Let $(X, \Omega, \mu)$ be a probability space, $H$ be a Hilbert space. It is known that $L^2(X) \otimes H \cong L^2(X, H)$ canonically, assuming $H$ is separable. Here, $L^2(X, H)$ is defined as the space of all functions $f: X \rightarrow H$ s.t.
- It is measurable, where $H$ is equipped with the Borel algebra generated by its norm topology;
- $\int_X \|f(x)\|^2 d\mu(x) < \infty$.
Two such functions are identified if they only differ on a set of measure zero.
It is not hard to see that the first criterion is equivalent to the statement that $x \mapsto <h, f(x)>$ is measurable for all $h \in H$, again assuming $H$ is separable. (To put it succinctly, measurability is equivalent to weak measurability.)
As pointed out by @geetha290krm in the comment, without the assumption that $H$ is separable, we can’t really expect all of this to work in the current form. Vector Measures by Diestel and Uhl defines measurability in a stronger sense, which in our case would be equivalent to weak measurability plus the range being essentially contained in a separable subspace. They also provided an example showing weak measurability does not imply separable range. The example goes as follows: consider $l^2((0, 1))$ with $(0, 1)$ equipped with the counting measure. It has an orthonormal basis $(e_t)$ indexed by $t \in (0, 1)$. Then the function $f: (0, 1) \rightarrow l^2((0, 1)), f(t) = e_t$ is weakly measurable when the domain is equipped with the Lebesgue measure, but the range is clearly not contained in any separable subspace (even excluding a measure zero set).
Assuming Diestel and Uhl‘s definition, I can show that $L^2(X) \otimes H \cong L^2(X, H)$. So I consider most of my original question answered. Just out of curiosity though, here are two questions that Diestel and Uhl doesn’t seem to answer:
- Diestel and Uhl‘s definition is certainly well-behaved for our purpose. However, the function in their counterexample as provided above is not measurable in the sense that the inverse image of any open set is measurable, so it doesn’t really exclude the possibility that their definition is equivalent to that. Is that true? That is, is the inverse image of any open set being measurable equivalent to strong measurability as defined by Diestel and Uhl? (Also, I realized that it’s not clear, without separability, how to even prove that measurable functions, in the sense that the inverse image of any open set is measurable, actually forms a linear space. Specifically, are they closed under addition?)
- While we generally only identify two measurable functions if they coincide outside a set of measure zero, in the non-separable case with weakly measurable functions, it seems also natural to identify two functions if their compositions with any fixed bounded linear functional coincide outside a set of measure zero, i.e., in our case, two functions $f, g: X \rightarrow H$ are identified if for any $h \in H$, $<h, f(x)> = <h, g(x)>$ $\mu$-a.e. Under this definition, the counterexample before will simply be identified with 0. Is it true, then, that any weakly measurable function can be identified with a strongly measurable one under this definition? (It can be proved that two strongly measurable functions are identified under this new definition iff they actually coincide outside a set of measure zero. So this definition seems well-behaved.)