Pointwise Convergence While Supports Escape
On a space of finite measure, almost-everywhere convergence implies convergence in measure. On an infinite-measure space, that implication can fail: a sequence may converge to zero at every fixed point even though its nonzero values occupy sets of the same positive measure at every stage. The key is that the sets move farther away rather than shrinking.
Throughout the main examples, take \((\mathbb{R},\mathcal{L},m)\), the complete measure space with Lebesgue measure. It is sigma-finite but not finite: \(\mathbb{R}\) is a countable union of bounded intervals of finite measure, while \(m(\mathbb{R})=\infty\). All functions below are real-valued and measurable. Convergence pointwise means convergence at every real number; convergence almost everywhere permits failure on a Lebesgue-null set.
For indicators, the error set has a particularly direct form. If \(f_n=\chi_{A_n}\) and the proposed limit is zero, then \(|f_n-0|\) equals \(1\) on \(A_n\) and \(0\) off \(A_n\). Thus for \(0<\varepsilon<1\), the error set is exactly \(A_n\). Its measure, not the location of \(A_n\), determines whether convergence in measure holds.
The Translated Unit-Interval Example
Worked Example: Unit Intervals Moving to the Right
On \((\mathbb{R},\mathcal{L},m)\), let \(f_n=\chi_{[n,n+1)}\), for integers \(n\ge1\). Determine its pointwise limit and test convergence in measure to that limit.
Fix \(x\in\mathbb{R}\). If \(n>x\), then every \(y\in[n,n+1)\) satisfies \(y\ge n>x\), so \(x\notin[n,n+1)\). Every integer \(n>x\) therefore has \(f_n(x)=0\). This proves that \(f_n(x)\to0\) for every \(x\in\mathbb{R}\), not merely almost everywhere.
Now fix \(\varepsilon>0\). If \(0<\varepsilon<1\), then \(|f_n(x)|>\varepsilon\) holds exactly when \(f_n(x)=1\). Hence the error set is \([n,n+1)\), and its Lebesgue measure is \((n+1)-n=1\). If \(\varepsilon\ge1\), the strict inequality \(|f_n(x)|>\varepsilon\) never holds, since \(|f_n(x)|\) is either \(0\) or \(1\); the error set is empty and has measure \(0\). In particular, at \(\varepsilon=\tfrac12\), the error-set measures are all \(1\), so they do not tend to zero. Therefore \(f_n\) does not converge in measure to zero.
Check: at every \(n\), the support interval has length \((n+1)-n=1\), so \(m(\{|f_n|>\tfrac12\})=1\), as computed. The pointwise limit is indeed zero, but its zero value does not make the error sets small in measure.
Proof. Fix \(x\in\mathbb{R}\). Since \(a_n\to+\infty\), there is an index \(N\) such that \(a_n>x\) for every \(n\ge N\). For those \(n\), \(x\notin[a_n,a_n+1)\), and therefore \(f_n(x)=0\). This holds for every \(x\), proving pointwise convergence everywhere. For any fixed \(0<\varepsilon<1\), the strict error set \(\{|f_n|>\varepsilon\}\) is exactly \([a_n,a_n+1)\). Its measure is \((a_n+1)-a_n=1\) for each \(n\). Since these measures do not tend to zero, the definition of convergence in measure fails. \(\square\)
The proposition needs no monotonicity of the translations: only \(a_n\to+\infty\). Each point is eventually left behind, but the interval’s length remains one. The finite-measure implication from almost-everywhere convergence to convergence in measure cannot therefore be extended to all sigma-finite spaces merely by replacing “finite” with “sigma-finite.”
Two Further Checks on the Geometry
Worked Example: Translations to the Left
Let \(g_n=\chi_{[-n,1-n)}\) on \((\mathbb{R},\mathcal{L},m)\). Verify that the same failure occurs when the intervals escape toward negative infinity.
Fix \(x\in\mathbb{R}\). For every integer \(n>1-x\), we have \(1-n<x\). Every point of \([-n,1-n)\) is strictly less than \(1-n\), so \(x\notin[-n,1-n)\), and \(g_n(x)=0\). Thus \(g_n\to0\) pointwise everywhere.
For \(0<\varepsilon<1\), the error set \(\{|g_n|>\varepsilon\}\) is \([-n,1-n)\). Its measure is \((1-n)-(-n)=1\). For \(\varepsilon\ge1\), the strict error set is empty because \(g_n\) only takes the values zero and one. Taking, for example, \(\varepsilon=\tfrac12\) gives error-set measure \(1\) for every \(n\), so there is no convergence in measure.
