Uniformly Asymptotically Negligible (2): Connect to the Characteristic Function

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Let $\{X_{nj}\}$, $n=1,2,...$, $j=1,2,...,k_n$, be a double array of random variables and $\{f_{nj}(t)\}$ be their ch.f.'s.  $$\forall\,\varepsilon>0,\,\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\mathscr{P}\{|X_{nj}|>\varepsilon\}=0\iff\forall\,t\in\mathbb{R},\,\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\left|f_{nj}(t)-1\right|=0.$$

$\bullet$ Proof.

$(\Longrightarrow)$ Note that
(1) $\left|e^{itx}-1\right|\leq\left|e^{itx}\right|+1\leq1+1=2$;
(2) By the Taylor expansion $\cos{tx}=1-\frac{(tx)^2}{2!}+\cdots\leq1-\frac{(tx)^2}{2}$, $$\begin{array}{rl}\left|e^{itx}-1\right|&=\left|\cos{tx}-1+i\sin{tx}\right|=\sqrt{(\cos{tx}-1)^2+\sin^2{tx}}\\
&=\sqrt{2-2\cos{tx}}=\sqrt{2}\sqrt{1-\cos{tx}}\leq|tx|.\end{array}$$We have  $$\begin{array}{rl}\left|f_{nj}(t)-1\right|
&\leq\int_\mathbb{R}\left|e^{itx}-1\right|\,dF_{nj}(x)\\
&=\int_{|x|>\varepsilon}\left|e^{itx}-1\right|\,dF_{nj}(x)+\int_{|x|\leq\varepsilon}\left|e^{itx}-1\right|\,dF_{nj}(x)\\
&\leq\int_{|x|>\varepsilon}2\,dF_{nj}(x)+|t|\int_{|x|\leq\varepsilon}|x|\,dF_{nj}(x)\\
&\leq2\mathscr{P}\{|X_{nj}|>\varepsilon\}+\varepsilon|t|. \\ \end{array}$$Thus, for all $t\in\mathbb{R}$ and let $\varepsilon\rightarrow0$, $$\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\left|f_{nj}(t)-1\right|\leq\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\left(2\mathscr{P}\{|X_{nj}|>\varepsilon\}+\varepsilon|t|\right)\rightarrow0.$$

$(\Longleftarrow)$  First we introduce the following lemma.

[Lemma]  Let $f(t)$ be a ch.f. w.r.t. $\mu$, we have for each $A>0$,  $$\mu[-2A,2A]\geq A\left|\int_{-A^{-1}}^{A^{-1}}f(t)\,dt\right|-1.$$

Back to the proof.  By the above lemma, we have $$\mu[-A,A]^c\leq 1-\left[\frac{A}{2}\left|\int_{-\frac{2}{A}}^{\frac{2}{A}}f(t)\,dt\right|-1\right]=2-\frac{A}{2}\left|\int_{-\frac{2}{A}}^{\frac{2}{A}}f(t)\,dt\right|.$$Now, we turn to use the $\varepsilon$ sign, $$\begin{array}{rl}\mathscr{P}\{|X_{nj}|>\varepsilon\}&\leq 2-\frac{\varepsilon}{2}\left|\int_{-\frac{2}{\varepsilon}}^{\frac{2}{\varepsilon}}f_{nj}(t)\,dt\right|=\frac{\varepsilon}{2}\int_{|t|\leq\frac{2}{\varepsilon}}\,dt-\frac{\varepsilon}{2}\left|\int_{|t|\leq\frac{2}{\varepsilon}}f_{nj}(t)\,dt\right|\\ &\leq\frac{\varepsilon}{2}\int_{|t|\leq\frac{2}{\varepsilon}}\left|1-f_{nj}(t)\right|\,dt. \end{array}$$Thus, for all $\varepsilon>0$, we have $$\begin{array}{rl}0\leq\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\mathscr{P}\{|X_{nj}|>\varepsilon\}&\leq\frac{\varepsilon}{2}\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\int_{|t|\leq\frac{2}{\varepsilon}}\left|1-f_{nj}(t)\right|\,dt\\ & = \frac{\varepsilon}{2}\int_{|t|\leq\frac{2}{\varepsilon}}\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\left|1-f_{nj}(t)\right|\,dt=0 \end{array}$$by the Bounded Convergence Theorem.

$\Box$