Three Distinct Ways a Family Can Misbehave
A sequence can have large values on very small sets, oscillate without settling down, or move its mass farther and farther out in space. These behaviours are different, and they obstruct different conclusions. A tail-integral test detects spikes; convergence in measure detects persistent oscillation; spatial tightness detects mass that escapes on an infinite measure space. No one test replaces the others.
Throughout, functions are real-valued and measurable. On a finite measure space \((X,\mathcal{M},\mu)\), the tail-integral test for a family \(\mathcal{F}\subset L^1(\mu)\) examines \(\sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu\) at a common threshold \(K\). On a sigma-finite space, local control and spatial tightness must also be considered. Convergence in measure means that \(\mu(\{|f_n-f|>\varepsilon\})\to0\) for every \(\varepsilon>0\); it is not the same as almost-everywhere, \(L^1\), or uniform convergence.
The tail integral weights a tail event by the size of the function on that event. Probability tightness of real-valued random variables asks only whether the probabilities of their tail events are small. Spatial tightness, by contrast, asks whether the integrals can be confined to one finite-measure region of the underlying space.
Spikes: Small Probability, Uncontrolled Tail Integral
Worked Example: Tall Spikes on a Probability Space
Let \(X=[0,1]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra, and let \(\mu=m\), so that \(\mu(X)=1\). For \(n\ge1\), define the nonnegative measurable function \(f_n=n\chi_{(0,1/n)}\). Its integral and its convergence in measure can both be computed exactly.
The function equals \(n\) on an interval of measure \(1/n\), and equals zero elsewhere. Therefore \(\int_X f_n\,dm=n(1/n)=1\). For every \(\varepsilon>0\), if \(n>\varepsilon\), then \(\{|f_n|>\varepsilon\}=(0,1/n)\), of measure \(1/n\); if \(n\le\varepsilon\), the set is empty. Thus \(f_n\to0\) in measure. It also tends to zero pointwise at every \(x\in[0,1]\): for \(x>0\), eventually \(1/n\le x\), and at \(x=0\) every function is zero.
Now fix the common tail threshold \(K\ge0\). If \(n\le K\), the strict event \(\{|f_n|>K\}\) is empty. If \(n>K\), this event is \((0,1/n)\), and its tail integral is \(n(1/n)=1\). There are integers \(n>K\), so \(\sup_{n\ge1}\int_{\{|f_n|>K\}}|f_n|\,dm=1\) for every finite \(K\). The family is not uniformly integrable. Since \(\int|f_n-0|\,dm=1\) for every \(n\), convergence in measure here does not give convergence in \(L^1\).
Check: the tail-event probability is \(\sup_n m(\{|f_n|>K\})=1/(\lfloor K\rfloor+1)\), which tends to zero. So the laws of these random variables are tight, even though their tail integrals remain equal to one. Probability tightness does not imply uniform integrability.
The distinction is that probability tightness controls how often a variable is large, while uniform integrability controls the contribution to its expectation from those large values. In the example, the event probability shrinks like \(1/n\), while the value on the event grows like \(n\).
Proof. For each \(\varepsilon>0\), the measure of \(\{|f_n|>\varepsilon\}\) is \(1/n\) when \(n>\varepsilon\) and zero otherwise, so it tends to zero. Also, \(\|f_n-0\|_1=\int_0^1 f_n\,dm=n\,m((0,1/n))=1\), so the \(L^1\) distance does not tend to zero. Finally, for every finite \(K\ge0\), choosing any integer \(n>K\) gives \(\int_{\{|f_n|>K\}}|f_n|\,dm=n(1/n)=1\). The supremum over \(n\) is therefore one for every \(K\), which violates the defining tail condition. \(\square\)
Oscillation: Uniform Integrability Without Convergence
Worked Example: Dyadic Oscillation on a Finite Space
On \(X=[0,1)\) with Lebesgue measure, define \(r_n(x)=(-1)^{\lfloor 2^n x\rfloor}\) for \(n\ge1\). Each function is measurable, takes only the values \(-1\) and \(1\), and has absolute value one everywhere. Hence, for \(K\ge1\), the strict tail event \(\{|r_n|>K\}\) is empty for every \(n\), so its tail integral is zero. For \(0\le K<1\), the event is all of \(X\), and its integral is one. In particular, the family is uniformly integrable.
Nevertheless, this sequence does not converge in measure to any measurable real-valued function. To see the persistent oscillation, let \(m>n\). The function \(r_n\) is constant on each interval of length \(2^{-n}\). Within each such interval, \(r_m\) alternates sign over \(2^{m-n}\) equal subintervals. Since \(2^{m-n}\) is even, exactly half of those subintervals have sign opposite to \(r_n\). Thus the set where \(r_m\ne r_n\) has measure \(1/2\). On that set, \(|r_m-r_n|=2\), and elsewhere the difference is zero. Therefore, for every \(m>n\), \(\int_0^1|r_m-r_n|\,dm=2(1/2)=1\), and \(m(\{|r_m-r_n|>1\})=1/2\).
Check: each \(r_n\) has integral zero: its \(2^n\) equal subintervals have alternating signs, with \(2^{n-1}\) of each sign. But these zero integrals do not make the sequence converge: the pairwise distance calculation shows that the functions remain separated.
