Uniform Integrability and Weak Compactness in L1

Why \(L^1\) Needs Tail Control

In \(L^1\), a family can have bounded norms and still fail to have weakly convergent subsequences: its mass may concentrate on smaller and smaller sets. Uniform integrability rules out that concentration. On a finite measure space, it is exactly the condition that distinguishes relatively weakly compact subsets of \(L^1\) from merely bounded ones.

Throughout, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu(X)<\infty\). Functions are real-valued and measurable, identified when they agree \(\mu\)-a.e. The space \(L^1(\mu)\) consists of those equivalence classes \(f\) for which \(\|f\|_1=\int_X|f|\,d\mu<\infty\). A family \(F\subset L^1(\mu)\) is relatively weakly compact if its closure in the weak topology \(\sigma(L^1,L^\infty)\) is compact. Here weak convergence means convergence when tested against every element of \(L^\infty(\mu)\), using the duality \(\langle f,h\rangle=\int_Xfh\,d\mu\).

Here is why the two conditions are equivalent. If the tail condition holds and \(\sup_F\|f\|_1=C<\infty\), choose \(K\) so that every tail integral is less than \(\varepsilon/2\). Then \[ \int_E|f|\,d\mu \le K\mu(E)+\int_{\{|f|>K\}}|f|\,d\mu. \] Taking \(\delta=\varepsilon/(2K)\) when \(K>0\) gives the absolute-continuity condition. If \(K=0\), one may increase it to any positive value. Conversely, suppose the norms are bounded by \(C\) and the integrals are uniformly absolutely continuous. For \(K>0\), Markov's inequality gives \(\mu(\{|f|>K\})\le C/K\). Given \(\varepsilon>0\), choose \(\delta\) from absolute continuity, then choose \(K\) large enough that \(C/K<\delta\). The integral over the strict tail is then less than \(\varepsilon\), uniformly in \(f\).

The Dunford–Pettis Criterion

Proof. First suppose \(F\) is uniformly integrable. It is norm bounded, say \(\|f\|_1\le C\) for every \(f\in F\). Let \(J:L^1(\mu)\to L^1(\mu)^{**}\) be the canonical embedding, defined by \(Jf(\Lambda)=\Lambda(f)\) for \(\Lambda\in L^1(\mu)^*=L^\infty(\mu)\). By Banach–Alaoglu, the weak-star closure \(Q\) of \(J(F)\) in \(L^1{}^{**}\) is compact; it is contained in the ball of radius \(C\).

We show that every \(z\in Q\) belongs to \(J(L^1(\mu))\). For \(A\in\mathcal{M}\), define \(\nu(A)=z(\chi_A)\). This is a finitely additive set function. Moreover, for every \(A\), the weak-star closure definition implies \[ |\nu(A)|\le\sup_{f\in F}\left|\int_A f\,d\mu\right| \le\sup_{f\in F}\int_A|f|\,d\mu. \] If \(A_n\) are pairwise disjoint measurable sets and \(A=\bigcup_{n\ge1}A_n\), finiteness of \(\mu\) gives \(\mu(\bigcup_{n>N}A_n)\to0\). Uniform absolute continuity therefore gives \(\nu(\bigcup_{n>N}A_n)\to0\). Finite additivity applied to \(A=(\bigcup_{n\le N}A_n)\cup(\bigcup_{n>N}A_n)\) now proves countable additivity.

For any finite measurable partition \(A_1,\ldots,A_r\) of \(X\), choose \(s_i\in\{-1,1\}\) so that \(s_i\nu(A_i)=|\nu(A_i)|\). The simple function \(h=\sum_{i=1}^r s_i\chi_{A_i}\) satisfies \(\|h\|_\infty\le1\), so \[ \sum_{i=1}^r|\nu(A_i)|=z(h)\le\|z\|\le C. \] Thus \(\nu\) has finite total variation. The displayed bound also implies \(\nu\ll\mu\): if \(\mu(A)=0\), every integral over \(A\) vanishes, so \(\nu(A)=0\). By the Radon–Nikodym theorem for finite signed measures, there is a real \(g\in L^1(\mu)\) such that \(\nu(A)=\int_Ag\,d\mu\) for every \(A\in\mathcal{M}\).

For every measurable simple \(h\), linearity gives \(z(h)=\int_Xgh\,d\mu\). A bounded measurable \(h\) is a uniform limit of measurable simple functions. Since \(z\) is continuous on \(L^\infty\) and \(g\in L^1\), passing to that uniform limit gives \(z(h)=\int_Xgh\,d\mu\) for every \(h\in L^\infty\). This says \(z=Jg\). Consequently \(Q\subset J(L^1)\). The weak-star topology on \(L^1{}^{**}\), restricted to \(J(L^1)\), is precisely the weak topology on \(L^1\), because \(L^1{}^*=L^\infty\). Hence \(J^{-1}(Q)\) is weakly compact and contains \(F\). This proves relative weak compactness.

For the converse, we use the standard Dunford–Pettis lemma: on any measure space, if a sequence in \(L^1\) converges weakly, then the sequence together with its limit is uniformly integrable. This is the sequence form of the weak-convergence criterion in \(L^1\); its proof uses a disjointization argument and is not reproduced here. Since the weak closure of a relatively weakly compact family is weakly compact, the Eberlein–Šmulian theorem gives a weakly convergent subsequence from every sequence in that closure. The lemma then implies that every such sequence and its limit is uniformly integrable. If the whole closure were not uniformly integrable, one could choose a sequence in it violating the uniform tail or absolute-continuity bound; applying the lemma to a weakly convergent subsequence contradicts that violation. Thus the closure, and therefore \(F\), is uniformly integrable. \(\square\)

Examples: Tails and Concentration

Using the Criterion

The practical test has two parts: verify a common \(L^1\) bound, then control tails at the same threshold \(K\) across the entire family. A bound on each individual function is not enough if the bound depends on the function. Likewise, testing only a fixed function as \(K\to\infty\) is not the uniform tail condition: the supremum over \(f\in F\) must be inside the limit.

On a finite measure space, the equivalent small-set formulation is often convenient. It asks for one \(\delta\) that works for every \(f\in F\): whenever \(\mu(E)<\delta\), the integral \(\int_E|f|\,d\mu\) is small uniformly in \(f\). This explains the role of finiteness of \(\mu(X)\) in the tail-to-small-set argument: the uniform \(L^1\) bound makes the measure of high-level sets small by Markov's inequality. The criterion is specific to \(L^1\); boundedness in \(L^1\) alone, as the spike example shows, does not imply weak compactness.