), exactly one of the intervals \(I_{k,j}\) contains \(x\), while the other intervals in that block do not. Thus the sequence has value \(1\) for infinitely many indices and value \(0\) for infinitely many indices. This also illustrates that convergence in measure does not require pointwise convergence.

Worked Example: Small Atoms on a Finite-Measure Space

Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be all subsets of \(X\), and define \(\mu(\{j\})=2^{-j}\). Countable additivity gives \(\mu(X)=\sum_{j=1}^{\infty}2^{-j}=1\), so this is a finite measure space. Set \(f_n=\chi_{\{n\}}\) and \(f=0\). At each fixed \(j\), \(f_n(j)=0\) for every \(n>j\), so \(f_n(j)\to0\) pointwise. For \(0<\varepsilon<1\), the strict-threshold error set is \(\{n\}\), and its measure is \(2^{-n}\to0\). For \(\varepsilon\ge1\), the error set is empty. Thus convergence in measure holds.

For every \(n\), however, \(|f_n(n)-f(n)|=1\). With the fixed threshold \(\varepsilon=1/2\), no index has error less than \(\varepsilon\) at every point, so the convergence is not uniform. Check: \(\mu(\{1\})=1/2\), \(\mu(\{2\})=1/4\), and \(\mu(\{3\})=1/8\), while the error height at the corresponding point is \(1\) in each case. The space has finite total measure, but its individual point masses have no positive lower bound.

When the Converse Does Hold

The small-atom example suggests a condition that prevents nonempty error sets from having arbitrarily small measure. In particular, on a finite set with positive mass at each point, convergence in measure does imply uniform convergence. Here “finite set” means that \(X\) has finitely many elements; this is a stronger, different condition than merely having finite total measure.

Proposition: Suppose \(X\) is a nonempty finite set, \(\mathcal{M}\) is its power set, and \(\mu(\{x\})>0\) for every \(x\in X\). If measurable real-valued functions \(f_n,f\) converge in measure on \((X,\mathcal{M},\mu)\), then they converge uniformly on \(X\).

Proof. Since \(X\) is finite and every singleton has positive measure, the number \(\delta=\min_{x\in X}\mu(\{x\})\) exists and satisfies \(\delta>0\). Fix a desired uniform error \(\eta>0\). Convergence in measure, applied with threshold \(\eta/2\), implies that for all sufficiently large \(n\),

\mu(\{|f_n-f|>\eta/2\})<\delta.

If this error set were nonempty, it would contain some point \(x\), and then its measure would be at least \(\mu(\{x\})\ge\delta\), a contradiction. Therefore the error set is empty for all sufficiently large \(n\). At every \(x\in X\) we then have \(|f_n(x)-f(x)|\le\eta/2<\eta\), which is uniform convergence. \(\square\)

The proposition pinpoints what fails in the countable example: there is no positive lower bound for the masses of nonempty sets, since the singleton masses \(2^{-n}\) tend to zero. On a general finite-measure space, a nonempty error set can likewise have arbitrarily small measure, so convergence in measure alone does not force uniform control at every point.

A Common Quantifier Mistake

A tempting but invalid argument is: “The error set has small measure, so it must be empty; therefore the error is uniformly small.” Small measure does not imply emptiness. For the dyadic-interval example, at each index \(n=2^k+j\) the error set at threshold \(1/2\) has measure \(2^{-k}>0\), even though these measures tend to zero as the level \(k\) increases. At every such index, the error still equals \(1\) somewhere. Convergence in measure says that the measures of these sets tend to zero; it does not say that any one of them has measure zero or that the sets eventually disappear.

The implication proved above works for the opposite reason: uniform convergence makes the error set exactly empty for all sufficiently large indices at each fixed threshold. Keeping track of this distinction—small sets versus empty sets—is the key to comparing the two modes of convergence.

Check Your Understanding

For each question, keep the threshold fixed while taking the limit in \(n\), and distinguish “small measure” from “empty.”

Why does the proof that uniform convergence implies convergence in measure not require \(\mu(X)<\infty\)?
For \(f_n=n\chi_{(0,1/n)}\) on \([0,1]\), compute \(m(\{|f_n|>2\})\) for \(n=1,2,3\), and determine its limit.
In the dyadic-interval example, what is the error-set measure at threshold \(1/2\) for an index in the block corresponding to level \(k\)?
Why does \(f_n=\chi_{\{n\}}\) converge pointwise to zero on \(\mathbb{N}\), even though \(\sup_{j\in\mathbb{N}}|f_n(j)|=1\) for every \(n\)?
Which positive lower bound on singleton masses is used in the finite-set proposition, and where is it used in the proof?