A tight factor times a vanishing factor

The power set on $\mathrm{\Omega }$ is a sigma-algebra by [F1], hence gives a measurable space by [F2]. Assigning mass $1\mathrm{/}2$ to each singleton defines a measure: any pairwise disjoint sequence contains at most two nonempty sets, and the mass of their union is the sum of their masses. Its total mass is one, so it is a probability measure by [F3]–[F4]. The identity map $Z$ is measurable because every preimage of a Borel set is a subset of $\mathrm{\Omega }$; thus it is a real random variable by [F5]–[F6].

For every $n$, $\mathrm{\mid }{Z}_n\mathrm{\mid }=1$ everywhere, so choosing $M=1$ makes $P(\mathrm{\mid }{Z}_n\mathrm{\mid }>M)=0$ and proves $Z_n=O_P(1)$ by [F7]. For each $\epsilon >0$, $n^{-1\mathrm{/}2}\le \epsilon$ for all sufficiently large $n$, so $P(\mathrm{\mid }{a}_n\mathrm{\mid }>\epsilon )=0$ eventually; hence $a_n=o_P(1)$ by [F7].

The multiplication map $B(u,v)=uv$ satisfies $\mathrm{\mid }B(u,v)\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }$ for all real $u,v$, so it has the required bilinear norm bound with constant one. Each $R_n=a_n{Z}_n$ is measurable because the power-set sigma-algebra on $\mathrm{\Omega }$ contains every preimage. Thus [F8] applies to $a_n=o_P(1)$ and $Z_n=O_P(1)$ and gives $R_n=o_P(1)$. The identity $\sqrt{n}{R}_n=Z_n$ then gives $\sqrt{n}{R}_n=O_P(1)$ from step 1.2. For the finite check $n=4$, $R_4$ takes the values $\pm 1\mathrm{/}2$ and $\sqrt{4}{R}_4$ takes the values $\pm 1$.