Is there a version of Fischer-Riesz theorem for Banach space? $( \Omega,F, P )$: a measurable space equipped with a finite measure 
$(B , \Vert \cdot \Vert) $ : a Banach space with $\mathcal{B}$ as its borelian $\sigma$-algebra
$p$ : a constant bigger than $1$
Define $L^p(\Omega, B )$ the vector space that contain all $( F, \mathcal{B})$-measurable function $f$ such that :
$ \vert \Vert  f \Vert \vert = \sqrt[p]{ \int_{\Omega} \Vert f \Vert ^p \cdot dP } < \infty$ 
I'm looking for a version of Riesz-Fischer theorem which affirms that:
Proposition: 
$\left( L^p(\Omega, B ) , \vert \Vert \cdot \Vert \vert \right)$ is a Banach space
With some quick calculations, I have the feeling that this proposition is quite easy to be proved. But as we all know, it's always better to have a reliable reference.
So my question is: "Is the above proposition true? And does anyone have references to this matter?" 
 A: These are called Bochner spaces.  Under mild assumptions (see Gerald's post), they are Banach spaces. 
It is sufficient to assume that $B$ is separable, or that $L^p(\Omega, B)$ is defined to include only functions with almost every value in a separable subspace.  Without some assumptions, it is possible that your $L^p(\Omega, B)$ is not even a vector space.
Given such assumptions, then at least one of the standard proofs that $L^p$ is complete goes through basically without change:
Let $f_n$ be Cauchy in this norm.  Pass to a subsequence so that $|\|f_n - f_{n+1}\|| \le 4^{-n}$.  By Chebyshev's inequality, we then have $P(\|f_n - f_{n+1}\| \ge 2^{-n}) \le 2^{-pn}$.  Then the Borel-Cantelli lemma implies that for almost every $\omega \in \Omega$, we have $\|f_n(\omega) - f_{n+1}(\omega)\| \le 2^{-n}$ for all but finitely many $n$.  In particular, for such $\omega$, the sequence $\{f_n(\omega)\}$ is Cauchy in $B$, so it converges to some $f(\omega) \in B$.  
Now you have that $f$ is the a.e. limit of the $f_n$.  Let $\epsilon > 0$.  Since $f_n$ is Cauchy in $|\|\cdot\||$-norm, choose $N$ so large that $|\|f_n - f_m\|| < \epsilon$ for all $n,m > N$.  Letting $m \to \infty$ and using Fatou's lemma on the integrals $\int_\Omega \|f_n - f_m\|\,dP$, conclude that $|\|f_n - f\|| < \epsilon$ as well.  Thus the subsequence $f_n$ converges to $f$ in norm.  Now use the Cauchy assumption one more time to see that the original sequence converges to $f$ as well.
I think that Evans's PDE book has some basic results about these spaces.  There should be lots of other functional analysis texts that discuss them in more detail.
A: A beautiful treatment of vector valued $L^p$ spaces is in the book:
J. Diestel, J. J. Uhl, Vector measures. Mathematical Surveys, No. 15. American Mathematical Society, Providence, R.I., 1977.
A: With the definitions in the OP, this is false.  It is OK if the Banach space $B$ is separable and $(\Omega,\mathcal F, P)$ is an arbitrary probability space.  It is OK if the Banach space $B$ is arbitrary and $(\Omega,\mathcal F,P)$ is a perfect measure space.  But for arbitrary $B$ and $(\Omega,\mathcal F, P)$, it can fail.  It can fail in many different ways.  
(A theorem of Charles Stegall: if $(\Omega,\mathcal F,P)$ is a perfect probability space, $B$ is a metric space, and $f : \Omega \to B$ is $(\mathcal F, \mathcal B)$-measurable, then there is a set $\Omega_1 \subseteq \Omega$ of measure $1$ such that $f(\Omega_1)$ is separable.)
Here is the simplest way in which it may fail.  Write $\mathcal B = \mathrm{Borel}(B)$. Let $L^p(\Omega,B)$ be the set of all functions $f : \Omega \to B$ such that $f$ is $(\mathcal F, \mathcal B)$-measurable, and
$$
\int_\Omega \|f(\omega)\|^p\;dP(\omega) < \infty .
$$
It is possible that there are $f,g \in L^p(\Omega,B)$ such that $f+g \notin L^p(\Omega,B)$ because $f+g$ is not even $(\mathcal F , \mathcal B)$-measurable.
Example I
Let $T$ be a discrete space with cardinal $\frak{a} > 2^{\aleph_0}$.  Let $B = l^2(T)$, that is, a Hilbert space with orthonormal basis of cardinal $\frak{a}$.  For each $t \in T$ let $e_t \in l^2(T)$ be defined by: $e_t(t) = 1$ and $e_t(s) = 0$ if $t\ne s$.  This system of "unit vectors" is an orthonormal basis of the space $B$.
Let $\Omega = T \times T$ be the Cartesian square.  Let $\mathrm{Borel}(T)$ be the Borel sigma-algebra on $T$, which is of course the power set of $T$.
Let the sigma-algebra $\mathcal{F} = \mathrm{Borel}(T) \otimes \mathrm{Borel}(T)$, the product sigma-algebra.  The reason for requiring that $\mathrm{card}(T) > 2^{\aleph_0}$ is so that the diagonal
$$
\Delta := \{(t,t) \in \Omega : t \in T\},
$$
although closed, is not in $\mathcal F$.  See HERE.  
We do not care what the probability measure $P$ is.  (In an extreme case it could even be the point mass at a single point.)
Finally we are ready.  Define $f : \Omega \to B$ by
$$
f\big((u,v)\big) = e_u,
$$
That is: Given $\omega = (u,v)$ in $\Omega$, we take its first component, and use the corresponding unit vector.  Similarly, define $g : \Omega \to B$ by
$$
g\big((u,v)\big) = -e_v,
$$
using the second component and a minus sign.  
I claim that $f, g \in L^p(\Omega,B)$ but $f+g$ is not.  
First: $f$ is $(\mathcal F, \mathcal B )$-measurable.  Indeed, if
$Q \in B$ is Borel, then $f^{-1}(Q) \in \mathcal F$ because
$f^{-1}(Q) =  \widetilde{Q} \times T \in \mathcal F$ where
$\widetilde{Q} = \{t \in T : e_t \in Q\}$.
So $f$ is $(\mathcal F, \mathcal B )$-measurable.  Similarly
$g$ is $(\mathcal F, \mathcal B )$-measurable.  
Next,
$$
\int_\Omega \|f(\omega)\|^p\,dP(\omega) = 1 < \infty.
$$
(Regardless of what the probability measure $P$ is, the integral of the constant $1$ is $1$.)
So $f \in L_p(\Omega,B)$.  Similarly, $g \in L_p(\Omega,B)$.  
Now we claim the sum $f+g$ is not measurable.  Indeed, even more, we claim that $\{\omega\in \Omega : f(\omega)+g(\omega) = 0\} \notin\mathcal F$.  (Since $\{0\}$ is closed, this shows $f+g$ is not measurable.)  Indeed,
$$
\{\omega : f(\omega) + g(\omega) = 0\} = 
\{(u,v) : e_u-e_v = 0\} =
\{(u,v) : u=v\} = \Delta.
$$
As noted above, $\Delta \notin \mathcal F$
End of Example I 
