Convergence in Measure Through Error Sets

Errors at a Fixed Threshold

Convergence in measure asks whether the part of the space where a function differs substantially from its proposed limit becomes small. “Substantially” is measured using a fixed positive threshold: the threshold is held constant while the sequence index tends to infinity. The sets on which the error exceeds that threshold are called error sets. Calculating these sets directly makes the definition concrete, and it also shows why the measure space matters: an error set of small width has small Lebesgue measure, while an unbounded error set can have infinite measure.

Throughout the definition and results below, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; it need not be finite, sigma-finite, or complete. The functions $f_n$ and $f$ are measurable and real-valued, and statements of convergence in measure concern their error sets. We use the strict inequality $>$ in those sets.

Definition: For $\varepsilon>0$, the error set at index $n$ and threshold $\varepsilon$ is $$ E_n(\varepsilon)=\{x\in X:|f_n(x)-f(x)|>\varepsilon\}. $$ The sequence $(f_n)$ converges in measure to $f$ if, for every fixed $\varepsilon>0$, $$ \lim_{n\to\infty}\mu(E_n(\varepsilon))=0. $$ The limit is a limit of nonnegative extended real numbers; the definition does not require $\mu(X)$ to be finite.

Because $f_n$ and $f$ are measurable real-valued functions, $x\mapsto |f_n(x)-f(x)|$ is measurable, so $E_n(\varepsilon)\in\mathcal{M}$. The order of quantifiers is essential: choose any one $\varepsilon>0$, keep it fixed, and then let $n$ tend to infinity. The definition does not require the error to be small everywhere, nor does it require one index to work for every threshold simultaneously.

How Thresholds Organize the Error Sets

Lemma: For a fixed $n$, if $0<s<t$, then $E_n(t)\subseteq E_n(s)$. Consequently, $\mu(E_n(t))\le\mu(E_n(s))$.

Proof. Suppose $x\in E_n(t)$. By the definition of the error set, $|f_n(x)-f(x)|>t$. Since $t>s$, this implies $|f_n(x)-f(x)|>s$, so $x\in E_n(s)$. Thus $E_n(t)\subseteq E_n(s)$. Monotonicity of the measure $\mu$ gives $\mu(E_n(t))\le\mu(E_n(s))$, including when either measure is infinite. $\square$

This inclusion is useful for keeping track of thresholds: a larger allowed error produces a smaller error set. It also gives a countable way to test convergence in measure.

Theorem (Countable threshold criterion): On an arbitrary measure space $(X,\mathcal{M},\mu)$, for measurable real-valued $f_n$ and $f$, convergence of $f_n$ to $f$ in measure is equivalent to $$ \lim_{n\to\infty}\mu(E_n(1/k))=0 \quad\text{for every integer }k\ge1. $$

Proof. If $f_n$ converges in measure to $f$, then the definition applied to the positive threshold $\varepsilon=1/k$ gives the displayed limit for each integer $k\ge1$.

Conversely, suppose that the displayed limits hold for every $k\ge1$. Fix an arbitrary $\varepsilon>0$. Choose an integer $k\ge1$ so large that $1/k<\varepsilon$. If $x\in E_n(\varepsilon)$, then $|f_n(x)-f(x)|>\varepsilon>1/k$, so $x\in E_n(1/k)$. Therefore $E_n(\varepsilon)\subseteq E_n(1/k)$, and $$ 0\le\mu(E_n(\varepsilon))\le\mu(E_n(1/k)). $$ The right-hand side tends to zero by the assumption for this $k$. Hence $\mu(E_n(\varepsilon))\to0$. Since $\varepsilon>0$ was arbitrary, the definition gives convergence in measure. $\square$

Computing Error Sets

Worked Example: Powers on a Finite Interval

Let $X=[0,1]$, let $\mathcal{M}$ be the Lebesgue sigma-algebra on $X$, and let $\mu$ be Lebesgue measure restricted to $X$. This is a finite measure space with $\mu(X)=1$. Set $f_n(x)=x^n$ and $f(x)=0$. For each $x\in[0,1]$, these functions are measurable and real-valued. We calculate the error set for a fixed $\varepsilon>0$.

