A Metric for Convergence in Measure on Finite Spaces

Truncating the Error to Build a Distance

Convergence in measure is expressed through the size of error sets: for every fixed threshold, the measure of the set where the error exceeds that threshold must tend to zero. A useful way to encode all thresholds in one number is to cap the pointwise error at $1$, then integrate. Small errors remain small, while large errors contribute at most $1$ per point. On a finite measure space this construction gives a metric, and convergence in that metric is equivalent to convergence in measure.

Throughout, $(X,\mathcal{M},\mu)$ is a measure space with $\mu(X)<\infty$; no completeness assumption is needed. All functions considered are measurable, real-valued functions on $X$. Two such functions are equivalent if they are equal $\mu$-a.e. We write $[f]$ for the equivalence class of $f$. Convergence in measure of $f_n$ to $f$ means that for every $\varepsilon>0$, $\mu(\{|f_n-f|>\varepsilon\})\to0$.

Definition: For measurable real-valued $f,g$, define $$ d([f],[g])=\int_X \min(1,|f-g|)\,d\mu. $$ The integrand is measurable and bounded by $1$, so this integral is finite because $\mu(X)<\infty$.

The cap is applied before integration. Thus this distance is not generally the $L^1$ distance $\int_X|f-g|\,d\mu$: differences larger than $1$ contribute exactly $1$ per point, rather than their full size. We first verify that the formula is well-defined on equivalence classes and satisfies the metric axioms.

Why the Formula Is a Metric

Theorem: On the set of equivalence classes of measurable real-valued functions modulo equality $\mu$-a.e., $d$ is a metric. In particular, it is finite, nonnegative, symmetric, satisfies the triangle inequality, and $d([f],[g])=0$ if and only if $f=g$ $\mu$-a.e.

Proof. If $f=f'$ $\mu$-a.e. and $g=g'$ $\mu$-a.e., then $f-g=f'-g'$ outside the union of two null sets. Consequently the integrands defining $d([f],[g])$ and $d([f'],[g'])$ agree $\mu$-a.e., so their integrals agree. The definition therefore does not depend on the representatives. The integrand is nonnegative and bounded by $1$, so $d\ge0$ and $d\le\mu(X)<\infty$. Since $|f-g|=|g-f|$, symmetry follows.

For nonnegative real numbers $a,b$, we have $$ \min(1,a+b)\le \min(1,a)+\min(1,b). $$ Indeed, if either $a\ge1$ or $b\ge1$, the right side is at least $1$, while the left side is at most $1$. If both are less than $1$, the right side is $a+b$, which is at least the left side. Now $|f-h|\le|f-g|+|g-h|$, and the function $t\mapsto\min(1,t)$ is nondecreasing. Applying the displayed inequality pointwise gives $$ \min(1,|f-h|)\le\min(1,|f-g|)+\min(1,|g-h|). $$ Integrating proves $d([f],[h])\le d([f],[g])+d([g],[h])$.

It remains to check separation. If $f=g$ $\mu$-a.e., then the integrand is zero $\mu$-a.e., so $d([f],[g])=0$. Conversely, suppose $d([f],[g])=0$. For every integer $k\ge1$, let $E_k=\{|f-g|>1/k\}$. On $E_k$, the integrand is at least $1/k$; hence $$ 0=d([f],[g])\ge \frac{1}{k}\mu(E_k), $$ so $\mu(E_k)=0$. Since $\{|f-g|>0\}=\bigcup_{k=1}^{\infty}E_k$, countable subadditivity gives $\mu(\{|f-g|>0\})=0$. Thus $f=g$ $\mu$-a.e., as required. $\square$

Worked Example: Computing the Distance on a Finite Space

Let $X=\{a,b,c\}$, let $\mathcal{M}$ be all subsets of $X$, and assign masses $\mu(\{a\})=1/2$, $\mu(\{b\})=1$, and $\mu(\{c\})=3/2$. Define $f(a)=0,\ f(b)=2,\ f(c)=5$ and $g(a)=1/2,\ g(b)=0,\ g(c)=3$. Compute $d([f],[g])$.

