Is $C^{\infty}(M) \subseteq L^2(M, \text{loc})$?

Question

Let $M$ be a Riemannian manifold. Is it true that every smooth function on $M$ is also in $L^2(M, \text{loc})$? If so, could you give me some hint as to how to prove it or suggest a reference where I could find a proof? I tried to show that for any coordinate patch $U$ and any smooth $f$ we have $f|_U \in L^2(U)$, but I couldn't justify this.

Answer 1

$C(M)$ functions are locally bounded. Remember that manifolds are locally compact, and compact sets have finite measure.

What you tried to show is not true: you could take $M={\bf R}$ and $f=(x\mapsto x^2)\in C^\infty(M)$. $M$ is a single coordinate patch and $f=f\vert_M$ is not integrable in any way.