Here $|f_n(x)-f(x)|=x^n$. If $0<\varepsilon<1$, then $x^n>\varepsilon$ is equivalent to $x>\varepsilon^{1/n}$, since the function $x\mapsto x^n$ is increasing on $[0,1]$. Thus $$ E_n(\varepsilon)=(\varepsilon^{1/n},1], \qquad \mu(E_n(\varepsilon))=1-\varepsilon^{1/n}. $$ For fixed $0<\varepsilon<1$, $\varepsilon^{1/n}\to1$, so $\mu(E_n(\varepsilon))\to0$. If $\varepsilon\ge1$, then $x^n\le1\le\varepsilon$ for every $x\in[0,1]$, and the strict inequality $x^n>\varepsilon$ never holds. In that case $E_n(\varepsilon)=\varnothing$ and its measure is zero. These cases prove convergence in measure to zero.

Check: take $n=4$ and $\varepsilon=1/16$. The fourth root of $1/16$ is $1/2$, so $E_4(1/16)=(1/2,1]$, whose measure is $1-1/2=1/2$. Substituting into the formula gives $1-(1/16)^{1/4}=1-1/2=1/2$, as required.

Worked Example: A Narrow Triangular Error on the Real Line

Let $X=\mathbb{R}$, $\mathcal{M}=\mathcal{L}$, and $\mu=m$, Lebesgue measure. This measure space is sigma-finite but not finite. Define the measurable real-valued functions $$ f_n(x)=\max(1-n|x|,0),\qquad f(x)=0. $$ For $0<\varepsilon<1$, the error is positive only when $|x|<1/n$. Within that interval, $f_n(x)>\varepsilon$ is equivalent to $1-n|x|>\varepsilon$, hence to $|x|<(1-\varepsilon)/n$. Therefore $$ E_n(\varepsilon)=\left(-\frac{1-\varepsilon}{n},\frac{1-\varepsilon}{n}\right), \qquad m(E_n(\varepsilon))=\frac{2(1-\varepsilon)}{n}. $$ This measure tends to zero for each fixed $0<\varepsilon<1$. For $\varepsilon\ge1$, $f_n(x)\le1$ everywhere, so the strict inequality $f_n(x)>\varepsilon$ never holds; the error set is empty. Thus $f_n$ converges in measure to zero on $\mathbb{R}$.

Check: for $n=5$ and $\varepsilon=1/4$, the radius is $(1-1/4)/5=(3/4)/5=3/20$. The error set is $(-3/20,3/20)$, with measure $3/20-(-3/20)=6/20=3/10$. The general formula gives $2(1-1/4)/5=2(3/4)/5=3/10$, the same value.

For every fixed $x\ne0$, $f_n(x)=0$ once $n\ge1/|x|$, whereas $f_n(0)=1$ for every $n$. The persistent error at the single point $0$ does not prevent convergence in measure, because a singleton has Lebesgue measure zero. The calculation above establishes the convergence directly through the error sets.

Worked Example: Pointwise Decay with Infinite Error Sets

Let $X=[0,\infty)$ with the Lebesgue sigma-algebra and restricted Lebesgue measure. This space has infinite measure and is sigma-finite. Define the measurable real-valued functions $f_n(x)=x/(n+x)$ and $f(x)=0$. For each fixed $x\ge0$, $f_n(x)\to0$. We now test convergence in measure rather than infer it from this pointwise calculation.