The absolute differences at $a,b,c$ are respectively $1/2,2,2$. After truncation, their values are $1/2,1,1$. The integral on this finite space is the sum of each value times the mass of its point, so $$ d([f],[g])=\frac12\cdot\frac12+1\cdot1+1\cdot\frac32 =\frac14+1+\frac32=\frac{11}{4}. $$ Check: the masses sum to $1/2+1+3/2=3$, and the truncated integrand is at most $1$, so $d\le3$. The computed value $11/4$ satisfies this bound. Reversing $f$ and $g$ leaves all three absolute differences unchanged, confirming symmetry in this calculation.

Metric Convergence and Convergence in Measure

Theorem: Suppose $(X,\mathcal{M},\mu)$ is a finite measure space and $f_n,f$ are measurable real-valued functions. Then $$ d([f_n],[f])\longrightarrow0 \quad\text{if and only if}\quad f_n\longrightarrow f\text{ in measure}. $$

Proof. First suppose $d([f_n],[f])\to0$. Fix any threshold $\varepsilon>0$ and set $c=\min(1,\varepsilon)>0$. On $\{|f_n-f|>\varepsilon\}$, we have $\min(1,|f_n-f|)\ge c$. Integrating this pointwise bound gives $$ c\,\mu(\{|f_n-f|>\varepsilon\})\le d([f_n],[f]). $$ The right side tends to zero and $c$ is fixed and positive, so the error-set measure tends to zero. This holds for every $\varepsilon>0$, proving convergence in measure.

Conversely, suppose $f_n\to f$ in measure. If $\mu(X)=0$, every measurable subset of $X$ has measure zero, and $d([f_n],[f])=0$ for every $n$. Now suppose $0<\mu(X)<\infty$. Fix a target $\eta>0$, and choose $\delta=\eta/(2\mu(X))>0$. Partition $X$ into the sets where $|f_n-f|\le\delta$ and where $|f_n-f|>\delta$. On the first set, the truncated error is at most $\delta$; on the second, it is at most $1$. Therefore $$ d([f_n],[f]) \le \delta\,\mu(X)+\mu(\{|f_n-f|>\delta\}) =\frac{\eta}{2}+\mu(\{|f_n-f|>\delta\}). $$ By convergence in measure, there is an index after which the last measure is less than $\eta/2$. For those indices, $d([f_n],[f])<\eta$. Since this works for every $\eta>0$, the metric distance tends to zero. $\square$

The two estimates behind the proof also explain the role of the truncation. A small metric distance forces every fixed-threshold error set to have small measure. In the other direction, choose a small threshold $\delta$: the contribution below that threshold is controlled by $\delta\mu(X)$, and convergence in measure controls the remaining error set. Finiteness of $\mu(X)$ makes the first contribution small when $\delta$ is small.

Worked Example: A Spike Sequence

On $[0,1]$ with Lebesgue measure $m$, let $f_n=n\chi_{(0,1/n)}$ and $f=0$, for integers $n\ge1$. Compute the metric distance and verify convergence in measure.

On $(0,1/n)$, the absolute error is $n$, so its truncated value is $\min(1,n)=1$. Outside that interval the error is zero. Since $m((0,1/n))=1/n$, $$ d([f_n],[0])=\int_0^1\min(1,|f_n|)\,dm =1\cdot\frac1n+0\cdot\left(1-\frac1n\right)=\frac1n. $$ Thus $d([f_n],[0])\to0$. For any fixed $\varepsilon>0$, the strict error set $\{|f_n|>\varepsilon\}$ is empty if $n\le\varepsilon$, and is $(0,1/n)$ if $n>\varepsilon$. In the latter case its measure is $1/n$, which tends to zero as $n\to\infty$. Hence $f_n\to0$ in measure. Check: for $n=1,2,4$, the metric distances are respectively $1,1/2,1/4$, as given by $1/n$.

