This is motivated from an exercise in real analysis:
Prove that $C([0,1])$ is not dense in $L^\infty([0,1])$.
My first question is how $C([0,1])$ is identified as a subset of $L^\infty([0,1])$? (I think one would never say something like "$A$ is (not) dense in $B$" if $A$ is not even a subset of $B$. )
First of all, $L^\infty([0,1])$ is defined as a quotient space, but $C([0,1])$ is a set of functions:
$$
C([0,1]):=\{f:[0,1]\to{\Bbb R}\mid f \ \text{is continuous}\}. \tag{1}
$$
I think one should also take $C([0,1])$ as
$$
C([0,1]):=\{f:[0,1]\to{\Bbb R}|f\sim g \ \text{for some g where g is continuous on}\ [0,1]\} \tag{2}
$$
where $f\sim g$ if only if $f=g$ almost everywhere. But I've never read any textbook (PDE, measure theory, or functional analysis, etc) that defines $C([0,1])$ (or more generally $C(X)$ where $X\subset{\Bbb R}$ is compact) in this way before.
Second question: Could anyone come up with a reference with such definition?
[EDITED:]The original title doesn't reflect my point. I've changed it accordingly.
[EDITED:] Some thoughts after reading the comments and answers:
When one regards $C([0,1])$ as a subset of $L^\infty([0,1])$, (1) is not correct, and (2) would be not correct either. The final version I can come up with is
$$
C([0,1]):=\{f:[0,1]\to{\Bbb R}|f\sim g \ \text{for some g where g is continuous on}\ [0,1]\}\big/\sim.
\tag{3}
$$
Best Answer
You can actually identify $C([0,1])$ and $C([0,1])/\sim$ because, two continuous fonctions who agree almost everywhere are equal.
Indeed, let $f,g \in C([0,1])$ be such that $A = \{x\in [0,1]\mid f(x) \neq g(x)\}$ is negligible. Then $A$ must have an empty interior, so its complementary is dense in $[0,1]$. The function $h = f-g$ is continuous, hence $h([0,1]) = h(\overline{[0,1]\setminus A}) \subset \overline{h([0,1]\setminus A)} = \{0\}$. This proves that $f=g$.
If you want to be really rigorous, it would be better to say that the natural injection $C([0,1]) \hookrightarrow \mathcal{L}^\infty([0,1])$ factorizes with $\sim$, so that it induces an injection $C([0,1]) \hookrightarrow L^\infty([0,1])$. That way, $C([0,1])$ is identified with the image of this injection.