Why can we interchange the order of limits here?

Question

Assumptions:

$\mu$ is a probability distribution on $(X,\mathcal{X})$, $f \in L^1(X,\mu)$, $\{\xi_{N,i}\}_{1 \le i \le M_N}$ is a triangular array, conditionally independent given $\{\mathcal{F}^N\}_N$, a filtration on $(X,\mathcal{X})$.

Goal:

My goal is to show for any $\epsilon> 0$: $$ M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge M_N \epsilon \}} \mid \mathcal{F}^N] \overset{p}{\to} 0 $$ as $N \to \infty$ (which sends $M_N \to \infty$) and all I know is that for any $C \ge 0$ $$ M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge C \}} \mid \mathcal{F}^N] \overset{p}{\to} \mu(|f(\xi_{N,1})|1_{\{ |f(\xi_{N,1})| \ge C \}}). $$

Work so far:

Pick any positive $C, \epsilon$. If $N$ is large to guarantee $C < \epsilon M_N$ then

\begin{align*} & M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge M_N \epsilon \}} \mid \mathcal{F}^N] \\ &\le M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})| 1_{\{ |f(\xi_{N,i})| \ge C \}} \mid \mathcal{F}^N] \\ &\overset{p}{\to} \mu[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge C \}}] \end{align*}

as $N \to \infty$. Also, by dominated convergence and because $f$ is integrable, $\lim_{C \to \infty} \mu(|f|1[|f|\ge C]) = \mu( \lim_{C \to \infty} |f|1[|f|\ge C]) = 0$.

If we could interchange the order of limits then

\begin{align*} &\text{plim}_{N \to \infty} \lim_{C \to \infty} M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge M_N \epsilon \}} \mid \mathcal{F}^N]\\ &= \lim_{C \to \infty} \text{plim}_{N \to \infty} M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge M_N \epsilon \}} \mid \mathcal{F}^N]\\ &= 0. \end{align*} But can we? Or is something else going on? This is all from page 303 in chapter 9.2 of Inference in Hidden Markov Models.

Answer 1

$M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge M_N \epsilon \}} \mid \mathcal{F}^N]$ does not have a $C$ in it, so neither does its limit!

\begin{align*} &\text{plim}_{N \to \infty} M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge M_N \epsilon \}} \mid \mathcal{F}^N]\\ &=\lim_{C \to \infty} \text{plim}_{N \to \infty} M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge M_N \epsilon \}} \mid \mathcal{F}^N]\\ &\le \lim_{C \to \infty}\text{plim}_{N \to \infty} M_N^{-1} \sum_{i=1}^{M_N} E[ |f(\xi_{N,i})| 1_{\{ |f(\xi_{N,i})| \ge C \}} \mid \mathcal{F}^N] \\ &= \lim_{C \to \infty} \mu[ |f(\xi_{N,i})|1_{\{ |f(\xi_{N,i})| \ge C \}}] \\ &= 0. \end{align*}