TITLE: $L^1$ space with values in a Banach Space
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: 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.

REPLY [2 votes]: 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.