Indicator Errors Are Set Differences
For indicator functions, the size of a pointwise error can take only two values: zero or one. This makes the error set especially simple to calculate, but the strict inequality in its definition matters. In particular, an error of exactly one is not counted at threshold one.
Let \((X,\mathcal{M},\mu)\) be a measure space; no finiteness, completeness, or sigma-finiteness assumption is needed for the facts in this section. For \(E\in\mathcal{M}\), the indicator \(\chi_E:X\to\{0,1\}\) is measurable. For real-valued measurable functions \(f_n,f\), convergence in measure means that for every \(\varepsilon>0\), the measures \(\mu(\{|f_n-f|>\varepsilon\})\) tend to zero. This definition makes sense even when \(\mu(X)=\infty\).
The Exact Error-Set Rule
Proof. If \(x\in A\,\triangle\,B\), then exactly one of \(x\in A\) and \(x\in B\) holds. Thus one indicator is one and the other is zero, so \(|\chi_A(x)-\chi_B(x)|=1\). If \(x\notin A\,\triangle\,B\), then either \(x\in A\cap B\) or \(x\in A^c\cap B^c\); in either case both indicators agree and their absolute difference is zero. When \(0<\varepsilon<1\), the value one is strictly greater than \(\varepsilon\), while zero is not, so the error set is exactly \(A\,\triangle\,B\). When \(\varepsilon\ge1\), neither possible value, zero nor one, is strictly greater than \(\varepsilon\). The error set is therefore empty. Taking its measure gives the two stated cases, including when \(\mu(A\,\triangle\,B)=\infty\). \(\square\)
The measure of a symmetric difference can be computed by adding the measures of its disjoint parts. If both parts have finite measure, this gives the finite sum \(\mu(A\,\triangle\,B)=\mu(A\setminus B)+\mu(B\setminus A)\). The disjointness is important: adding \(\mu(A)\) and \(\mu(B)\) instead would count their common part twice.
Proof. Suppose first that \(\mu(A_n\,\triangle\,A)\to0\). At each fixed threshold \(0<\varepsilon<1\), the proposition gives \(\mu(\{|\chi_{A_n}-\chi_A|>\varepsilon\})=\mu(A_n\,\triangle\,A)\), which tends to zero. At each \(\varepsilon\ge1\), the error set is empty for every \(n\), so its measure is zero. These are all positive thresholds, and hence the indicators converge in measure. Conversely, if the indicators converge in measure, use the particular threshold \(\varepsilon=1/2\). The proposition gives \(\mu(A_n\,\triangle\,A)=\mu(\{|\chi_{A_n}-\chi_A|>1/2\})\to0\). This proves the other direction. \(\square\)
Worked Calculations
Worked Example: Shrinking Intervals on the Real Line
Take \((X,\mathcal{M},\mu)=(\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure, and let \(A_n=(0,1/n)\). Let the target be \(A=\varnothing\). Compute the error-set measure for every \(\varepsilon>0\) and decide whether \(\chi_{A_n}\) converges in measure to the zero function.
For \(x\in A_n\), the indicator difference \(|\chi_{A_n}(x)-\chi_{\varnothing}(x)|\) is one. For \(x\notin A_n\), it is zero. Applying the strict-threshold rule gives
For each fixed \(0<\varepsilon<1\), \(1/n\to0\); for \(\varepsilon\ge1\), the error measure is already zero. Thus the sequence converges in measure. Check: at \(n=4\) and \(\varepsilon=1/2\), the error set is \((0,1/4)\) and its measure is \(1/4\). At the same \(n\), threshold \(\varepsilon=1\) gives the empty error set and measure zero, because \(1>1\) is false.
Worked Example: Translated Intervals and Both Sides of the Difference
Again use \((\mathbb{R},\mathcal{L},m)\). Let \(A=(0,1)\), and for \(n\ge2\) let \(A_n=(1/n,1+1/n)\). Compute \(m(A_n\,\triangle\,A)\) and determine convergence in measure of \(\chi_{A_n}\) to \(\chi_A\).
