Let $(M, g)$ be a compact manifold with a smooth boundary. There is a compact connected manifold without boundary $(N, g)$ having the same dimension as $M$ so that $(M, g)$ is isometrically embedded in $(N, g) .$
I can prove this theorem for a special case if $M\subset \mathbb R^n$.
Consider some cube $N=$ $[-K, K]^{n}$ with $M \subset N^{\text {int }}$, and extend $g$ smoothly as a $2 K$ -periodic positive definite symmetric matrix function in $N$.
But in general how to prove a given statement?
Any hint/help will be appreciated.
Let $N_1$ be the double of $M$. Since $M$ is compact, so is $N_1$. Since $M$ is compact, it has only finitely many component. Then by forming finitely many connected sums (away from $M \subset N_1$), we assume that $M$ is embedded in a connected compact manifold $N$ without boundary with the same dimension.
Extending the metric $g$ from $M$ to $N$ is standard: for each $x\in M$, since $g$ is smooth, there is some neighborhood $U_x$ of $x$ in $N$ and a smooth metric $g_x$ on $U_x$ so that $g_x(y) = g(y)$ for all $y\in M\cap U_x$. Let $\{\varphi_0\} \cup \{\varphi_x : x\in M\}$ be a partition of unity subordinated to $\{ U_0 = N\setminus M\} \cup \{ U_x: x\in M\}$. Let $g_0$ be any smooth metric on $U_0$. Then
$$ h = \varphi_0 g_0 + \sum_{x\in M }\varphi _x g_x$$ is a smooth metric on $N$ and $h = g$ on $M$.