If $0<\varepsilon<1$, then, because $n+x>0$, $$ \frac{x}{n+x}>\varepsilon \quad\Longleftrightarrow\quad x>\varepsilon(n+x) \quad\Longleftrightarrow\quad (1-\varepsilon)x>\varepsilon n \quad\Longleftrightarrow\quad x>\frac{n\varepsilon}{1-\varepsilon}. $$ Consequently $E_n(\varepsilon)=(n\varepsilon/(1-\varepsilon),\infty)$, which has infinite Lebesgue measure for every $n$. If $\varepsilon\ge1$, then $x/(n+x)<1\le\varepsilon$ for all finite $x\ge0$, so the error set is empty. For every threshold in $(0,1)$, however, its measure does not tend to zero; it is infinite at every index. Therefore $f_n$ does not converge in measure to zero on this space.

Check: at $\varepsilon=1/2$, the inequality reduces to $x>n$, so $E_n(1/2)=(n,\infty)$ and $m(E_n(1/2))=\infty$ for every $n$. In particular, the required limit of these measures is not zero.

Convergence in Measure and Uniform Convergence

Uniform convergence controls the error at every point, while convergence in measure controls the measure of the set where the error exceeds a fixed threshold. The distinction can be stated directly. Uniform convergence of real-valued functions means that for every $\varepsilon>0$, there is an $N$ such that $|f_n(x)-f(x)|<\varepsilon$ for every $x\in X$ and every $n\ge N$.

Proposition: On any measure space $(X,\mathcal{M},\mu)$, uniform convergence of measurable real-valued functions $f_n$ to $f$ implies convergence in measure.

Proof. Fix $\varepsilon>0$. By uniform convergence, there is an $N$ such that $|f_n(x)-f(x)|<\varepsilon$ for every $x\in X$ whenever $n\ge N$. Thus $E_n(\varepsilon)=\varnothing$ for every $n\ge N$, and $\mu(E_n(\varepsilon))=0$ for those indices. It follows that $\mu(E_n(\varepsilon))\to0$. Since this holds for each $\varepsilon>0$, the sequence converges in measure. No finiteness assumption on $\mu(X)$ is needed. $\square$

The converse need not hold. In the triangular example, $f_n(0)=1$ for every $n$, so the errors do not tend uniformly to zero, but the error-set measures tend to zero at every fixed positive threshold. Convergence in measure therefore describes the size of the region of large error, not the largest error value.

A Common Threshold Mistake

A frequent error is to reverse the inequality when solving for the error set. In the powers example, solving $x^n>\varepsilon$ for $0<\varepsilon<1$ gives $x>\varepsilon^{1/n}$, not $x<\varepsilon^{1/n}$. The incorrect inequality would produce $[0,\varepsilon^{1/n})$, with measure $\varepsilon^{1/n}$, which tends to $1$ rather than $0$. The correction is to use the fact that $x\mapsto x^n$ is increasing on $[0,1]$: values above the threshold occur to the right of its root. The correct error set $(\varepsilon^{1/n},1]$ has measure $1-\varepsilon^{1/n}$, which tends to zero.

The strict threshold also matters. In this example the endpoint $x=\varepsilon^{1/n}$ satisfies $x^n=\varepsilon$, so it is not in $E_n(\varepsilon)$. For Lebesgue measure on the interval, including or excluding that single endpoint would not change the measure, but on a general measure space a single point can have positive measure. The definition uses $>$, so membership must be decided using that strict inequality before computing the measure.

Check Your Understanding

For each question, begin by fixing a positive threshold and writing its error set with the strict inequality from the definition.

On $[0,1]$ with restricted Lebesgue measure, let $f_n(x)=x^{2n}$ and $f(x)=0$. Compute $E_n(\varepsilon)$ and its measure for $0<\varepsilon<1$.
For the triangular functions on $\mathbb{R}$, what is $E_n(1)$? Explain why the strict inequality determines the answer.
On $[0,\infty)$ with Lebesgue measure, solve $x/(n+x)>1/3$ for $x$, and state the measure of the resulting error set.
Prove the countable threshold criterion using thresholds $1/k$ rather than an arbitrary positive sequence of thresholds.
Give one reason convergence in measure does not assert that the errors are small at every point.