A uniform L¹ approximation by Gaussian-convergent sequences forces a Gaussian limit

The truncation closure of a central limit theorem. Let $(Y_n)$ be a sequence of real random variables on a probability space, and suppose it is approximated, uniformly in $n$, by a family of auxiliary sequences $(W^K_n)_n$ indexed by $K$:

Suppose moreover that each auxiliary sequence satisfies a central limit theorem, $W^K_n \Rightarrow N(0, v_K)$ as $n \to \infty$, and that the variances converge, $v_K \to v$. Then $Y_n \Rightarrow N(0, v)$.

Discussion. This is the standard "$3\varepsilon$" or approximation lemma that closes a truncation argument. Its content is that convergence in distribution, although not itself an $L^1$ notion, is stable under $L^1$ perturbations that are small uniformly in $n$. The uniformity is essential: without it one would have to diagonalise, choosing $K = K(n)$, and the conclusion could fail. Note also that the auxiliary variables $W^K_n$ need not be measurable in any strong sense — the almost-everywhere measurability contained in each hypothesis $W^K \Rightarrow N(0,v_K)$ is enough.

Where this is used. For the central limit theorem for a square-integrable observable $f$ of a uniformly ergodic Markov chain, one sets $f_K =$ the truncation of $f$ at level $K$, $Y_n = S_n(f)/\sqrt n$ and $W^K_n = S_n(f_K)/\sqrt n$. The bounded-observable theorem provides the CLT for each $W^K$; the $O(n)$ bound on the variance of partial sums gives $\mathbb E|Y_n - W^K_n| \le 2\sqrt N \, \|f - f_K\|_{L^2(\pi)}$, uniformly in $n$, which tends to $0$ by dominated convergence; and the Cauchy property of $(v_K)$, obtained from the characteristic function, supplies the limiting variance. This lemma then delivers the theorem.

Proof. Weak convergence on $\mathbb R$ is tested by bounded Lipschitz functions. Fix such an $f$, with Lipschitz constant $L$ and range of diameter at most $C$; the latter bounds $|f|$, so all the integrals below converge. Given $\varepsilon > 0$, choose $K$ so large that both $\bigl|\int f\,dN(0,v_K) - \int f\,dN(0,v)\bigr| < \varepsilon/3$ — possible because centred Gaussians depend weakly continuously on the variance — and $L\,\delta_K < \varepsilon/3$. For that fixed $K$, the hypothesis $W^K_n \Rightarrow N(0,v_K)$ gives $\bigl|\mathbb E f(W^K_n) - \int f\,dN(0,v_K)\bigr| < \varepsilon/3$ for all large $n$. Finally, for every $n$, the Lipschitz bound and monotonicity of the integral give $\bigl|\mathbb E f(Y_n) - \mathbb E f(W^K_n)\bigr| \le L\,\mathbb E|Y_n - W^K_n| \le L\,\delta_K < \varepsilon/3$. Adding the three estimates completes the proof.