The following statements suggests $B(H)$-valued Egoroff's theorem when $H$ is a separable Hilbert space. Probably it will be hold even if a von Neumann algebra $M$ whose predual is separable is replaced. I have no clear proof for this claim. Any comment on this claim?
Theorem enter link description here. Let $H$ be a separable Hilbert space. Then Borel sigma algebra coming from the weak and strong operator topology coincide.
Operator-valued Egoroff's Theorem. Assume that $H$ is a separable Hilbert space and $(\Omega,M,\mu)$ is a finite measure space. Let $\varphi_n:\Omega\to B(H)$ be a sequence of operator-valued measurable functions on $\Omega$ converging pointwise-strongly to $\varphi$. Then for any arbitrary positive trace class operator $\omega$ and positive real numbers $\epsilon$ and $\delta$, there exist a measurable set $ E\subseteq\Omega$ and a projection $p\in B(H)$ satisfying $ \mu(E)<\epsilon$ and $ \omega(1-p)<\delta$,
such that
$$ \sup_{_{t\in E^c}}\|(\varphi_n(t)-\varphi(t))p\|\longrightarrow 0.$$
Proof. Note that $\omega=\sum_{i=1}^\infty\alpha_i e_i\otimes e_i$ where $\{\alpha_i\}_{_{i\in \mathbb{N}}}\in \ell^1$ ($\alpha_i\geq0$) and $\{e_i\}_{_{i\in \mathbb{N}}}$ is an orthonormal set of $H$. Let $N\in \mathbb{N}$ with $\sum_{i=N+1}^\infty\alpha_i<\delta.$ Without loss of generality, we may assume that $\{e_i\}_{_{i\in \mathbb{N}}}$ is an orthonormal basis for $H$. It is concluded $\omega(1-p)<\delta$ where $p=\sum_{i=1}^N e_i\otimes e_i$. For any $1\leq i\leq N$, we consider the following sequence of positive functions:
$$g_i^n(t)=\|(\varphi_n(t)-\varphi(t))e_i\|
=(\sum_{j=1}^\infty|\langle(\varphi_n(t)-\varphi(t))e_i,e_j\rangle|^2)^{\frac{1}{2}}$$
All functions $g_i^n$'s are measurable. Moreover, for every $1\leq i\leq N$ the sequence
$\{g_i^n\}_{n\in\mathbb{N}}$ is pointwise convergent to zero, since $\varphi_n\stackrel{p.s}{\longrightarrow}\varphi$. By the classical Egoroff's
theorem, there exists a measurable set $ E_i\subseteq\Omega$ with $\mu(E_i)<\frac{\epsilon}{2^i}$ such that $g_i^n$ converging uniformly to 0 on $E_i^c$, for $1\leq i\leq N$. It means that
$$ \sup_{_{t\in E_i^c}}\|(\varphi_n(t)-\varphi(t))e_i\|\longrightarrow 0.$$
By taking $E=\bigcup_{i=1}^N E_i$, we have then $\mu(E)<\epsilon$. One may see that$$\sup_{_{t\in E^c}}\|(\varphi_n(t)-\varphi(t))p\|\longrightarrow 0.$$
