Local Convergence With Spatial Tightness

From Local Information to Global L1 Convergence

On an infinite measure space, convergence on every finite-measure region does not by itself control what happens far away. Even when a sequence is well behaved on each such region, its integrals may remain concentrated in narrower and narrower spikes, or may keep moving out of every fixed region. A global \(L^1\) conclusion therefore needs two kinds of control: local uniform integrability to rule out spikes, and spatial tightness to rule out mass escaping to infinity.

Throughout, \((X,\mathcal{M},\mu)\) is a sigma-finite measure space, and \(f_n,f:X\to\mathbb{R}\) are measurable real-valued functions. Sigma-finiteness means that \(X\) is a countable union of measurable sets of finite measure. All integrals of absolute values below are Lebesgue integrals and may initially be infinite. Convergence in measure on a finite-measure set \(K\) means that for every \(\alpha>0\), \(\mu(K\cap\{|f_n-f|>\alpha\})\to0\).

In the small-set condition, the same \(\delta\) must work for every \(n\). Spatial tightness also uses the same set \(K\) for every member of the sequence. The complement \(K^c\) is where the tail integrals are measured.

The Local-to-Global Criterion

Proof. We first obtain a subsequence that converges to \(f\) almost everywhere on \(X\). By sigma-finiteness, choose increasing measurable sets \(K_j\) of finite measure with \(\bigcup_{j=1}^{\infty}K_j=X\). For each \(k\), local convergence in measure allows us to choose an index \(n_k>n_{k-1}\) such that, for each \(j\le k\),

The union of these sets over \(j\le k\) has measure less than \(2^{-k}\). For a fixed \(j\), the sum over \(k\ge j\) of the measures of the corresponding exceptional sets is finite. The measure of the union of those sets for \(k\ge N\) is at most the tail of this convergent sum, which tends to zero as \(N\to\infty\). Therefore almost every point of \(K_j\) belongs to only finitely many exceptional sets. At each such point, \(|f_{n_k}-f|\le1/k\) eventually, so \(f_{n_k}\to f\). Taking the countable union over \(j\) proves convergence almost everywhere on \(X\).

Fix any finite-measure \(K\). Fatou’s lemma, applied to the nonnegative functions \(|f_{n_k}|\chi_K\), gives \(\int_K|f|\,d\mu\le\liminf_k\int_K|f_{n_k}|\,d\mu<\infty\), where the final bound follows from local uniform integrability. If \(A\subseteq K\) and \(\mu(A)<\delta\), the same argument applied on \(A\) gives \(\int_A|f|\,d\mu\le\liminf_k\int_A|f_{n_k}|\,d\mu\). Hence the small-set bound for the sequence also holds for \(f\), with a non-strict bound; choosing the original bound smaller than any desired tolerance gives the strict bound when needed.

We now show convergence in \(L^1\) on \(K\). If \(\mu(K)=0\), the integral of \(|f_n-f|\) on \(K\) is zero. Otherwise fix \(\varepsilon>0\). Choose \(\eta>0\) with \(\eta\mu(K)<\varepsilon/3\). Local uniform integrability, and the small-set bound just obtained for \(f\), give a \(\delta>0\) such that whenever \(A\subseteq K\) and \(\mu(A)<\delta\), both \(\int_A|f_n|\,d\mu<\varepsilon/3\) for every \(n\) and \(\int_A|f|\,d\mu<\varepsilon/3\). Set \(A_n=K\cap\{|f_n-f|>\eta\}\). Convergence in measure on \(K\) gives \(\mu(A_n)<\delta\) for all sufficiently large \(n\). On \(K\setminus A_n\), \(|f_n-f|\le\eta\); on \(A_n\), \(|f_n-f|\le|f_n|+|f|\). Thus

So \(\int_K|f_n-f|\,d\mu\to0\) for every finite-measure \(K\).

It remains to control the part outside \(K\). Given \(\varepsilon>0\), spatial tightness supplies a finite-measure \(K\) with \(\sup_n\int_{K^c}|f_n|\,d\mu<\varepsilon/2\). For the almost-everywhere convergent subsequence constructed above, Fatou’s lemma on \(K^c\) gives \(\int_{K^c}|f|\,d\mu\le\liminf_k\int_{K^c}|f_{n_k}|\,d\mu\le\varepsilon/2\). In particular, \(f\) is integrable outside \(K\). It is integrable on \(K\) by the preceding local argument, so \(f\in L^1(\mu)\). Each \(f_n\) is integrable on \(K\) by local uniform integrability and on \(K^c\) by the spatial tail bound, so each \(f_n\in L^1(\mu)\) as well.

For this \(K\), the triangle inequality yields \(\int_{K^c}|f_n-f|\,d\mu<\varepsilon\) for every \(n\). The local result gives \(\int_K|f_n-f|\,d\mu\to0\). Hence \(\limsup_n\int_X|f_n-f|\,d\mu\le\varepsilon\). Since the argument applies for every \(\varepsilon>0\), the nonnegative integrals converge to zero. \(\square\)

Worked Examples: Applying and Testing the Conditions

What Each Hypothesis Controls

These examples separate the roles of the assumptions. Local convergence in measure describes the sequence on each finite-measure region, but says nothing by itself about the integral of a spike or about mass located far away. Local uniform integrability controls the first issue by making integrals small on sufficiently small sets, uniformly in \(n\). Spatial tightness controls the second by ensuring that one finite-measure set captures all but a uniformly small amount of every integral.

A useful way to apply the theorem is to choose a tolerance, first select a finite-measure region using spatial tightness, and then use convergence in measure and local uniform integrability on that region. The theorem concerns convergence in \(L^1\), meaning \(\int_X|f_n-f|\,d\mu\to0\); it does not assert pointwise convergence of the full sequence. In the proof, an almost-everywhere convergent subsequence is used only to pass local integrability and the tail bound to the limit function.