Where Is the Integral Located?
For a family of integrable functions on an infinite measure space, smallness of large function values does not say whether the integral is concentrated in a bounded region, or even near any one finite-measure set. Conversely, a family can have all its integral confined to one finite-measure set while concentrating that integral on sets where the function values become arbitrarily large. These are distinct forms of tail control.
Let \((X,\mathcal{M},\mu)\) be a sigma-finite measure space, meaning that \(X\) is a countable union of measurable sets of finite measure. The space need not have finite measure or be complete. Let \(\mathcal{F}\) be a family of real- or complex-valued measurable functions in \(L^1(\mu)\); thus each \(f\in\mathcal{F}\) satisfies \(\int_X |f|\,d\mu<\infty\). Integrals below depend only on the functions up to equality \(\mu\)-a.e.
Spatial tightness controls the integral outside a finite-measure region. It does not mean that the functions vanish there, nor does it require \(E\) to be bounded in any geometric sense. On \(\mathbb{R}^n\) with Lebesgue measure, bounded measurable sets are one possible choice, but the definition permits any finite-measure measurable set.
By contrast, the large-value tail condition asks whether $$ \sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu\longrightarrow 0 \quad\text{as }K\longrightarrow\infty. $$ The set in this integral depends on the values of each individual function, not on where in \(X\) those values occur. The strict inequality \(|f|>K\) is part of this convention.
Two Independent Kinds of Tail Control
Worked Example: Spatially Tight Spikes Without Large-Value Tail Control
On \((\mathbb{R},\mathcal{L},m)\), let \(f_n(x)=n\chi_{(0,1/n)}(x)\), for integers \(n\ge1\). These are nonnegative measurable functions, and direct integration gives $$ \int_{\mathbb{R}}|f_n|\,dm=n\,m((0,1/n))=n\cdot\frac1n=1. $$
For any \(\varepsilon>0\), choose the single finite-measure set \(E=[0,1]\). Since \(f_n=0\) on \(E^c\) for every \(n\), we have \(\sup_n\int_{E^c}|f_n|\,dm=0<\varepsilon\). Thus the family is spatially tight.
Now fix any \(K>0\). If \(n\le K\), then \(\{|f_n|>K\}\) is empty. If \(n>K\), then \(\{|f_n|>K\}=(0,1/n)\), because the value on that interval is exactly \(n>K\). Consequently, $$ \int_{\{|f_n|>K\}}|f_n|\,dm= \begin{cases} 0,&n\le K,\\ n\cdot(1/n)=1,&n>K. \end{cases} $$ There is always an integer \(n>K\), so the supremum over \(n\) equals \(1\) for every \(K>0\). Large-value tail control fails. Check: the total integral is \(1\) for each \(n\), and for \(n>K\) the entire integral lies above the strict threshold \(K\), also giving \(1\).
Worked Example: Large-Value Tail Control Without Spatial Tightness
On \((\mathbb{R},\mathcal{L},m)\), set \(I_j=[j,j+1)\) and \(g_j=\chi_{I_j}\), for integers \(j\ge1\). Each \(g_j\) is measurable and $$ \int_{\mathbb{R}}|g_j|\,dm=m(I_j)=1. $$ For every \(K\ge1\), the strict-level set \(\{|g_j|>K\}\) is empty: the only values of \(|g_j|\) are \(0\) and \(1\). Thus the supremum of the large-value tail integrals is \(0\) for every \(K\ge1\).
Nevertheless, the family is not spatially tight. Let \(E\in\mathcal{L}\) have finite measure. The intervals \(I_j\) are pairwise disjoint, so for every positive integer \(N\), $$ \sum_{j=1}^N m(E\cap I_j)\le m(E). $$ It follows that \(m(E\cap I_j)\to0\): otherwise some positive number would be a lower bound for infinitely many terms, making the partial sums unbounded. In particular there is a \(j\) with \(m(E\cap I_j)<1/2\). For this \(j\), $$ \int_{E^c}|g_j|\,dm=m(I_j\setminus E)=1-m(E\cap I_j)>1/2. $$ So no finite-measure \(E\) makes all the outside integrals less than \(\varepsilon=1/2\). Check: the integral over \(I_j\) is \(1\), split into the disjoint parts \(I_j\cap E\) and \(I_j\setminus E\); the displayed outside integral is exactly the remaining part.
The examples separate location from amplitude. In the first, all functions are supported inside one finite-measure set, yet their mass lies at increasingly high values. In the second, the functions never exceed \(1\), yet their supports move so that no one finite-measure set captures most of every integral. Neither property should be substituted for the other.
Using a Fixed Sigma-Finite Exhaustion
Sigma-finiteness provides an increasing sequence \(X_1\subseteq X_2\subseteq\cdots\) of measurable sets with \(\mu(X_n)<\infty\) and \(\bigcup_nX_n=X\): take finite unions of a countable finite-measure cover. A useful test asks whether the integrals outside these particular sets tend to zero. For an arbitrary spatially tight family, this test need not follow automatically; a small portion of a chosen finite-measure set can still carry a large integral for some family member.
Proof. If the displayed limit is zero, then for any \(\varepsilon>0\) there is an \(n\) for which \(\sup_f\int_{X_n^c}|f|\,d\mu<\varepsilon\). The set \(X_n\) has finite measure, so it witnesses spatial tightness.
