Convergence in Measure of Indicators and Symmetric Differences

Indicators Turn Function Errors into Set Differences

For general functions, convergence in measure is tested by measuring the sets on which the functions differ by more than a chosen positive threshold. For indicators, the pointwise difference has only two possible values: zero or one. This makes the error set especially concrete: below the threshold one, it is exactly the symmetric difference of the sets. The resulting equivalence is useful whenever a problem about functions can be expressed in terms of membership in measurable sets.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; it is not assumed to be finite, sigma-finite, or complete. All sets mentioned are in $\mathcal{M}$, and $\chi_E:X\to\{0,1\}$ denotes the real-valued indicator of $E$: it is one on $E$ and zero on $E^c$. For real-valued measurable functions $f_n,f$, convergence in measure means that for every $\varepsilon>0$, $\mu(\{|f_n-f|>\varepsilon\})\to0$. This definition allows the measures of individual error sets to be infinite; convergence requires that they eventually become small in the stated limit.

Definition: The symmetric difference of sets $A,B\in\mathcal{M}$ is $A\,\triangle\,B=(A\setminus B)\cup(B\setminus A)$. It consists exactly of the points that belong to one of $A,B$, but not both.

The Exact Error-Set Identity

Proposition: For every $A,B\in\mathcal{M}$ and every $x\in X$, $$ |\chi_A(x)-\chi_B(x)|=\chi_{A\,\triangle\,B}(x). $$ Consequently, for every $\varepsilon$ with $0<\varepsilon<1$, $$ \{x:|\chi_A(x)-\chi_B(x)|>\varepsilon\}=A\,\triangle\,B. $$

Proof. If $x\in A\cap B$, both indicators are one, so their absolute difference is zero, and $x\notin A\,\triangle\,B$. If $x\in A^c\cap B^c$, both indicators are zero, so the absolute difference is again zero, and $x\notin A\,\triangle\,B$. If $x\in A\setminus B$, the indicators are one and zero, respectively, so the absolute difference is one and $x\in A\,\triangle\,B$. If $x\in B\setminus A$, the absolute difference is also one and $x\in A\,\triangle\,B$. These four cases exhaust $X$, proving the pointwise identity. For $0<\varepsilon<1$, zero is not strictly greater than $\varepsilon$, while one is; therefore the strict-threshold error set consists exactly of the points where the absolute difference is one. $\square$

Theorem: Let $E_n,E\in\mathcal{M}$ in an arbitrary measure space $(X,\mathcal{M},\mu)$. The real-valued indicators $\chi_{E_n}$ converge in measure to $\chi_E$ if and only if $$ \mu(E_n\,\triangle\,E)\longrightarrow 0. $$

Proof. Suppose first that $\chi_{E_n}\to\chi_E$ in measure. The definition of convergence in measure applies at the particular threshold $\varepsilon=\tfrac12$. Since $0<\tfrac12<1$, the proposition gives $$ \mu(E_n\,\triangle\,E)=\mu(\{x:|\chi_{E_n}(x)-\chi_E(x)|>\tfrac12\})\longrightarrow 0. $$ For the converse, suppose $\mu(E_n\,\triangle\,E)\to0$, and fix any $\varepsilon>0$, as required by the definition of convergence in measure. If $0<\varepsilon<1$, the proposition says that the error set at threshold $\varepsilon$ is $E_n\,\triangle\,E$, whose measure tends to zero by assumption. If $\varepsilon\ge1$, the absolute difference of two indicators is always either zero or one. Neither value is strictly greater than $\varepsilon$, so the error set is empty for every $n$ and its measure is zero. Thus the error-set measures tend to zero for every $\varepsilon>0$, proving convergence in measure. No finiteness assumption on $\mu(X)$ was used. $\square$

Computing the Symmetric Difference

Worked Example: Nested Intervals with a Vanishing Error

On $(\mathbb{R},\mathcal{L},m)$, let $E=(0,1)$ and $E_n=(0,1+1/n)$ for $n\ge1$. Determine whether $\chi_{E_n}\to\chi_E$ in measure and compute the error-set measures.

Since $E\subset E_n$, their symmetric difference is $E_n\setminus E$. The interval $E_n$ includes the point $1$, while $E$ does not; the right endpoint $1+1/n$ is excluded. Hence $$ E_n\,\triangle\,E=[1,1+1/n). $$ Its Lebesgue measure is its length, so $m(E_n\,\triangle\,E)=(1+1/n)-1=1/n$. For any fixed $0<\varepsilon<1$, the exact identity gives $$ m(\{x:|\chi_{E_n}(x)-\chi_E(x)|>\varepsilon\})=1/n\longrightarrow0. $$ For $\varepsilon\ge1$, the error set is empty because the difference of the indicators is never strictly greater than one. Thus the indicators converge in measure. Check: at $n=2$, the symmetric difference is $[1,3/2)$, of measure $3/2-1=1/2$, which agrees with $1/n=1/2$.

Worked Example: Equal Set Measures Do Not Ensure Convergence

On $(\mathbb{R},\mathcal{L},m)$, let $E=(0,1/2)$. Define $E_n=(0,1/2)$ when $n$ is even and $E_n=(1/2,1)$ when $n$ is odd. Each set has measure $1/2$. Does $\chi_{E_n}$ converge in measure to $\chi_E$?

For even $n$, $E_n=E$, so $E_n\,\triangle\,E=\varnothing$ and its measure is zero. For odd $n$, the two intervals are disjoint: both exclude $1/2$, and their union is their symmetric difference. Its measure is $$ m((0,1/2)\cup(1/2,1))=1/2+1/2=1. $$ Thus the symmetric-difference measures are zero for even $n$ and one for odd $n$; they do not tend to zero. Equivalently, at the fixed threshold $\varepsilon=1/2$, the error-set measure is one for every odd $n$, so the indicators do not converge in measure. Check: although $m(E_n)=m(E)=1/2$ for every $n$, the error-set measures along odd indices remain $1$. Equality of set measures alone does not control where the sets lie.

