Problem: Given $(a_n)_{n \geq 0}$ a bounded sequence of reals and $\xi_n$ an i.i.d sequence of RV with $P(\xi_n =1)=P(\xi_n = -1) = 0.5$. Considering $$ S_n:= \sum_{k=0}^n \xi_k a_k $$ Claim: $\sum a_k^2 = \infty$ then $ T_c= \inf \lbrace n \geq 0 :|S_n| \geq c \rbrace$ for $c>0$ is finite a.s.
My approach: I want to use martingale convergence theorem in order to verify the claim but I struggle with the final reasoning.
I know that from a previous exercise I have that $S_n$ converges in $L^2$ $\iff$ $\sum a_k^2 < \infty$ (*)
Certainly $S_n$ is a martingale with respect to the natural filtration induced by $S_n$ for all $n \geq 0$ and therefore $|S_n|$ is a submartingale since $|.|$ is convex. Also since $T_c$ is a stopping time it follows that $|S_{n \wedge T_c}|$ is still a submartingale, but by the very definition of $T_c$ we have $$ |S_{n \wedge T_c}| \in [0, c + |\xi a_k|] = [0,c + \underbrace{|a_k|}_{\leq d}] \subset [0, 2 \max (c,d)], \ \forall n\in \mathbb{N} $$ Therefore we have $ \sup_n \mathbb{E} (| S_{n \wedge T_c} | ) < \infty$ and conclude by the martingale convergence theorem that $$ |S_{n \wedge T_c}| \to |S_{\infty \wedge T_c}| \text{ almost surely as } n \to \infty $$ Furthermore we have that the limit is in fact in $L^1$ and therefore finite almost surely.
Question: I now need an argument that lets me conclude that the above can only be if $T_c < \infty$ almost surely and it seems to be the time to make use of (*), but I don't see how.