$L^1$ space with values in a Banach Space I have been reading a bit about the Bochner integral and now I'm wondering the following:
For the theory to be "nice", one would expect that
$$L^1([0, \tau], L^1([0, \tau])) \cong L^1([0,\tau] \times [0, \tau]).$$
Is this the case? How do we prove this (or where can I find more about this), since if we take $u$ in the set in the LHS, then we can evaluate $(u(t))(x)$ and we want to map this to some $u(t,x)$. The problem I see is that $L^1$-functions are equivalence classes modulo sets of measure zero. So for each $t\in [0,\tau]$ we have different sets of measure zero than the ones in $[0,\tau]^2$. How do we get around this?
 A: There is no need to restrict to intervals, the same holds true for measure spaces $X$ and $Y$ and Banach spaces $E$ in general:

$L^{1}(X \times Y, E) \cong L^{1}(X,L^{1}(Y,E))$

Edit: Assume that $X$ and $Y$ are $\sigma$-finite and complete for the sake of simplicity. See also the edit further down.
By definition each Bochner-integrable function can be approximated by simple functions. In other words, the functions of the form $[A]e$ for some measurable subset $A \subset X$ of finite measure and $e \in E$ (or out of some dense subspace of $E$, if you prefer) generate a dense subspace of $L^{1}(X,E)$. Therefore you can reduce to the case of defining a bijection $[A]\cdot ([B] \cdot e) \leftrightarrow [A \times B]\cdot e$ and observe that this is a bijection on sets generating dense subspaces of $L^{1}(X, L^{1}(Y,E))$ and $L^{1}(X \times Y, E)$ and extend it linearly. That it is well-defined and an isometry is a special case of Fubini, that it is surjective in both directions follows from density.
The general fact underlying this is the canonical isomorphism

$L^{1}(Y,E) \cong L^{1}(Y) \hat{\otimes} E$ (projective tensor product)

and the isomorphism $L^{1}(X \times Y) \cong L^{1}(X) \hat{\otimes} L^{1}(Y)$ for $L^{1}$-spaces + associativity of the tensor product (and all this is proved in exactly the same way).
Edit: For the (non-​$\sigma$​-finite) general case the situation is quite a bit more subtle and is discussed carefully in the exercise section 253Y of Fremlin's Measure Theory, Volume II. Here are the essential points:


*

*For a general measure space $Y$ one can prove that $L^{1}(Y,E) \cong L^1(Y)\hat{\otimes} E$ (see 253Yf (vii)).

*For a pair of measure spaces $(X,\mu)$ and $(Y,\nu)$ let $(X \times Y, \lambda)$ be the complete locally determined product measure space as defined by Fremlin, 251F. The rather deep theorem 253F in Fremlin then tells us that $L^1(X \times Y, \lambda) \cong L^1(X,\mu) \hat{\otimes} L^1(Y,\nu)$.


Piecing these two results together and making use of the associativity of the projective tensor product we get
$$\begin{align*}L^1(X \times Y, E) &
\cong L^1(X \times Y) \hat{\otimes} E
\cong \left(L^1(X) \hat{\otimes} L^1(Y)\right) \hat{\otimes} E & &(\text{using 1. and 2., respectively})\\\ & \cong L^1(X) \hat{\otimes} \left(L^1(Y) \hat{\otimes} E\right) \cong L^1 (X) \hat{\otimes} L^1(Y,E) & &\text{(using associativity and 1.)} \\\ & \cong L^1(X,L^1(Y,E)) & & (\text{using 1. again})\end{align*}$$
as asserted in an earlier version of this answer.

Finally, a remark on user3148's cautionary counterexample. There is an isomorphism $(L^{1}(X,E))^{\ast} \cong L_{w^{\ast}}^{\infty}(X,E^{\ast})$ where the latter space is defined by weak$^{\ast}$-measurably in the sense of Gelfand-Dunford. So in this sense we have $L_{w^{\ast}}^{\infty}(X \times Y, E^{\ast}) \cong L_{w^{\ast}}^{\infty}(X, L_{w^{\ast}}^{\infty}(Y,E^{\ast}))$ simply by duality theory.
A: You can define the obvious isomorphism $T$ on representatives, which is isometric by Fubini's theorem.  Isometric maps defined on representatives are well defined on equivalence classes, because if $u_1$ and $u_2$ are representatives of an element of the LHS, then $\|Tu_1-Tu_2\|=\|T(u_1-u_2)\|=\|u_1-u_2\|=0$. 
It is a useful fact that a linear map initially defined on some type of representative can often be seen to be well defined from the fact that it is continuous.  In this case, all the work is hidden in Fubini's theorem.
A: Since I am not able to comment yet, I write this as an answer.
Be prepared to face the fact that $L^\infty([0,T],L^\infty([0,T])) \neq L^\infty([0,T]\times[0,T])$. To see this, observe that the function
$$
f(x,y) = \begin{cases} 1 & x>y\\ 0 & \text{else}\end{cases}
$$
is not an element of the first space.
