I read this paper, in Corollary 1 the author claims that $$\underset{\pi \in [0, 1]}{\sup}\ W_T(\pi) \overset{p}{\to} \infty$$ as $T \to \infty$. Where $W_T(\pi)$ is Wald statistics but I think it should not matter for my question.
The first step of the proof (formula A.35 in the paper) goes like this: Take any positive $c$,
$$\lim_{T \to \infty} \mathbb{P}(\underset{\pi \in [0, 1]}{\sup}\ W_T(\pi) < c) \le \lim_{\varepsilon \to 0}\sup\ \lim_{T \to \infty} \mathbb{P}(\underset{\pi \in [\varepsilon, 1-\varepsilon]}{\sup}\ W_T(\pi) < c)$$ and then he proves that the expression on the RHS is zero. For me it is unclear where this first inequality comes from. I think, using the monotone convergence theorem we can get $$\lim_{T \to \infty} \mathbb{P}(\underset{\pi \in [0, 1]}{\sup}\ W_T(\pi) < c) = \lim_{T \to \infty} \lim_{\varepsilon \to 0} \mathbb{P}(\underset{\pi \in [\varepsilon, 1-\varepsilon]}{\sup}\ W_T(\pi) < c)$$ but I do not think that in general we can say that $$\lim_{T \to \infty} \lim_{\varepsilon \to 0} \mathbb{P}(\underset{\pi \in [\varepsilon, 1-\varepsilon]}{\sup}\ W_T(\pi) < c) \le \lim_{\varepsilon \to 0}\sup\ \lim_{T \to \infty} \mathbb{P}(\underset{\pi \in [\varepsilon, 1-\varepsilon]}{\sup}\ W_T(\pi) < c)$$ take for example $a_{T, \varepsilon} = 1\ \{ \frac{1}{\varepsilon} > T \}$ then $$1 = \lim_{T \to \infty} \lim_{\varepsilon \to 0} a_{T, \varepsilon} > \lim_{\varepsilon \to 0}\sup\ \lim_{T \to \infty} a_{T, \varepsilon} = 0$$
Do you have any ideas what is implicitly assumed in the proof to justify this step? Or am I missing something?
Thanks so much!
$\newcommand{\bP}{\mathbb{P}}$ I think the argument is simpler. Note that, if $A \implies B$, then $\bP(A) \leq \bP(B)$ (I use statements in place of events, I am sure you are fine with it).
Now fix $\varepsilon > 0$. Then clearly $$ \sup_{\pi \in [0,1]} W_{T}(\pi) < c \quad \implies \quad \sup_{\pi \in [\varepsilon,1-\varepsilon]} W_{T}(\pi) < c, $$ because the RHS is just a weaker statement. Hence, $$ \bP\bigg(\sup_{\pi \in [0,1]} W_{T}(\pi) < c\bigg) \leq \bP\bigg(\sup_{\pi \in [\varepsilon,1-\varepsilon]} W_{T}(\pi) < c \bigg), $$ which implies $$ \lim_{T\to\infty} \bP\bigg(\sup_{\pi \in [0,1]} W_{T}(\pi) < c\bigg) \leq \lim_{T\to\infty} \bP\bigg(\sup_{\pi \in [\varepsilon,1-\varepsilon]} W_{T}(\pi) < c \bigg), $$ (assuming that the limits exist). Since this statement holds for each $\varepsilon > 0$, it also holds for $\limsup_{\varepsilon\to 0}$: $$ \limsup_{\varepsilon\to 0} \lim_{T\to\infty} \bP\bigg(\sup_{\pi \in [0,1]} W_{T}(\pi) < c\bigg) \leq \limsup_{\varepsilon\to 0} \lim_{T\to\infty} \bP\bigg(\sup_{\pi \in [\varepsilon,1-\varepsilon]} W_{T}(\pi) < c \bigg). $$ But the LHS does not depend on $\varepsilon$, so you arrive at the desired statement.
Comment: In your last example you do not have $a_{T,0} \leq a_{T,\varepsilon}$ for all $\varepsilon,T$ (rather the other way around), so the argument would not go through here.