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Let $M,N$ be smooth compact Riemannian manifolds. If I am not mistaken the definition of the $L^p$-space $L^p(M;N)$ is the following:

Fix a smooth (isometric?) embedding $i:N \to \mathbb{R}^D$.

$$ L^p(M;N) = \{f:M \to N \, \text{ is measurable* }| \, \, i \circ f \in L^p(M;\mathbb{R}^D) \},$$

We say that $f_n \to f$ in $L^p(M;N)$ if $\| i\circ f-i \circ f_n\|_{L^p(M;\mathbb{R}^D)} \to 0$.

Question: Why $L^p(M;N)$ is independent of the embedding $i$?

It is easy to see that the embedding does not affect the set $L^p(M;N)$:

Indeed, for any embedding $i:N \to \mathbb{R}^d$, every measurable map $M \to N$ is in $L^p(M;N)$. (compactness implies $i(N)$ is bounded and $\operatorname{Vol}(M) < \infty$).

The question is about the convergence:

If $i,j:N \to \mathbb{R}^d$ are embeddings, why

$$ \| i\circ f-i \circ f_n\|_{L^p(M;\mathbb{R}^D)} \to 0 \Rightarrow $$ $$ \| j\circ f-j \circ f_n\|_{L^p(M;\mathbb{R}^D)} \to 0$$

*By saying that $f:M \to N$ is measurable I mean that for any open subset $O \subseteq N,f^{-1}(O)$ is "Lebesgue-measurable" in $M$ (i.e belongs to the completion of the Borel sigma-algebra of $M$, w.r.t to the Riemannian density of $M$).

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    Note that (unlike in Sobolev spaces between manifolds) there is no need here to embed $N$, since $L^p(X;Y)$ is a well-defined metric space for any measure space $X$ and metric space $Y$. In particular, the independence you ask for is a corollary of the equivalence of extrinsic and intrinsic metrics in an embedded compact Riemannian manifold. – 2017-01-28

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Since $N$ is compact and $i$ is an embedding, the extrinsic distance $|i(x)-i(y)|$ is uniformly comparable to the intrinsic distance $d_N(x,y)$; that is, there is a constant $C_i$ independent of $x,y$ such that

$$ C_i^{-1} d_N(x,y) \le |i(x)-i(y)| \le C_i d_N(x,y)$$

for all $x,y\in N$. (For a proof of this fact see e.g. this question.)

Thus we have $$\Vert i \circ f- i \circ f_n\Vert_p \le C_i \Vert d_N(f,f_n)\Vert_p \le C_i C_j \Vert j\circ f - j \circ f_n \Vert_p$$ and vice versa, so the metrics are equivalent.

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    Thanks! I was actually aware of that other answer, but somehow it did not occur to me to use it... (I was trying to prove this with "bare-hands" and got nowhere). This gives me further motivation to study it (I must admit it was not entirely clear when I tried to understand it...). – 2017-01-16