Verify the completion of continuous compactly supported functions

[psie]

TL;DR Summary

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 $$\|a - b\| = \|a - c + c - d + d - b\| \leq \|a - c\| + \|c - d\| + \|d - b\|.)$$

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 $$\|a - b\| = \|a - c + c - d + d - b\| \leq \|a - c\| + \|c - d\| + \|d - b\|.)$$

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 $$\|a - b\| = \|a - c + c - d + d - b\| \leq \|a - c\| + \|c - d\| + \|d - b\|.)$$

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 $\|[f] - [g]\|_p = \|f - g\|_p$.