On the density of spaces in $L^p$

Question

This is a long question. You would really give me a hand by explaining the details and possibly giving me bibliographic references.

Let $\Omega\subseteq \mathbb{R}^n$ be an open set. We denote with $C(\Omega)$ the space of continuous functions on $\Omega$ and with $C_0(\Omega)$ the space of continuous functions with compact support. Define $$\rVert\cdot\lVert_1\colon C_0(\Omega)\to \mathbb{R}_+,\quad \rVert f \lVert_1 =\int_\Omega |f|\;d\lambda_n$$ where $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n.$ The space $\left(C_0(\Omega), \rVert \cdot\lVert_1\right)$ is not complete.

We know that $C_0(\Omega)$ is dense in $L^p(\Omega)$ for $p\in [1,\infty)$

Question 1 Why is it sufficient to show that $C_0(\Omega)$ is dense in $L^p(\Omega)$ to conclude that $L^p(\Omega)$ is the $p$-norm completion of $C_0(\Omega)$?

We denote now with $C(\overline{\Omega})$ the vector space of bounded and uniformly continuous functions on $\Omega$. We define $$\lVert \cdot\rVert_\infty\colon C(\overline{\Omega})\to \mathbb{R}_+\quad \lVert f \rVert=\sup |f(x)|.$$ The space $\left(C(\overline{\Omega}), \lVert \cdot\rVert_\infty)\right)$ is a Banach space.

Question 2. Why is $C_0(\Omega)$ not dense in $L^\infty(\Omega)$? The justification that is provided by my text for this fact is the following: $C_0(\Omega)\subseteq C(\overline{\Omega})$ and $C(\overline{\Omega})$ is a closed subspace of $L^\infty(\Omega)$. Why can we conclude from this that $C_0(\Omega)$ not dense in $L^\infty(\Omega)$?

Question 3. Why $C(\overline{\Omega})$ is a closed subspace of $L^\infty(\Omega)$?

We denote now with $\mathcal{R}(I)$ the space of Riemann integrable function on $I=[a,b]$. We note that $$C(I)\subseteq \mathcal{R}(I)\subseteq L^p(\Omega)\quad\forall p\in [1,\infty]$$

Question 4. Why the second inclusion hold?

Question 5 Why $\mathcal{R}(I)$ is dense in $L^p(\Omega)\quad \forall p\in [1,\infty)$?

Question 6 Why $\mathcal{R}(I)$ is not dense in $L^\infty(\Omega)$? Again the reasoning is as follows: $\mathcal{R}(I)$ is a closed subspace of $L^\infty(\Omega)$. Why this is true? And why from this can we conclude that $\mathcal{R}(I)$ is not dense in $L^\infty(\Omega)$?

Thanks in advance!

Answer 1

1: It's technically not. The other context-specific thing you need to show is that $L^p$ itself is complete. The only other thing to be shown is the more abstract result that there is only one complete metric space (up to isometry) containing a given metric space.

2 and 3: The uniform limit of a sequence of continuous functions, if it exists, is continuous. But there are $L^\infty$ functions which cannot be modified on a null set to make a continuous function. The rest of it is more or less just using the ideas of #1: if $A \subseteq C \subsetneq X$ and $C$ is closed then the closure of $A$ in $X$ must be contained in $C$, thus it cannot be all of $X$.

4: Riemann integrable functions are bounded, on a finite measure space $L^p \subseteq L^q$ whenever $p>q$.

5: $\mathcal{R}(I)$ contains all the step functions...showing that the completion of the step functions in the $L^p$ norm is $L^p$ takes a bunch of work but has been written down many times.

6: The reason that this result is sufficient to show that $\mathcal{R}(I)$ is not dense in $L^\infty(I) $ is the same as it was for show that $C_0$ is not dense in $L^\infty$. To get this result you need to show that the uniform limit of Riemann integrable functions is Riemann integrable. See https://math.stackexchange.com/a/3332696/83396 for that. (This was just the first MSE answer I found about this. The question in this link wasn't actually about this.)