In my work with Thomas Bothner on the elliptic Ginibre ensemble we had to prove a theorem of the following kind. We had a sequence of probability densities ρn:C→R\rho_n : \mathbb{C} \to \mathbb{R} defined on the complex plane, converging pointwise to an elliptic law ρ\rho as n→+∞n \to +\infty. What we would like to prove is that

∫Xρn(z) d2z⟶∫Xρ(z) d2z\int_X \rho_n(z) \, \mathrm{d}^2 z \longrightarrow \int_X \rho(z) \, \mathrm{d}^2 z

for any measurable set X⊂CX \subset \mathbb{C}. The “standard” way of obtaining a result like this is to use Lebesgue’s Dominated Convergence Theorem. This theorem states the following.

Theorem (Dominated Convergence): Let (Ω,Σ,μ)(\Omega,\Sigma,\mu) be a measure space and let {fn}n∈N\{ f_n \}_{n \in \mathbb{N}} be a sequence of (real or complex valued) measurable functions on Ω\Omega converging pointwise μ\mu-almost everywhere to ff. Suppose there exists a g∈L1(Ω,μ)g \in L^1(\Omega,\mu) such that

∣fn(x)∣≤g(x) for almost every x∈Ω and ∀n∈N\begin{aligned} \lvert f_n(x) \rvert \leq g(x) & & \text{ for almost every } x \in \Omega \text{ and } \forall n \in \mathbb{N} \end{aligned}

then

∫Ωfn dμ⟶∫Ωf dμ\int_\Omega f_n \, \mathrm{d}\mu \longrightarrow \int_\Omega f \, \mathrm{d}\mu

as n→+∞n \to +\infty. △\triangle

The difficulty with applying this theorem is to find a suitable “dominating function” g∈L1(Ω,μ)g \in L^1(\Omega,\mu). In this post I will present a simple lemma that is able to avoid the need to find such a gg. This lemma appears somewhat implicitly in our paper from Equations 3.12 to 3.14. I am drawing attention to it since it seems like it would be useful to simplify various proofs; for example, this lemma would have greatly simplified the proofs containing in my paper with Mezzadri and Simm on products of truncated orthogonal matrices.

Lemma: Let (Ω,Σ,μ)(\Omega, \Sigma, \mu ) be a measure space and let {fn}n∈N\{ f_n \}_{n \in \mathbb{N}} be a sequence of nonnegative measurable functions fn:Ω→[0,+∞]f_n : \Omega \to [0, +\infty] which converges pointwise to f(x):=lim⁡n→∞fn(x)≥0f(x) := \lim_{n \to \infty} f_n(x) \geq 0.

Assume

∫Ωfn dμ=1∀n≥1\begin{aligned} &\int_\Omega f_n \, \mathrm{d}\mu =1 & &\forall n \geq 1 \end{aligned}

and

∫Ωf dμ=1\int_\Omega f \, \mathrm{d}\mu =1

Then for any measurable subset X⊂ΩX \subset \Omega the sequence ∫Xfn dμ\int_X f_n \, \mathrm{d}\mu is convergent and converges to ∫Xf dμ\int_X f \, \mathrm{d}\mu. △\triangle

Proof: If XX is measurable then so is Ω∖X\Omega \setminus X. Thus by the Fatou lemma we have

A:=lim inf⁡n→∞∫Xfn dμ−∫Xf dμ≥0,B:=lim inf⁡n→∞∫Ω∖Xfn dμ−∫Ω∖Xf dμ≥0.\begin{aligned} A&:= \liminf_{n\to \infty}\int_{X} f_n \, \mathrm{d}\mu - \int_{X} f \, \mathrm{d}\mu \geq 0,\\ B&:= \liminf_{n\to \infty}\int_{\Omega \setminus X} f_n \, \mathrm{d}\mu - \int_{\Omega \setminus X} f \, \mathrm{d}\mu \geq 0. \end{aligned}

Thus A+B≥0A+B \geq 0. But A+B≤0A+B \leq 0 by our assumptions. Hence A+B=0A+B = 0 and hence A=B=0A=B=0. Thus lim inf⁡n→∞∫Xfn dμ=∫Xf dμ\liminf_{n\to \infty}\int_{X} f_n \, \mathrm{d}\mu = \int_{X} f \, \mathrm{d}\mu.

Now consider a subsequence nkn_k such that ∫Xfnk dμ→lim sup⁡n→∞∫Xfn dμ\int_X f_{n_k} \, \mathrm{d}\mu \to \limsup_{n \to \infty} \int_X f_{n} \, \mathrm{d}\mu as k→∞k \to \infty. We then have the equalities

lim sup⁡n→∞∫Xfn dμ=lim⁡k→∞∫Xfnk dμ=lim inf⁡k→∞∫Xfnk dμ=∫Xf dμ.\begin{aligned} \limsup_{n \to \infty} \int_X f_{n} \, \mathrm{d}\mu = \lim_{k \to \infty}\int_X f_{n_k} \, \mathrm{d}\mu = \liminf_{k \to \infty}\int_X f_{n_k} \, \mathrm{d}\mu = \int_X f \, \mathrm{d}\mu. \end{aligned}

Thus lim sup⁡n→∞∫Xfn dμ=lim inf⁡n→∞∫Xfn dμ=∫Xf dμ\limsup_{n \to \infty} \int_X f_{n} \, \mathrm{d}\mu = \liminf_{n \to \infty} \int_X f_{n} \, \mathrm{d}\mu = \int_X f \, \mathrm{d}\mu. □\square