I do not have a good title for this. Anyone has a better title idea, feel free to edit. Below is my question:
Suppose you have a sequence $b_n\overset{d}{\to}N(0,1)$, and also you have another sequence $a_n\overset{a.s.}{\to}0$ if $|b_n|<c$ where $0<c<\infty$. Does this imply that $a_n\overset{p}{\to}0$?
I know this is somewhat a weird setup. I encounter this while I was trying to show $a_n\overset{a.s.}{\to}0$ and $b_n$ is some term in the Taylor expansion of $a_n$ and my proof need $|b_n|$ to be bounded by some constant in order to show $a_n\overset{a.s.}{\to}0$. I guess maybe I can show a weaker result, that is, $a_n\overset{p}{\to}0$. I have a sketch of proof but I do not know if it is correct.
For any $\epsilon>0$, and for any $0<c<\infty$,
$$pr\bigg(|a_n|>\epsilon\bigg)<pr\bigg(|b_n|<c\bigg)pr\bigg(|a_n|>\epsilon\bigg| |b_n|<c\bigg)+pr\bigg(|b_n|>c\bigg)$$
For any fixed $c$, the first term goes to zero since $a_n\overset{a.s.}{\to}0$ given $|b_n|<c$, and applying the fact that $b_n\overset{d}{\to}N(0,1)$ we have
$$\lim_{n\to\infty} pr\bigg(|a_n|>\epsilon\bigg)\leq\lim_{n\to\infty}pr\bigg(|b_n|>c\bigg)=\Phi(|x|>c),$$
where $\Phi(\cdot)$ is the cdf for standard normal. Then, letting $c\to\infty$, we have
$$\lim_{n\to\infty} pr\bigg(|a_n|>\epsilon\bigg)=0.$$
Is my reasoning correct? I am not confident with the "letting $c\to\infty$" part of proof.
Thanks in advance for your time.