The points in \(A\) but not \(A_n\) form \(A\setminus A_n=(0,1/n]\). The points in \(A_n\) but not \(A\) form \(A_n\setminus A=[1,1+1/n)\). These sets are disjoint, and each has Lebesgue measure \(1/n\). Hence
For \(0<\varepsilon<1\), the error-set measure is \(2/n\), which tends to zero. For \(\varepsilon\ge1\), it is zero. Therefore the indicators converge in measure. Check: at \(n=4\), the two difference intervals each have length \(1/4\), so the symmetric difference has measure \(1/2\), agreeing with \(2/n=2/4=1/2\). The endpoints do not change these lengths; they are singletons of Lebesgue measure zero.
Worked Example: Changing Finite Supports with Weighted Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be all subsets of \(X\), and define \(\mu(E)=\sum_{k\in E}2^{-k}\). This is a finite measure, since \(\mu(X)=\sum_{k=1}^{\infty}2^{-k}=1\); it is complete and sigma-finite. Set \(A_n=\{n,n+1\}\) and \(A=\varnothing\). Compute the error measures and test convergence in measure.
Here \(A_n\,\triangle\,A=A_n\), since \(A\) is empty. For \(0<\varepsilon<1\), the error set is \(\{n,n+1\}\), so its measure is
This tends to zero as \(n\to\infty\). For \(\varepsilon\ge1\), the error set is empty. Thus the indicators converge in measure. Check: for \(n=2\), the support is \(\{2,3\}\), whose measure is \(2^{-2}+2^{-3}=1/4+1/8=3/8\); the formula gives \(3\cdot2^{-3}=3/8\).
The measure matters. If the same sets \(A_n=\{n,n+1\}\) are considered under counting measure \(\mu_{\mathrm{count}}\), then for \(0<\varepsilon<1\) the error measure is \(\mu_{\mathrm{count}}(\{n,n+1\})=2\) for every \(n\), so convergence in measure to zero fails. Under counting measure the points do not become lighter as their indices increase.
Thresholds, Checks, and a Common Mistake
For indicator functions, every threshold in \((0,1)\) selects exactly the points where the indicators differ, so all such thresholds give the same error set. The endpoint \(\varepsilon=1\) belongs with thresholds greater than one, not with thresholds below one. This is a consequence of the strict condition \(|f_n-f|>\varepsilon\), rather than \(|f_n-f|\ge\varepsilon\).
List \(A_n\setminus A\) and \(A\setminus A_n\) separately. Their union is the symmetric difference, and they are disjoint.
Use the measure specified on the space: interval lengths for Lebesgue measure, point weights for weighted counting measure, and point counts for counting measure.
For \(0<\varepsilon<1\), the error measure is the symmetric-difference measure. For \(\varepsilon\ge1\), it is zero.
For convergence in measure, the error measure must tend to zero for every fixed \(\varepsilon>0\). It is not enough to let the threshold vary with \(n\).
Common mistake. A mistaken calculation may say that for \(\varepsilon=1\), the error set of two indicators is their symmetric difference because the indicator difference there equals one. But the defining condition is \(1>1\), which is false. For example, in the shrinking-interval calculation with \(A_n=(0,1/n)\), the error set at \(\varepsilon=1\) is empty and has measure zero, not \(1/n\). The correction is to use the symmetric difference only when \(0<\varepsilon<1\); at \(\varepsilon=1\), as at every \(\varepsilon>1\), the error set is empty.
The practical advantage of the symmetric-difference calculation is that it replaces a pointwise function comparison with set bookkeeping. The measure of the symmetric difference records precisely the points where the two supports disagree. The measure space still matters: the same changing finite sets can have measures tending to zero under a weighted measure and remain of constant measure under counting measure. Exact thresholds, exact supports, and the correct measure together determine the conclusion.
Check Your Understanding
In each question, remember that the error set is defined using a strict inequality.
- For measurable \(A,B\), compute \(\{|\chi_A-\chi_B|>\varepsilon\}\) separately for \(0<\varepsilon<1\) and \(\varepsilon\ge1\).
- On \((\mathbb{R},\mathcal{L},m)\), take \(A_n=(0,2/n)\) and target \(\varnothing\). Find the error-set measure for each threshold range and decide whether the indicators converge in measure.
- On \(\mathbb{N}\) with \(\mu(\{k\})=3^{-k}\), let \(A_n=\{n\}\) and \(A=\varnothing\). Compute the symmetric-difference measure and its limit.
- Under counting measure on \(\mathbb{N}\), do the indicators of \(A_n=\{n\}\) converge in measure to zero? Justify your answer at threshold \(1/2\).
- Explain why an error of exactly one is not included at threshold \(\varepsilon=1\).