If $\omega$ is a faithful normal state on a von Neumann algebra $R$, and $P_1, P_2, \ldots$ are projection operators in $R$ such that $\omega(P_n) \rightarrow 0$ as $n \rightarrow \infty$, must the $P_n$ converge to $0$ in the strong operator topology?
(Assuming $R$ has a separating vector, $R$'s faithful normal states are those induced by separating vectors, and the question is equivalent to: If $P_n v \rightarrow 0$ holds for some vector $v$ that is separating for $R$, must it hold for all vectors $v$?)
If $\omega$ is a faithful, normal state then the GNS representation $(H_\omega,\pi_\omega)$ is a faithful, normal representation of $R$. Now the storng operator topology depends on the representation (in contrast to the ultraweak topology) but this is not an issue for us: normal (c.p.) maps are continuous with respect to the ultrastrong topologies and, since $p_n$ are projections (thus self-adjoint) we have that $p_n\to0$ strongly iff $\pi_\omega(p_n)\to0$ strongly.
So it suffices to show that $$\pi_\omega(p_n)\xrightarrow{SOT}0$$ But $(H_\omega,\pi_\omega)$ is a cyclic representation for $M$ and we have the canonical cyclic vector $\xi_\omega$. It is very easy to see that $H_\omega=\overline{\{x\xi_\omega:x\in\pi_\omega(M)'\}}$, i.e. that $\xi_\omega$ is also a cyclic vector for the commutant of $\pi_\omega(M)$.
Let $\eta\in H_\omega$. We will show that $\lim_{n\to\infty}\langle\pi_\omega(p_n)\eta,\eta\rangle=0$ and from this we will conclude that $\lim_{n\to\infty}\pi_\omega(p_n)\eta=0$, because $$\lim_{n\to\infty}\|\pi_\omega(p_n)\eta\|^2=\lim_n\langle\pi_\omega(p_n)\eta,\pi_\omega(p_n)\eta\rangle=\lim_n\langle\pi_\omega(p_n)^2\eta,\eta\rangle=\lim_n\langle\pi_\omega(p_n)\eta,\eta\rangle=0.$$ Since $\eta$ was arbitrary, this will prove that $\pi_\omega(p_n)\to0$ strongly.
Now if $\varepsilon>0$ we find $x\in\pi_\omega(M)'$ so that $\|\eta-x\xi_\omega\|<\varepsilon$.
Now $$0\leq\langle\pi_\omega(p_n)x\xi_\omega,x\xi_\omega\rangle=\langle x^*x\pi_\omega(p_n)\xi_\omega,\xi_\omega\rangle\leq\langle \|x^*x\|\cdot1_{H_\omega}\cdot\pi_\omega(p_n)\xi_\omega,\xi_\omega\rangle=\|x\|^2\cdot\omega(p_n)\to0 $$
With a little housekeeping (one needs to write $\eta=\eta-x\xi_\omega+x\xi_\omega$ and do some Cauchy-Schwarz es), this will show that $\langle\pi_\omega(p_n)\eta,\eta\rangle$ becomes arbitrarily small while $n$ gets large, proving the claim.
I am posting this because I find Takesaki's book hard to read sometimes. I think this is more elementary than the approach presented there, but of course it is essentially the same.