Some comments on convergence in distribution

This is a note for myself.

Proposition. Let $(X_n)_{n\geq 1}$ be a sequence of random variables. Suppose that, for each $\epsilon > 0$, there exist random variables $(Y^{\epsilon}_n)_{n\geq 1}$, $(Z^{\epsilon}_n)_{n\geq 1}$, and $Y^{\epsilon}$ such that $$X_n = Y^{\epsilon}_n + Z^{\epsilon}_n, \qquad Y^{\epsilon}_n \xrightarrow[n\to\infty]{\text{d}} Y^{\epsilon}, \qquad \limsup_{n\to\infty} \mathbb{E}[|Z^{\epsilon}_n|] \leq \epsilon.$$ Then $(X_n)_{n\geq 1}$ converges in distribution as $n\to\infty$. Moreover, $(Y^{\epsilon})_{\epsilon > 0}$ also converges in distribution to the same limit as $\epsilon \to 0^{+}$.

Proof. By the assumption, it is clear that $(X_n)_{n\geq 1}$ is tight. So it suffices to prove that there is a unique subsequential limit. To this end, note that $$\bigl| \varphi_{X_n}(\xi) - \varphi_{Y^{\epsilon}}(\xi) \bigr| \leq \bigl| \varphi_{Y^{\epsilon}_n}(\xi) - \varphi_{Y^{\epsilon}}(\xi) \bigr| + \mathbb{E}[|\xi Z^{\epsilon}_n|].$$ If $X$ is a subsequential limit of $(X_n)_{n\geq 1}$, then taking limsup along the subsequence $(n_k)_{k\geq 1}$ for which $X_{n_k}$ converges in distribution to $X$, we get $$\bigl| \varphi_{X}(\xi) - \varphi_{Y^{\epsilon}}(\xi) \bigr| \leq |\xi|\epsilon.$$ Since this is true for any $\epsilon > 0$, it follows that $Y^{\epsilon} \to X$ in distribution as $\epsilon \to 0^+$. This shows that $X$ is uniquely determined, and we are done. $\square$

Proposition. Let $\Omega$ be a Radon space equipped with the Borel probability measure $\mathbb{P}$. Let $(X_n)_{n\geq 1}$ and $Y$ be random variables on $\Omega$ and $\mu$ be the law of $Y$. Suppose that $(X_n)_{n\geq 1}$ converges in distribution under $\mathbb{P}(\cdot \mid Y = y)$ for $\mu$-almost every $y$. Then $(X_n)_{n\geq 1}$ converges in distribution.

Proof. The assumption tells that $$\varphi_y(\xi) := \lim_{n\to\infty} \mathbb{E}[e^{i\xi X_n} \mid Y = y]$$ exists as characteristic function for $\mu$-almost every $y$. Then by the Bounded Convergence Theorem, $$\varphi(\xi) := \lim_{n\to\infty} \mathbb{E}[e^{i\xi X_n}] = \int_{\mathbb{R}} \lim_{n\to\infty} \mathbb{E}[e^{i\xi X_n} \mid Y = y] \, \mu(\mathrm{d}y) = \int_{\mathbb{R}} \varphi_y(\xi) \, \mu(\mathrm{d}y)$$ exists. Now, by the Bounded Convergence Theorem again, for any $(\xi_k)_{k\geq 1}$ with $\xi_k \to 0$, $$\lim_{k \to \infty} \varphi(\xi_k) = \int_{\mathbb{R}} \lim_{k \to \infty} \varphi_y(\xi_k) \, \mu(\mathrm{d}y) = 1.$$ Therefore the desired conclusion follows from Lévy's Continuity Theorem.