Is there a standard $L^2$ norm for multi-valued function $f:\mathbb R^n \to \mathbb R^n$? Equipping $\mathbb R^n$ with the usual product Lebesgue measure,
what is the standard $L^2$ norm for the function $f :\mathbb R^n \to \mathbb R^n$ define by
\begin{align}
f(x) &=\left(f_1(x), f_2(x), \ldots, f_n(x) \right),
\end{align}
where $x =\left(x_1,x_2, \ldots, x_n\right)$
Combining standard Euclidean norm and usual $L^2$ norm, I would say
$$
\| f \| _ 2 = \sqrt{ \sum_{i=1}^n \int \left| f_i(x) \right|^2 dx }. \tag{1}
$$
But I couldn't find any reference confirming my intuition, so I'm wondering if there is another canonical way to define (1)
 A: I noticed this question and respectfully disagree with the accepted answer. I consider Bochner integrals to be an excessively complicated tool for the case at hand. 
I would proceed as follows. Assume that a measure space $\Omega$ and a finite dimensional vector space $V$ with scalar product $\langle, \rangle$, are given. For functions $f\colon \Omega\to V$, the property of measurability (in the sense that the preimages of open sets of $V$ are measurable) is equivalent to the measurability of the components functions of $f(x)$ with respect to one (hence all) of the linear bases of $V$. (Notice that the concept of measurability does not depend on the scalar product). 
In particular, if $f$ and $g$ are measurable, then $\langle f(x), g(x)\rangle$ is a measurable function defined in $\Omega$ and taking values in $\mathbb{K}$ (real or complex field): this can be seen by taking an orthonormal basis $e_1\ldots e_n$ of $V$ and observing that 
$$
\langle f(x), g(x)\rangle=\sum_{j=1}^n \langle f(x), e_j\rangle \langle g(x), e_j\rangle,$$
so $\langle f(x), g(x)\rangle$ is a linear combination of measurable functions. But then, for measurable $f, g$ one only needs the ordinary Lebesgue integral to define the scalar product 
$$
\langle f, g\rangle_{L^2(\Omega; V)}=\int_{\Omega} \langle f(x), g(x)\rangle\, dx$$
and the corresponding norm
$$
\lVert f\rVert_{L^2(\Omega;V)}^2=\langle f, f\rangle_{L^2(\Omega;V)}.$$
Restricting ourselves to measurable functions for which the latter norm is finite we obtain the $L^2(\Omega; V)$ space. 
The topology of this space depends only on the choice of a measure on $\Omega$: indeed, if we had chosen another scalar product on $V$, we would have obtained an equivalent norm on $L^2(\Omega; V)$, because all norms on $V$ are equivalent. In this sense, this construction is canonical.
This is all pretty obvious, I just wanted to stress that NO Bochner integral theory is needed here. That machinery becomes indispensable only when $V$ is taken to be infinite dimensional. 
A: I believe the standard way to do this is to use the Bochner integral. Suppose $V$ is some Banach space and we have a function $[0,1]\rightarrow V$ that we want to integrate. The Lesbegue theory is not applicable here, because it uses the ordering of the range in a crucial way, and $V$ is in general not ordered. Recall we define a Lesbegue integral by taking a certain supremum of simple functions. It is not clear how to extend this to $V$-valued functions, since we can no longer take supremums of such functions (because do not have have access to an order on $V$). 
The Wikipedia page on Bochner integration is probably a good place to start. I don't know any great references offhand. Hopefully someone else can supply you with some. 
