Yes – the factor that appears in the homogeneous direct integral L∞[0,1] ⊗̄ M completely determines the isomorphism class of the integral, no matter how big (or non–separable) the factors are. Hence, if

  L∞[0,1]  ⊗̄  M   ≅   L∞[0,1]  ⊗̄  N                                    (1)

then M ≅ N. A convenient way to see this is to use the standard central– (or direct–integral–) decomposition of a von Neumann algebra over its centre. Separability of the predual is not needed; what is used is that the centre is the separable abelian algebra L∞[0,1].

Step 1. The isomorphism sends the centre onto the centre
Let ϑ be an isomorphism that realises (1). The centre of A := L∞[0,1] ⊗̄ M is L∞[0,1]; the same is true for B := L∞[0,1] ⊗̄ N. Because ϑ(x) commutes with everything in B whenever x commutes with everything in A, we have

    ϑ(L∞[0,1]) = Z(B) = L∞[0,1].

(In particular, the image of the centre is exactly the centre, not just some abelian algebra.)

Step 2. Identify the centres
Choose an automorphism σ of L∞[0,1] which in measure–theoretic language corresponds to an invertible measure preserving map τ : [0,1] → [0,1] such that σ = ϑ|_{L∞}. Replacing ϑ by (σ^{-1} ⊗ id) ∘ ϑ we may – and will – assume from now on that ϑ acts as the identity on the centre:

    ϑ(f ⊗ 1) = f ⊗ 1, for every f ∈ L∞[0,1].                       (2)

Step 3. Direct–integral picture
Write the two algebras as constant fields of factors:

    A   =   ∫_[0,1] ⊕  M   dλ(t),
    B   =   ∫_[0,1] ⊕  N   dλ(t).

(The usual realisation is “operator–valued multiplication’’ on L²[0,1] ⊗ H_M and L²[0,1] ⊗ H_N, but only the abstract direct–integral form is used.)

Because ϑ fixes the centre (2), classical results on decompositions over diffuse centres (see Dixmier, Algèbres de von Neumann, §10.8; Takesaki II, Theorem IV.8.24) imply that ϑ is given fibre–wise by a measurable field of normal ***-isomorphisms

    ϑ(t) :  M  →  N, t ∈ [0,1],

in the sense that

    (ϑ(f ⊗ x))(t) = f(t) · ϑ(t)(x)        (λ–a.e. t).               (3)

Step 4. Fibre–wise injectivity and surjectivity
For each t, the map x ↦ ϑ(t)(x) is a normal *-homomorphism from the factor M into the factor N. Equation (3) shows that

• if ϑ(t) were not injective on a set of positive measure, then ϑ would not be injective; • if ϑ(t) were not surjective on a set of positive measure, then ϑ would not be surjective.

Since ϑ is an isomorphism, both pathologies have measure zero. Hence

    ϑ(t) is a *-isomorphism M → N for λ–almost every t.             (4)

Step 5. Extract one fibre
Because the set of t for which (4) holds has full measure, pick one such point t₀ ∈ [0,1]. Then ϑ(t₀) is an isomorphism of factors, i.e.

                     M  ≅  N.

That is exactly what we wanted to prove.

Remarks

1. Nothing in the argument uses separability of the preduals of M or N; only that the centre L∞[0,1] is the von Neumann algebra of a standard (σ-finite) measure space.

2. More generally, if Z is any diffuse abelian von Neumann algebra and M, N are factors, an isomorphism Z ⊗̄ M ≅ Z ⊗̄ N forces M ≅ N.

3. For type I factors one can see the necessity directly from the cardinal-valued dimension function of projections; but the argument above works uniformly for all types.

Consequently, the “exercise’’ is correct: L∞[0,1]  ⊗̄ M and L∞[0,1] ⊗̄ N are isomorphic if and only if the factors M and N themselves are isomorphic, with no separability hypothesis needed.