Worked Example: Convergence on an Infinite-Measure Space

On $(\mathbb{R},\mathcal{L},m)$, let $E=\mathbb{R}$ and $E_n=\mathbb{R}\setminus(0,1/n)$, for $n\ge1$. The space has infinite measure, as do $E$ and every $E_n$. Determine whether the indicators converge in measure.

The points that belong to exactly one of $E_n,E$ are precisely those in the removed open interval: $$ E_n\,\triangle\,E=(0,1/n). $$ The endpoints are not in this symmetric difference: $0$ and $1/n$ both belong to $E_n$ and to $E$. Therefore $m(E_n\,\triangle\,E)=1/n-0=1/n\to0$. By the theorem, $\chi_{E_n}\to\chi_E$ in measure, despite the infinite measures of the sets. Directly, for $0<\varepsilon<1$ the error set is $(0,1/n)$, of measure $1/n$; for $\varepsilon\ge1$ it is empty. Check: at $n=4$, the difference interval has length $1/4-0=1/4$, as given by $1/n$. This example also shows why the theorem does not require a finite measure space.

What Set-Measure Convergence Does and Does Not Say

If the sets themselves have finite measure, vanishing symmetric differences also implies convergence of their measures. The measure space need not be finite; it is enough to assume that the particular sets have finite measure.

Proposition: Let $E_n,E\in\mathcal{M}$ satisfy $\mu(E_n)<\infty$ and $\mu(E)<\infty$ in $(X,\mathcal{M},\mu)$. Then $$ |\mu(E_n)-\mu(E)|\le\mu(E_n\,\triangle\,E). $$

Proof. Write $A_n=E_n\setminus E$ and $B_n=E\setminus E_n$. The disjoint decompositions $E_n=(E_n\cap E)\cup A_n$ and $E=(E_n\cap E)\cup B_n$ give $$ \mu(E_n)-\mu(E)=\mu(A_n)-\mu(B_n). $$ Both measures on the right are finite because $A_n\subset E_n$ and $B_n\subset E$. For nonnegative real numbers $a,b$, $|a-b|\le a+b$; applying this with $a=\mu(A_n)$ and $b=\mu(B_n)$ gives $$ |\mu(E_n)-\mu(E)|\le\mu(A_n)+\mu(B_n)=\mu(E_n\,\triangle\,E). $$ This proves the inequality. $\square$

Consequently, if $\mu(E_n\,\triangle\,E)\to0$ and all these sets have finite measure, then $\mu(E_n)\to\mu(E)$. The converse is false, as the alternating-interval example shows: the measures are all $1/2$, but the symmetric differences do not become small. When $\mu(E)=\infty$, the displayed difference of set measures is not a finite real-valued quantity, so this consequence is not asserted.

Symmetric differences also give a useful bound for combining set operations. For any $A,B,C,D\in\mathcal{M}$, $$ (A\cup B)\,\triangle\,(C\cup D)\subseteq(A\,\triangle\,C)\cup(B\,\triangle\,D). $$ Indeed, if a point has the same membership status in $A,C$ and also the same status in $B,D$, then it has the same status in $A\cup B$ and $C\cup D$. Taking the contrapositive gives the inclusion. By subadditivity of $\mu$, its measure is at most $\mu(A\,\triangle\,C)+\mu(B\,\triangle\,D)$. Thus small symmetric differences for two pairs of sets control the error when their unions are formed. This is a set-level estimate; the exact indicator theorem remains the direct test for convergence in measure.

A Threshold Pitfall

A common mistake is to test indicators only at thresholds $\varepsilon\ge1$. Because the definition uses the strict inequality $|f_n-f|>\varepsilon$, this test always produces an empty error set for indicators, whether or not there is convergence. For example, in the alternating-interval example, at $\varepsilon=1$ every error set is empty, even though at $\varepsilon=1/2$ its measure is one for every odd $n$. The correction is to use a threshold strictly between zero and one: there the error set is exactly the symmetric difference. Conversely, when proving convergence from vanishing symmetric differences, thresholds $\varepsilon\ge1$ need no estimate beyond the fact that the indicator difference is at most one.

The equivalence is exact because indicators have a fixed jump size. It does not say that arbitrary real-valued functions have errors equal to symmetric differences: for general functions, the error set records where the pointwise difference exceeds a threshold, but it need not be a symmetric difference of two underlying sets. For indicators, the jump size makes that set identity available, and so replaces a family of threshold tests by one set-measure limit.

Check Your Understanding

Use the strict-threshold definition of convergence in measure in each question.

For $A,B\in\mathcal{M}$, identify $\{|\chi_A-\chi_B|>1/3\}$ and $\{|\chi_A-\chi_B|>1\}$.
Suppose $\mu(E_n\,\triangle\,E)\to0$. Explain why this implies convergence in measure at every $\varepsilon>0$, including $\varepsilon=1$.
On $\mathbb{R}$, take $E_n=(0,1+2/n)$ and $E=(0,1)$. Compute $m(E_n\,\triangle\,E)$ and decide whether the indicators converge in measure.
Can $m(E_n)=m(E)$ for every $n$ while $\chi_{E_n}$ fails to converge in measure to $\chi_E$? Give an example and calculate the relevant symmetric-difference measures.
Which hypotheses are needed to conclude $\mu(E_n)\to\mu(E)$ from $\mu(E_n\,\triangle\,E)\to0$, using the finite-set-measure inequality?