Proof. Since \(|r_n|=1\), for every \(K\ge1\) each strict tail event \(\{|r_n|>K\}\) is empty. Thus the tail integrals are all zero for \(K\ge1\), proving uniform integrability. Suppose, for contradiction, that \(r_n\to f\) in measure for some measurable real-valued \(f\). For any \(\varepsilon>0\), the triangle inequality gives \(\{|r_m-r_n|>\varepsilon\}\subseteq\{|r_m-f|>\varepsilon/2\}\cup\{|r_n-f|>\varepsilon/2\}\). As \(m,n\to\infty\), the measures of both sets on the right tend to zero by convergence in measure. Hence \((r_n)\) must be Cauchy in measure. But for every \(m>n\), \(m(\{|r_m-r_n|>1\})=1/2\), so it is not Cauchy in measure. This contradiction proves the claim. \(\square\)
This example separates two questions: uniform integrability controls the size of values in the tails, while convergence in measure asks whether the functions settle toward a limit. The finite-measure Vitali convergence theorem, established in the earlier material on Vitali convergence, says that uniform integrability together with convergence in measure to an integrable limit implies convergence in \(L^1\). Uniform integrability alone does not supply that convergence hypothesis.
Escaping Mass: The Infinite-Space Diagnostic
Worked Example: Unit Mass Moving Along the Real Line
Take \(X=\mathbb{R}\) with its Lebesgue sigma-algebra and Lebesgue measure \(m\), which is sigma-finite but not finite. Let \(g_n=\chi_{[n,n+1)}\), a nonnegative measurable function. For every \(n\), \(\int_{\mathbb{R}}g_n\,dm=1\). For each fixed \(x\in\mathbb{R}\), \(g_n(x)=0\) for all sufficiently large \(n\), so \(g_n\to0\) pointwise and hence almost everywhere.
However, global convergence in measure to zero fails: for \(0<\varepsilon<1\), the event \(\{|g_n|>\varepsilon\}\) is \([n,n+1)\), of measure one for every \(n\). Convergence in \(L^1(m)\) also fails, since \(\|g_n-0\|_1=m([n,n+1))=1\) for every \(n\). The tail-value integrals are harmless: for \(K\ge1\), the strict event \(\{|g_n|>K\}\) is empty. The obstruction is spatial tightness.
Indeed, let \(E\subset\mathbb{R}\) be any measurable set with \(m(E)<\infty\). The intervals \([n,n+1)\) are pairwise disjoint, so the nonnegative series \(\sum_{n=1}^{\infty}m(E\cap[n,n+1))\) is at most \(m(E)\) and is finite. Its terms therefore tend to zero. Consequently, \(\int_{E^c}g_n\,dm=m([n,n+1)\setminus E)=1-m(E\cap[n,n+1))\to1\). No finite-measure set \(E\) can capture all these integrals uniformly up to an arbitrarily small error.
Check: the total integral stays exactly one, while the mass inside any fixed finite-measure set tends to zero. This is escape in space, not escape to large function values.
Proof. For any measurable \(E\) of finite measure, the sets \(E\cap[n,n+1)\) are pairwise disjoint and their measures have finite sum bounded by \(m(E)\). Thus their measures tend to zero. Since \(\int_{E^c}g_n\,dm=1-m(E\cap[n,n+1))\), these outside integrals tend to one. In particular, for every such \(E\), \(\sup_n\int_{E^c}g_n\,dm\ge1\), so the spatial-tightness requirement fails, for example with \(\varepsilon=1/2\). For \(0<\varepsilon<1\), \(m(\{|g_n|>\varepsilon\})=1\) for every \(n\), so there is no convergence in measure to zero. Finally, \(\|g_n\|_1=1\) for every \(n\), so there is no \(L^1\) convergence to zero. \(\square\)
Classifying the Failure Before Choosing a Theorem
When a family behaves unexpectedly, first identify which quantity fails to be controlled. A useful diagnostic is to test the same threshold \(K\) for every member of the family, then separately test convergence and spatial localization. A small tail-event probability is not enough: the tail integral may stay large. A small tail integral is not enough to ensure convergence: an oscillating family can be uniformly integrable. On an infinite measure space, even pointwise convergence and bounded \(L^1\) norms do not rule out spatial escape.
| Family | Tail integrals | Convergence behaviour | Main obstruction |
|---|---|---|---|
| \(n\chi_{(0,1/n)}\) on \([0,1]\) | Equal to one at every finite \(K\) after taking the supremum | Converges in measure to zero, not in \(L^1\) | Spikes; failure of uniform integrability |
| \(r_n\) on \([0,1)\) | Zero for \(K\ge1\) | Does not converge in measure | Persistent oscillation |
| \(\chi_{[n,n+1)}\) on \(\mathbb{R}\) | Zero for \(K\ge1\) | Pointwise converges to zero, not globally in measure or \(L^1\) | Escaping mass; failure of spatial tightness |
On a finite measure space, the earlier Vitali theorem provides a direct route from uniform integrability and convergence in measure to \(L^1\) convergence, provided the limit is integrable. On an infinite space, the additional spatial question matters: local convergence does not by itself control the integral outside large finite-measure regions. The examples above are designed to keep these tests separate rather than to treat them as interchangeable versions of one condition.
Check Your Understanding
For each question, specify the measure space and distinguish tail-integral control, convergence in measure, and convergence in \(L^1\) where relevant.
- For \(h_n=n\chi_{(0,1/n)}\) on \([0,1]\), compute \(\int h_n\,dm\) and the supremum of its tail integrals at an arbitrary fixed threshold \(K\ge0\).
- Why does tightness of the laws of \(n\chi_{(0,1/n)}\) not imply uniform integrability? State which quantity is small and which remains equal to one.
- For the dyadic sequence \(r_n\), compute \(m(\{|r_m-r_n|>1\})\) when \(m>n\), and explain why this rules out convergence in measure.
- For \(g_n=\chi_{[n,n+1)}\) on \(\mathbb{R}\), show why pointwise convergence to zero does not imply global convergence in measure.
- What extra diagnostic, beyond tail-value integrals, detects the obstruction for \(g_n\) on \(\mathbb{R}\)?