Verify the completion of continuous compactly supported functions

[psie]

TL;DR

Let $\Omega\subset\mathbb R^n$. Consider the space $C_c(\Omega)$ of continuous, compactly supported functions equipped with the $L^p$ norm. Then in my lecture notes it is claimed the completion of this space is $L^p(\Omega)$. I'm trying to verify to myself that this is indeed the case by the definition given below.

Consider the attached definition of completion of a metric space.

It has already been stated in my notes that $L^p(\Omega)$ equipped with $\lVert\cdot\rVert_p$ is a Banach space, hence complete. So (c) is satisfied. Also, there is a theorem that states that $C_c(\Omega)$ is a dense subset of $L^p(\Omega)$. But I'm still having trouble verifying (a) and (b) in the definition. How could I construct a function between these metric spaces such that it is an isometry and its image is dense? What troubles me is that $L^p(\Omega)$ is a set of equivalence classes of functions, and hence the inclusion map $i:C_c(\Omega)\to L^p(\Omega)$ defined by $f\mapsto f$ would maybe not work.

 

[pasmith]

You can show by the triangle inequality that f_1 \sim f_2 \mbox{ and } g_1 \sim g_2 \quad\Rightarrow\quad \|f_1 - g_1\| \leq \|f_2 - g_2 \| \mbox{ and } \|f_2 - g_2\| \leq \|f_1 - g_1 \| and therefore \|f_1 - g_1\| = \|f_2 - g_2\| so that f \mapsto [f] is an isometry. (Use <br /> \|a - b\| = \|a - c + c - d + d - b\| \leq \|a - c\| + \|c - d\| + \|d - b\|.)

EDIT: If we don't have this result, we can't define L_p as a space of equivalence classes; the distance between two classes would depend on which representatives we choose.

 

[psie]

You can show by the triangle inequality that f_1 \sim f_2 \mbox{ and } g_1 \sim g_2 \quad\Rightarrow\quad \|f_1 - g_1\| \leq \|f_2 - g_2 \| \mbox{ and } \|f_2 - g_2\| \leq \|f_1 - g_1 \| and therefore \|f_1 - g_1\| = \|f_2 - g_2\| so that f \mapsto [f] is an isometry. (Use <br /> \|a - b\| = \|a - c + c - d + d - b\| \leq \|a - c\| + \|c - d\| + \|d - b\|.)

EDIT: If we don't have this result, we can't define L_p as a space of equivalence classes; the distance between two classes would depend on which representatives we choose.

This makes sense, but just to be certain. When you consider the function $f\mapsto [f]$, by $[f]$ you mean a representative function of the equivalence class, right?

The reason I'm asking is because the definition of isometry requires that it is a function $i:X\to Y$ between two metric spaces $X,Y$ which satisfies $$d_Y(i(x_1),i(x_2))=d_X(x_1,x_2)$$ for all $x_1,x_2\in X$. In this case, $d_Y=d_X=\lVert\cdot\rVert_p$.

 

[psie]

You can show by the triangle inequality that f_1 \sim f_2 \mbox{ and } g_1 \sim g_2 \quad\Rightarrow\quad \|f_1 - g_1\| \leq \|f_2 - g_2 \| \mbox{ and } \|f_2 - g_2\| \leq \|f_1 - g_1 \| and therefore \|f_1 - g_1\| = \|f_2 - g_2\| so that f \mapsto [f] is an isometry. (Use <br /> \|a - b\| = \|a - c + c - d + d - b\| \leq \|a - c\| + \|c - d\| + \|d - b\|.)

EDIT: If we don't have this result, we can't define L_p as a space of equivalence classes; the distance between two classes would depend on which representatives we choose.

I think I understand it now. We can define $\lVert [f]\rVert_{L^p}=\lVert f\rVert_{L^p}$ for some representative $f$ of $[f]$. This is well-defined, since for any other representative $g\in [f]$, we have $f=g$ a.e., which implies $\lVert f\rVert_{L^p}=\lVert g\rVert_{L^p}$.

 

[pasmith]

By [f] I mean the equivalence class of f.

In this case the isometry is obvious, since the distance between two equivalence classes does not depend on which representatives of each class one uses to measure it; it is by definition the case that \|[f] - [g]\|_p = \|f - g\|_p.