Continuous maps preserve convergence in probability

Fix $\epsilon ,\eta >0$. Choose a compact cube $K$ with $\mathbb{P}(Z\mathrm{\notin }K)<\eta$ and a compact cube $K^{\mathrm{\prime }}$ containing every point within distance $1$ of $K$. Uniform continuity of $f$ on $K^{\mathrm{\prime }}$ gives $\delta \in (0,1)$ such that points of $K^{\mathrm{\prime }}$ within $\delta$ have $f$-images within $\epsilon$.

If $Z\in K$ and $\mathrm{\parallel }{Z}_n-Z\mathrm{\parallel }<\delta$, then $Z_n\in K^{\mathrm{\prime }}$ and $\mathrm{\parallel }f(Z_n)-f(Z)\mathrm{\parallel }<\epsilon$. Thus the image bad-event probability is at most $\eta +\mathbb{P}(\mathrm{\parallel }{Z}_n-Z\mathrm{\parallel }\ge \delta )$, which is at most $\eta +\mathbb{P}(\mathrm{\parallel }{Z}_n-Z\mathrm{\parallel }>\delta \mathrm{/}2)$. Its limsup is at most $\eta$ by [L1]. Letting $\eta \downarrow 0$ proves the claim.

Apply [L2] and the claim to $(x,y)\mapsto x+y$ and $(x,y)\mapsto xy$. For division, first restrict to $\mathrm{\mid }y\mathrm{\mid }\ge r$ and then let $r\downarrow 0$; the limiting denominator is nonzero almost surely, and the zero-denominator convention for the approximating pair is contained in the remaining event.