Worked Example: Small Uniform Errors on a Finite Interval

On $[0,2]$ with Lebesgue measure, define $f_n(x)=1/n$ at every point and let $f(x)=0$. Find the distance and check the error-set definition directly.

For every $x$, the truncated error is $\min(1,1/n)=1/n$, since $n\ge1$. The interval has measure $2$, so $$ d([f_n],[0])=\int_0^2\frac1n\,dm=\frac{2}{n}. $$ For a fixed $\varepsilon>0$, the strict error set is all of $[0,2]$ when $1/n>\varepsilon$, and empty when $1/n\le\varepsilon$. For all sufficiently large $n$, $1/n\le\varepsilon$, so its measure is eventually zero. This verifies convergence in measure, and the computed distance also tends to zero. Check: at $n=1,2,4$, the distances are $2,1,1/2$, respectively, which agree with $2/n$.

Why the Finite-Measure Hypothesis Matters

The implication from metric convergence to convergence in measure used only the threshold estimate, not finiteness of the measure space. The converse used finiteness to control the small-error region by $\delta\mu(X)$. On an infinite measure space, a uniformly tiny error can occupy a set of very large measure, so its integral after truncation need not be small.

Worked Example: Convergence in Measure Without Metric Convergence

On $\mathbb{R}$ with Lebesgue measure $m$, let $f_n=(1/n)\chi_{[-n^2,n^2]}$ and $f=0$. Each $f_n$ is measurable and real-valued, but $m(\mathbb{R})=\infty$. Show that $f_n\to0$ in measure although the truncated integral distance does not tend to zero.

Fix $\varepsilon>0$. If $1/n\le\varepsilon$, then $|f_n(x)|\le\varepsilon$ for every $x$, so the strict error set $\{|f_n|>\varepsilon\}$ is empty. This inequality holds for all sufficiently large $n$, proving convergence in measure. For the metric integral, $1/n\le1$, so the truncated error equals $1/n$ throughout $[-n^2,n^2]$, whose measure is $2n^2$, and is zero outside. Therefore $$ \int_{\mathbb{R}}\min(1,|f_n|)\,dm =\frac1n\cdot 2n^2=2n. $$ This tends to infinity, not to zero. Check: for $n=1,2,3$, the interval lengths are $2,8,18$, and the truncated error heights are $1,1/2,1/3$; the corresponding integrals are $2,4,6$, in agreement with $2n$.

Common pitfall: confusing the truncated metric with the $L^1$ distance. For the spike sequence on $[0,1]$, a mistaken calculation might omit the cap and write $\int_0^1|f_n|\,dm=n\cdot(1/n)=1$, then conclude that the metric distance is always $1$. That computes the $L^1$ distance, not $d$. The metric integrand is $\min(1,|f_n|)$, which equals $1$, not $n$, on the support. Thus $d([f_n],[0])=1\cdot(1/n)=1/n$. The cap matters precisely when the pointwise error exceeds $1$.

This metric is useful because it replaces the family of error-set tests in convergence in measure with a single numerical distance. It does not turn convergence in measure into $L^1$ convergence: the spike example has $\int_0^1|f_n|\,dm=1$ for every $n$, while its truncated metric distance tends to zero. The metric records convergence in measure, not the total size of the uncapped error.

Check Your Understanding

Use the definition of $d$ and the threshold estimates to answer the following.

On a finite measure space with $\mu(X)=5$, suppose $d([f],[g])=0.06$. What bound does the threshold estimate give for $\mu(\{|f-g|>0.2\})$?
Explain why changing either representative on a null set does not change $d([f],[g])$.
On $[0,1]$, compute the metric distance from the zero function to $f=3\chi_{[1/4,1/2]}$.
For the functions $f_n=(1/n)\chi_{[-n^2,n^2]}$ on $\mathbb{R}$, why does convergence in measure hold even though the truncated integrals do not tend to zero?
For the spike sequence $f_n=n\chi_{(0,1/n)}$ on $[0,1]$, compare its $L^1$ distance from zero with its truncated metric distance.