every $t \notin M_{\ell}$.

Now we put $\tau'_n := \tau_{n,n}$ and $n' := 1/\tau'_n$, so that $\{n'\}$ is a subsequence of the strictly increasing sequence $\{m_n\}$ of positive integers from which we have started. Then it follows that for every $t \in [0,\infty) \setminus \cup_{\ell
=1}^{\infty} M_{\ell}\,$, the sequence $\{(I+
S(it,\tau'_{n}))^{-1}f\}$ converges to $(I+ itH_P)^{-1}Pf$ strongly in ${\cal H}$ as $\tau'_n \rightarrow 0$ for every $f \in
{\cal D}$, and therefore also in ${\cal H}$, because both $(I+
S(it,\tau_{\ell,n}))^{-1}$ and $(I+ itH_P)^{-1}P$ are bounded operators on ${\cal H}$ with the norms not exceeding one. We denote $M := \cup_{\ell =1}^{\infty} M_{\ell}$, which is, of course, again a set of Lebesgue measure zero. In this way we have found a subsequence $\{\tau'_n\}$ of $\{\tau_n = 1/m_n\}$ and an exceptional subset $M$ of $[0,\infty)$ such that $$\label{resolconv}
 (I+ S(it,\tau'_{n}))^{-1}f = (I+ S(it,1/n'))^{-1}f
    \longrightarrow (I+ itH_P)^{-1}Pf$$ strongly in ${\cal H}$ as $\tau'_n \rightarrow 0$ or $n'
\rightarrow \infty$ for every $f \in {\cal H}$ and for each fixed $t \notin M$; it is important that $M$ is independent of $f$.

\[L2\] For the sequence $\{n'\}$ specified above and every $f \in {{\mathcal H}}$ we have $$\label{prodconv}
 [P(1/n')\exp(-itH/n') P(1/n')]^{n'}f
 \longrightarrow  e^{-itH_P}Pf$$ as $\:n' \to\infty$ 