Let $M$ be a von Neumann algebra with separable predual. Let us assume that $M$ is of type II$_1$, meaning that it is finite but has no type I part. Let $\tau$ be a faithful normal tracial state on $M$.

First, I would like to choose some free ultrafilter $\omega$ on $\mathbb N$ and consider the tracial ultrapower $$ M^\omega = \ell^\infty(\mathbb N, M)/\{ (x_n)_n \mid \tau(x_n^*x_n)\to 0 \}. $$

Next, let $\alpha$ be an automorphism on $M$. It is called properly outer, if for every non-zero projection $p\in M^\alpha$, the induced automorphism on $pMp$ is not inner.

Note that each automorphism gives rise to $\alpha^\omega\in\operatorname{Aut}(M^\omega)$ via pointwise application, and restricts to an automorphism on the relative commutant $M^\omega\cap M'$.

**Theorem:** (Connes) If $M$ is the hyperfinite II$_1$-factor, then $\alpha$ is (properly) outer if and only if $\alpha^\omega$ is (properly) outer on $M^\omega\cap M'$.

My question:Is the above also true if $M$ is a hyperfinite type II$_1$ von Neumann algebra, i.e., not assumed to be a factor? So is an automorphism $\alpha$ properly outer if and only if $\alpha^\omega$ is properly outer on $M^\omega\cap M'$?

I am sure that somebody must have worked this out somewhere, but I cannot seem to find it. I would be grateful either for the argument showing/disproving this statement or a reference with the answer.

EDIT: I realized that it is not necessary for $\alpha$ to assume that it preserves the trace $\tau$. The question makes perfect without it.