Proving $\lim_{n\to\infty}\sup_{\lambda \in \Lambda}\big|\int_\mathbb X f_{n, \lambda}(x) \mu_n(d x)-\int_\mathbb X f_\lambda(x) \mu(d x)\big|=0$

Question

Could any one help me to prove the following?

Let $X$ be a Borel space, suppose that we have a family of uniformly bounded real Borel measurable functions $\{f_{n,\lambda}\}_{n\ge 0, \lambda\in J}$ and $\{f_{\lambda}\}_{\lambda\in J}$, if for any $x_n\to x\in X$ we have

\begin{array}{l} \lim\limits _{n \rightarrow \infty} \sup _{\lambda \in \Lambda}\left|f_{n, \lambda}\left(x_{n}\right)-f_{\lambda}(x)\right|=0 \\ \lim\limits _{n \rightarrow \infty} \sup _{\lambda \in A}\left|f_{\lambda}\left(x_{n}\right)-f_{\lambda}(x)\right|=0 \end{array} Then for any $\mu_n\to\mu$ weakly in $P(X)$,

$$ \lim\limits _{n \rightarrow \infty} \sup _{\lambda \in \Lambda}\left|\int_{\mathbb{X}} f_{n, \lambda}(x) \mu_{n}(d x)-\int_{\mathbb{X}} f_{\lambda}(x) \mu(d x)\right|=0 $$ First of all, I did not understand quite well the notation, $\{f_{n,\lambda}\}_{n\ge 0, \lambda\in J}$. Thanks for any help and example.

Note: The Borel space $X$ is Polish and you can assume $X=\mathbb R^n$.

Answer 1

By a theorem of Skorokhod, because $\mu_n\to\mu$ weakly, there is a probability space $(\Omega,\mathcal F,\Bbb P)$ and random elements $Y_n:\Omega\to X$ and $Y:\Omega\to X$ such that (i) $\lim_n Y_n(\omega) = Y(\omega)$ for each $\omega\in\Omega$, (ii) $Y_n$ has distribution $\mu_n$, and (iii) $Y$ has distibution $\mu$. By your hypothesis $$ \lim_n\sup_\lambda|f_{n,\lambda}(Y_n(\omega))-f_\lambda(Y(\omega))|=0, $$ for each $\omega\in\Omega$. Consequently, $$ \eqalign{ \sup_\lambda \left|\int_X f_{n,\lambda}(y)\,\mu_n(dy)-\int_X f_\lambda(y)\,\mu(dy)\right| &=\sup_\lambda\left|\Bbb E[f_{n,\lambda}(Y_n)-f_\lambda(Y)]\right|\cr &\le\sup_\lambda\Bbb E[\left|f_{n,\lambda}(Y_n)-f_\lambda(Y)\right|]\cr &\le\Bbb E[\sup_\lambda\left|f_{n,\lambda}(Y_n)-f_\lambda(Y)\right|],\cr } $$ and the last expectation tends to $0$ by the Bounded Convergence Theorem, by the earlier-noted pointwise convergence and the assumed uniform boundedness of the $f_{n,\lambda}$ and $f_\lambda$. (The second of your two "continuity" conditions is not needed.)

Answer 2

You can think $\{f_{n,\lambda}\}$ this sequence as a sequence of function with parameter $\lambda$, for example Normal distribution with parameter $\lambda$(mean) and variance $\frac{1}{n}$.

Now Consider the problem, $$ |\int_{\mathbb{X}} f_{n,\lambda}d\mu_n - \int_{\mathbb{X}} f_{\lambda}d\mu| \\ = |\int_{\mathbb{X}} f_{n,\lambda}d\mu_n - \int_{\mathbb{X}}f_{\lambda}d\mu_n +\int_{\mathbb{X}}f_{\lambda}d\mu_n-\int_{\mathbb{X}} f_{\lambda}d\mu| $$ Now from triangle inequality, $$ |\int_{\mathbb{X}} f_{n,\lambda}d\mu_n - \int_{\mathbb{X}} f_{\lambda}d\mu| \leq \int_{\mathbb{X}} |f_{n,\lambda} - f_{\lambda}|d\mu_n +|\int_{\mathbb{X}}f_{\lambda}d\mu_n-\int_{\mathbb{X}} f_{\lambda}d\mu| ......(*) $$ Now focus on second term of the right hand side of the inequality(*),

from your second condition $\lim_n \sup_{\lambda}|f_{\lambda}(x_n)-f_{\lambda}(x)|=0$, we say that function $f_{\lambda}$ is continuous and all function belongs to the uniformly bounded family. So by Portmanteau theorem, second term of the right hand side of the inequality(*) goes to $0$ as $n\rightarrow\infty$.

Now focus on the first term of the right hand side of the inequality(*),

$$ \int_{\mathbb{X}} |f_{n,\lambda} - f_{\lambda}|d\mu_n \leq \int_{\mathbb{X}} \sup_{\lambda}|f_{n,\lambda} - f_{\lambda}|d\mu_n $$ Now from your second condition $ \lim_n\sup_{\lambda}|f_{n,\lambda}(x_n)-f_{\lambda}(x)|=0$, we say that $$ \lim_n\sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|=0$$$$\implies \liminf_n \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|= \limsup_n \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)| = \lim_n \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)| =0 $$ Taking integral both-side with respect to $\mu$ we get $$ \int_{\mathbb{X}}\liminf_n \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|d\mu_n=\int_{\mathbb{X}} \limsup_n \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|d\mu =0 $$ Now using Fatou's lemma for varying measure, we get $$ 0=\int_{\mathbb{X}}\liminf_n \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|d\mu\leq\liminf_n\int_{\mathbb{X}} \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|d\mu_n$$$$\leq \limsup_n \int_{\mathbb{X}} \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|d\mu_n \leq \int_{\mathbb{X}} \limsup_n \sup_{\lambda}|f_{n,\lambda}(x)-f_{\lambda}(x)|d\mu=0 $$ So, $$ \lim_n\int_{\mathbb{X}} |f_{n,\lambda} - f_{\lambda}|d\mu_n =0 $$

I hope you can extend this by taking $\sup_{\lambda}$ in the inequality (*).