Conversely, fix \(\varepsilon>0\). Spatial tightness gives a measurable \(E\) of finite measure such that \(\sup_f\int_{E^c}|f|\,d\mu<\varepsilon/2\). Apply uniform absolute continuity with \(\eta=\varepsilon/2\), obtaining a \(\delta>0\). The sets \(E\setminus X_n\) decrease to the empty set. Since \(m\) is not assumed here and \(\mu(E)<\infty\), continuity from above for the measure \(\mu\) gives \(\mu(E\setminus X_n)\to0\). Choose \(n\) with \(\mu(E\setminus X_n)<\delta\). As \(X_n^c\subseteq E^c\cup(E\setminus X_n)\), monotonicity and subadditivity of the integral yield, for every \(f\in\mathcal{F}\), $$ \int_{X_n^c}|f|\,d\mu \le \int_{E^c}|f|\,d\mu+\int_{E\setminus X_n}|f|\,d\mu <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$ The same \(n\) works for every \(f\), so the supremum is at most the uniform bound obtained from these strict estimates, and the tails tend to zero. More explicitly, applying this argument with any smaller positive tolerance, such as \(\varepsilon/2\), gives a strict bound below \(\varepsilon\) for the supremum. This proves the limit assertion. \(\square\)
The hypothesis on small sets is essential to this route from an arbitrary finite-measure set to a fixed exhaustion: it controls the integral on the leftover set \(E\setminus X_n\). It is a statement about the integrals, not a claim that the functions themselves are absolutely continuous.
Worked Example: Testing an Exhaustion
On \((\mathbb{R},\mathcal{L},m)\), let \(\mathcal{F}=\{f\}\), where \(f=\chi_{[0,1]}\), and use \(X_n=[-n,n]\), \(n\ge1\). For every measurable \(A\), \(0\le\int_A|f|\,dm\le m(A)\). Thus if \(m(A)<\delta=\eta\), then \(\int_A|f|\,dm<\eta\); the integrals are uniformly absolutely continuous for this singleton family.
For each \(n\ge1\), \([0,1]\subseteq[-n,n]\), so \(f=0\) on \(X_n^c\) and \(\int_{X_n^c}|f|\,dm=0\). The exhaustion criterion holds exactly. Independently, \(E=[0,1]\) has measure \(1\) and \(\int_{E^c}|f|\,dm=0\), verifying spatial tightness directly. The total integral is also \(m([0,1])=1\), consistent with the split into the integral on \(E\) and its zero outside tail.
Stability and Limits of the Definition
Proof. Fix \(\varepsilon>0\). Choose finite-measure sets \(E_F,E_G\) such that \(\sup_{f\in\mathcal{F}}\int_{E_F^c}|f|\,d\mu<\varepsilon/2\) and \(\sup_{g\in\mathcal{G}}\int_{E_G^c}|g|\,d\mu<\varepsilon/2\). The union \(E=E_F\cup E_G\) has finite measure, and \(E^c\subseteq E_F^c\cap E_G^c\). The triangle inequality gives, for every \(f\in\mathcal{F}\) and \(g\in\mathcal{G}\), $$ \int_{E^c}|f+g|\,d\mu \le\int_{E^c}|f|\,d\mu+\int_{E^c}|g|\,d\mu \le\int_{E_F^c}|f|\,d\mu+\int_{E_G^c}|g|\,d\mu<\varepsilon. $$ Thus the one set \(E\) works uniformly for all sums. \(\square\)
Spatial tightness alone does not bound the \(L^1\) norms. On \((\mathbb{R},\mathcal{L},m)\), the family \(h_n=n\chi_{[0,1]}\) is spatially tight because its outside integral is zero for \(E=[0,1]\), but \(\int|h_n|\,dm=n\). It also has no uniform large-value tail control: for any \(K>0\), choose an integer \(n>K\); then \(\int_{\{|h_n|>K\}}|h_n|\,dm=n\). A spatial condition locates mass; it does not limit its total amount or its height.
Common pitfall. For the translated indicators \(g_j\), it is tempting to take a large interval \(E=[0,N]\) and conclude that the outside integrals are small. That only captures \(g_1,\ldots,g_N\): for every \(j>N\), \(I_j\subseteq E^c\) and \(\int_{E^c}|g_j|\,dm=1\). Spatial tightness requires one finite-measure set to control every family member, not a set that works for an increasing finite subfamily.
Check Your Understanding
For each question, keep the location of the mass separate from the size of the function values.
- For \(f_n=n\chi_{(2,2+1/n)}\) on \(\mathbb{R}\), verify spatial tightness and compute the supremum of the strict tail integrals over \(\{|f_n|>K\}\).
- Why does the family \(\{\chi_{[j,j+1)}:j\ge1\}\) have zero large-value tail integral for \(K\ge1\) but fail spatial tightness?
- State the extra hypothesis in the exhaustion proposition that permits passage from an arbitrary finite-measure set to a fixed exhaustion.
- Give an example of a spatially tight family whose \(L^1\) norms are unbounded.
- In the definition of spatial tightness, which quantifier must be uniform over the family: the choice of \(E\), or the choice of \(f\)?