Check: the interval endpoints give length \(1-n+n=1\), independently of \(n\); the direction of translation does not change the calculation.
A useful contrast is what happens when observations are restricted to a bounded region. This does not contradict the proposition: restricting the underlying measure space changes the sets whose measures are being tested.
Proof. Because \(E\) is bounded, there is a real number \(b\) such that \(x\le b\) for all \(x\in E\). Since \(a_n\to+\infty\), there is \(N\) such that \(a_n>b\) for every \(n\ge N\). For such \(n\), every point of \([a_n,a_n+1)\) is greater than \(b\), so \([a_n,a_n+1)\cap E=\varnothing\). The restricted indicator is therefore zero everywhere on \(E\) for all \(n\ge N\). For every \(\varepsilon>0\), its error set in \(E\) is then empty for \(n\ge N\), and its \(m|_E\)-measure is zero. This is convergence in measure by definition. \(\square\)
Worked Example: Restricting to a Bounded Window
Take \(E=[-2,2]\) with restricted Lebesgue measure, and define \(h_n=\chi_{[n,n+1)}|_E\) for \(n\ge1\). Compute the error-set measures for \(\varepsilon=\tfrac12\).
For \(n=1\), the intersection \([1,2)\cap[-2,2]\) is \([1,2)\), which has measure \(2-1=1\). Thus \(m|_E(\{|h_1|>\tfrac12\})=1\). For \(n=2\), the intersection is the singleton \(\{2\}\), which has Lebesgue measure \(0\). For every \(n\ge3\), the interval \([n,n+1)\) is disjoint from \(E\), so the error set is empty and has measure \(0\). In particular, all error-set measures are zero for \(n\ge2\), and \(h_n\to0\) in measure on \(E\). In fact, \(h_n\) is identically zero on \(E\) for every \(n\ge3\).
Check: the exceptional intersection at \(n=2\) contains only the endpoint \(2\), not an interval of positive length. Its measure is zero, consistent with the eventual-zero conclusion. On the full real line, by contrast, each corresponding translated interval has measure \(1\).
Why Pointwise Escape Does Not Control Measure
Pointwise convergence tests one point at a time: for a fixed \(x\), the question is whether the values eventually become close to the limit. Convergence in measure asks a different question: for a fixed tolerance, does the measure of all points where the error exceeds that tolerance tend to zero? The order of these quantifiers matters. A point can eventually fall outside every translated interval while each interval still has length one.
A common mistaken calculation is: “For every fixed \(x\), \(f_n(x)=0\) for all sufficiently large \(n\), so the error set is eventually empty; therefore its measure tends to zero.” The first claim is pointwise and permits the required index to depend on \(x\). It does not say that one index works for every \(x\). In the unit-interval example, for each \(n\) there are still points \(x\in[n,n+1)\) with error \(1\), and the set of all such points has measure \(1\). The correction is to compute \(m(\{|f_n|>\varepsilon\})\) at the same fixed threshold \(\varepsilon=\tfrac12\) for each \(n\), obtaining \(1\) each time.
This example isolates the role of the underlying space. The real line is sigma-finite, but its infinite measure allows a fixed amount of mass to move indefinitely far away. On a bounded observation window, the translated supports eventually miss the whole window; on the full line, they never lose their unit measure. The finite-measure hypothesis in the earlier implication from almost-everywhere convergence to convergence in measure is therefore substantive.
Check Your Understanding
Use the error-set definition, keeping the threshold fixed while \(n\) varies.
- For \(f_n=\chi_{[2n,2n+1)}\) on \(\mathbb{R}\), prove pointwise convergence to zero and compute \(m(\{|f_n|>\tfrac34\})\).
- For \(f_n=\chi_{[n,n+1)}\), what is the error set when \(\varepsilon=1\)? Explain why the strict inequality in the definition matters.
- Suppose \(a_n\to+\infty\). Which part of the proof that \(\chi_{[a_n,a_n+1)}\to0\) pointwise depends on monotonicity of \(a_n\), if any?
- For the intervals \([-n,1-n)\), compute the error-set measure at \(\varepsilon=\tfrac14\) and state whether convergence in measure holds.
- Explain why the sequence \(\chi_{[n,n+1)}|_{[-2,2]}\) can converge in measure on the bounded window without contradicting its behavior on \(\mathbb{R}\).