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 \(\rho_n : \mathbb{C} \to \mathbb{R}\) defined on the complex plane, converging pointwise to an elliptic law \(\rho\) as \(n \to +\infty\). What we would like to prove is that

\[\int_X \rho_n(z) \, d^2 z \longrightarrow \int_X \rho(z) \, d^2 z\]

for any measurable set \(X \subset \mathbb{C}\). 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 \((\Omega,\Sigma,\mu)\) be a measure space and let \(\{ f_n \}_{n \in \mathbb{N}}\) be a sequence of (real or complex valued) measurable functions on \(\Omega\) converging pointwise \(\mu\)-almost everywhere to \(f\). Suppose there exists a \(g \in L^1(\Omega,\mu)\) such that

\[\begin{aligned} \lvert f_n(x) \rvert \leq g(x) & & \text{ for almost every } x \in \Omega \text{ and } \forall n \in \mathbb{N} \end{aligned}\]

then

\[\int_\Omega f_n \, d\mu \longrightarrow \int_\Omega f \, d\mu\]

as \(n \to +\infty\). \(\triangle\)

The difficulty with applying this theorem is to find a suitable “dominating function” \(g \in L^1(\Omega,\mu)\). In this post I will present a simple lemma that is able to avoid the need to find such a \(g\). 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 \((\Omega, \Sigma, \mu )\) be a measure space and let \(\{ f_n \}_{n \in \mathbb{N}}\) be a sequence of nonnegative measurable functions \(f_n : \Omega \to [0, +\infty]\) which converges pointwise to \(f(x) := \lim_{n \to \infty} f_n(x) \geq 0\).

Assume

\[\begin{aligned} &\int_\Omega f_n \, d\mu =1 & &\forall n \geq 1 \end{aligned}\]

and

\[\int_\Omega f \, d\mu =1\]

Then for any measurable subset \(X \subset \Omega\) the sequence \(\int_X f_n \, d\mu\) is convergent and converges to \(\int_X f \, d\mu\). \(\triangle\)

Proof: If \(X\) is measurable then so is \(\Omega \setminus X\). Thus by the Fatou lemma we have

\[\begin{aligned} A&:= \liminf_{n\to \infty}\int_{X} f_n \, d\mu - \int_{X} f \, d\mu \geq 0\\ B&:= \liminf_{n\to \infty}\int_{\Omega \setminus X} f_n \, d\mu - \int_{\Omega \setminus X} f \, d\mu \geq 0 \end{aligned}\]

Thus \(A+B \geq 0\). But \(A+B \leq 0\) by our assumptions. Hence \(A+B = 0\) and hence \(A=B=0\). Thus \(\liminf_{n\to \infty}\int_{X} f_n \, d\mu = \int_{X} f \, d\mu\).

Now consider a subsequence \(n_k\) such that \(\int_X f_{n_k} \, d\mu \to \limsup_{n \to \infty} \int_X f_{n} \, d\mu\) as \(k \to \infty\). We then have the equalities

\[\begin{aligned} \limsup_{n \to \infty} \int_X f_{n} \, d\mu = \lim_{k \to \infty}\int_X f_{n_k} \, d\mu = \liminf_{k \to \infty}\int_X f_{n_k} \, d\mu = \int_X f \, d\mu \end{aligned}\]

Thus \(\limsup_{n \to \infty} \int_X f_{n} \, d\mu = \liminf_{n \to \infty} \int_X f_{n} \, d\mu = \int_X f \, d\mu\). \(\square\)