Centred Gaussians depend weakly continuously on their variance

The centred Gaussian law depends continuously on its variance, in the weak topology. If $v_K \to v$ in $[0,\infty)$ then $N(0, v_K) \Rightarrow N(0, v)$. Concretely: for every bounded Lipschitz $f : \mathbb R \to \mathbb R$,

Bounded Lipschitz test functions suffice to determine weak convergence (the bounded-Lipschitz metric metrizes it on a separable metric space), so the displayed statement is equivalent to $N(0,v_K) \Rightarrow N(0,v)$; it is stated in integral form because that is the form in which it is consumed. Note that no continuity at $v = 0$ needs to be excluded: $N(0,0)$ is the Dirac mass at $0$ and the statement remains true there.

Where this is used. In the central limit theorem for a square-integrable observable of a uniformly ergodic Markov chain, the observable is truncated at level $K$; the bounded case gives $S_n(f_K)/\sqrt n \Rightarrow N(0, v_K)$, and the truncated variances are shown to be Cauchy, say $v_K \to v$. The final $3\varepsilon$ argument then needs to replace $N(0,v_K)$ by $N(0,v)$ at a cost that vanishes with $K$, which is exactly this lemma.

Proof. Every centred Gaussian is a rescaled standard Gaussian: $N(0,t)$ is the pushforward of $N(0,1)$ under $y \mapsto \sqrt t\, y$. Changing variables,

the integrand being continuous and bounded, hence integrable. For two variances $t$ and $v$ the Lipschitz property gives the pointwise bound $|f(\sqrt t y) - f(\sqrt v y)| \le L\,|\sqrt t - \sqrt v|\,|y|$, so

and the standard Gaussian has a finite first absolute moment. Since $\sqrt\cdot$ is continuous, the right-hand side tends to $0$ as $t \to v$, and a squeeze finishes the proof.