Proof of $L^2(X, H)$ isomorphic to $L^2(X) \otimes H$ Let $(X, \mu)$ be a measure space and let $H$ be a Hilert space. Let $L^2(X, H)$ denotes the class of all functions $f: X \rightarrow H$ such that $\int ||f(x)||^2d\mu(x) < \infty$.How to prove that $L^2(X, H)$ isomorphic to $L^2(X) \otimes H$? Is it true alwys or do we need any assumtions?
 A: Let $\mathcal{S}_{X,H}\subseteq L^2(X,H)$ denote the space of simple functions, i.e. functions of the form $f(x)=\sum_{i=1}^n e_i 1_{A_i}$, where $e_i\in H$ and the $A_i$ are disjoint and have finite measure. The function $e_i 1_{A_i}$ is to be understood as the function takes value $e_i$ on $A_i$ and $0$ else.
Likewise, define $\mathcal{S}\subseteq L^2(X)$ be the usual simple functions.
Let $\otimes_{Vect}$ denote the algebraic tensor product, and define
$T: \mathcal{S}\otimes_{Vect}H\to \mathcal{S}_{X,H}$
to be the linear extension of 
$$
T( 1_{A}\otimes e)=e1_{A}
$$
Then, we see that for $g_1,g_2$ of the form $\lambda_j 1_{A_j}$ and $e,f\in H$, we have
$$
\langle T(g_1\otimes e),T(g_2\otimes f)\rangle_{L^2(X,H)}=\int_X 1_{A_1}(x)1_{A_2}(x) \langle e,f\rangle_H \textrm{d}\mu(x)=\mu(A_1\cap A_2)\langle e,f\rangle_H =\langle g_1,g_2\rangle_{L^2(X)}\langle e,f\rangle_H=\langle g_1\otimes e,g_2\otimes f\rangle_{L^2(x)\otimes H}, 
$$
and hence, $T$ is an inner-product preserving linear isomorphism. Furthermore, $\mathcal{S}\otimes_{vect} H$ is clearly dense in $L^2(X)\otimes H,$ since $\mathcal{S}$ is dense in $L^2(X)$. This implies that $L^2(X)\otimes H$ is isomorphic to the closure of $\mathcal{S}_{X,H}$ in $L^2(X,H)$, and this is the space of so-called strongly measurable $L^2(X,H)$-functions.
Hence, a necessary and sufficient condition is that $L^2(X,H)$ is isomorphic to its subspace of strongly measurable functions, and I'm sadly not an expert on this, but this is apparently not always the case.
