# $L^p$ spaces between Riemannian Manifolds

I've been working through Jost's Riemannian Geometry and Geometric Analysis and I've been having some trouble understanding the definition of the Sobolev space $H^{1,2}(M,N)$. In particular, I'm not sure how he defines the space $L^2(M,N)$.

Let $M$ and $N$ be Riemannian manifolds. Then the Sobolev space $H^{1,2}(M,N)$ is defined to be the set of functions $f\in L^2(M,N)$ with finite energy, $E_2(f)<+\infty$.

The first mention of the space $L^2(M,N)$ is in $\S 8.3$, but it's provided without definition. Jost refers to convergence in this space $L^2(M,N)$ in several statements of theorems (e.g. Lemma 8.3.2). It would make more sense to me if $N$ was a vector bundle over $M$ since we would have a norm in $N$, and then we could look at integrability of functions $M\to N$ in a way that is similar to Bochner integrability. But this doesn't seem to be the case.

One could alternatively use Nash's isometric embedding theorem to find an isometry $i:N\to\mathbb{R}^k$ for some $k\in\mathbb{N}$ and then say that $f\in L^2(M,N)$ if $f\in L^2(M,i(N))$, but this doesn't seem to be well-defined or canonical in any way (if $M$ is non-compact) since we can just shift $i(N)$ arbitrarily around in $\mathbb{R}^k$.

So my question really is: how are the $L^p(M,N)$ spaces defined for maps between Riemannian manifolds?

A function is in $L^p(M,N)$ if it is measurable (the preimage of every open set in $N$ is measurable in $M$) and the integral $$\int_M d(f(x),Q)^p\,dx$$ is finite for some $Q\in N$. Here $d$ is the intrinsic distance function of the manifold $N$.
Observe that this makes sense for every metric space $N$, the manifold structure is not really relevant.
If $M$ is has finite measure, then "for some $Q\in N$" is equivalent to "for all $Q\in N$", via the triangle inequality.
(To be precise, Korevaar and Schoen require $f$ to be Borel measurable.)
• I was looking at the definition in Korevaar, can't M also be taken to be a complete metric space (for the $L^p$ definition, but clearly not for the Sobolev one). – AIM_BLB May 19 at 9:14