# Is $C([0,1])$ a “subset” of $L^\infty([0,1])$?

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.

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 $$C([0,1]):=\{f:[0,1]\to{\Bbb R}|f\sim g \ \text{for some g where g is continuous on}\ [0,1]\}\big/\sim. \tag{3}$$

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Perhaps a topological, simple explanation would be that there exists at least one open non-empty subset $\,S\subset L^\infty[0,1]$ not containing any continuous function. – DonAntonio Mar 29 '13 at 15:13
each continuous function is its only representative, so there's a natural injection – suissidle Mar 29 '13 at 15:16
Verify that the map $C([0,1]) \to L^\infty([0,1])$ is isometric. In particular, the image of $C([0,1])$ in $L^\infty([0,1])$ is closed. Of course, the map is not onto. – Martin Mar 29 '13 at 15:18
The map sending a continuous function to its equivalence class a.e. in $L^\infty$. – Martin Mar 29 '13 at 15:26
Let me try again: Definition (1) is the correct one. There is an obvious bounded linear map $T\colon C[0,1] \to L^\infty[0,1]$ given by $Tf = [f]$, where $[f]$ is the equivalence class of $f$ modulo $\sim$. The image of $T$ is (3). The answers explain that $T$ is an isometric isomorphism from (1) onto (3), so you can actually identify them via $T$. – Martin Mar 30 '13 at 17:59

Actually, such identifications are done in the textbook about partial differential equations and Sobolev spaces. For example, we can see theorems like "$C_0^\infty(\Bbb R^d)$, the set of smooth functions with compact support, is dense in $W^{1,p}(\Bbb R^d)$ for all $1\leqslant p<\infty$". This means that we identify a test function $\phi$ to the class of functions which are almost everywhere equal to $\phi$, as what is done in the OP.

The characteristic function of $g=[1/2,1)$ cannot be approached in $L^\infty$ by such functions, because there would be an $f$ almost everywhere equal to a continuous function such that $\sup_{x\in [0,1]\setminus N}|g(x)-f(x)|<1/3$, where $N$ is a set of $0$ measure. In particular, $\sup_{x\in [0,1/2)\setminus N}|f(x)|<1/3$ and $\sup_{x\in [1/2,1)\setminus N}|1-f(x)|<1/3$.

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As I understand from your answer, one should have the indicator function $1_{{\Bbb Q}\cap[0,1]}\in C([0,1])$, though $1_{{\Bbb Q}\cap[0,1]}$ is obviously not continuous on $[0,1]$, right? – Jack Mar 29 '13 at 15:30
No, the characteristic function of $\Bbb Q$ is in the class of the null function, so it's not the problem. In the example I gave, the problem comes from a jump. – Davide Giraudo Mar 29 '13 at 15:34
Oh, what my question in the comment mean is $1_{{\Bbb Q}\cap[0,1]}=0$ almost everywhere and since $0\in C([0,1])$, one should identify $1_{{\Bbb Q}\cap[0,1]}$ as an element in $C([0,1])$. – Jack Mar 29 '13 at 15:39
Regarding your last claim: The indicator function of $[1/2,1)$ (on the interval $[0,1]$) is continuous a.e. The set of points where it's discontinuous is just $\{\tfrac12,1\}$. I think you meant to write that it is not a.e. equal to a continuous function. – kahen Mar 29 '13 at 15:54
@kahen Thanks, fixed now (I think). – Davide Giraudo Mar 29 '13 at 17:01

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.

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I think $C([0,1])/\sim$ is just $C([0,1])$ since for continuous function, $f=g$ if and only if $f=g$ almost everywhere. So the quotient space is not need. I edited accordingly in my original post. – Jack Mar 29 '13 at 16:04
You have functions in $C([0,1])$ but you only have classes of equivalence of function in $C([0,1])/\sim \subset L^\infty$, so you cannot say that they are equal but you can still identify them (with a canonical bijection). – Siméon Mar 29 '13 at 16:12
Thank you for your explanation. In your first sentence, (2) is used for $C([0,1])$, right? – Jack Apr 20 '13 at 13:20

This includes an answer to the original question posted, modified to answer the latest version of the question at time of writing. I have left the original answer because it includes an observation (and an example of how that observation is applied) that illustrates why people are a bit careless about differentiating between the equivalence classes and their representatives.

The norm on $L^\infty[0,1]$ is the essential supremum, so it 'ignores' changes on null sets. By $\|f\|$ below, I mean the essential supremum norm. I use $[f]$ to mean the equivalence class of $f$, the notation is potentially confusing, but context will disambiguate. By $f_1 \sim f_2$, I mean that $\{x | f_1(x) \neq f_2(x) \}$ is a null set.

One identifies $C[0,1]$ with a subset of $L^\infty[0,1]$ by taking equivalence classes, ie, we are really dealing with $\{[f] | f \in C[0,1] \}$, which is a subset of $L^\infty[0,1]$. (As an aside, continuity means that the identification $f \mapsto [f]$ is injective.)

(I use $m$ below to denote the Lebesgue measure, however the observation holds for any measure, of course. The subsequent demonstration of 'not being dense' does depend on the Lebesgue measure.)

Observation: Suppose $P$ is some property on $\mathbb{R}$ (or $\mathbb{C}$ as the case may be), and suppose $f_1 \sim f_2$. Then we have $m \{x | P(f_1(x)) \} = m \{x | P(f_2(x)) \}$. This follows since $f_1(x) = f_2(x)$ a.e. $x$. So even though $[f]$ is an equivalence class, we can think about $m \{x | P(\,[f]\,(x)) \}$ with the understanding that we really mean $m \{x | P(f(x)) \}$ for some representative $f \in [f]$. This is what allows us to be somewhat blasé about dealing with a function vs. its equivalence class.

The previous observation can be extended considerably, but loosely the idea is that the measure of the set of points that satisfies a 'nice' property is independent of the particular representations from the equivalence classes. By a 'nice' property, I mean a property whose truth value at $x$ depends only on the values of the representations at $x$.

Now consider $[1_{[\frac{1}{2},1]}]$, and $[c]$ where $c \in C[0,1]$.

I claim $\|[1_{[\frac{1}{2},1]}]-[c] \| \ge \frac{1}{2}$, and since $c$ was arbitrary, we see that $C[0,1]/ \sim$ cannot be dense in $L^\infty[0,1]$.

To see why the claim is true, we will prove the statement for specific representatives of $[1_{[\frac{1}{2},1]}], [c]$ (ie, $1_{[\frac{1}{2},1]}, c$ respectively) and then invoke the above observation to conclude.

Let $\gamma =c(\frac{1}{2})$. We have $|\gamma-1|+|\gamma| \ge 1$, and hence $\max(|\gamma-1|, |\gamma|) \ge \frac{1}{2}$. Continuity of $c$ implies that for any $\epsilon>0$, there is a $\delta >0$ such that $|c(x)-\gamma| < \epsilon$ whenever $x \in B(\frac{1}{2}, \delta)$. Noting that if $x \in B(\frac{1}{2}, \delta)$, we have $1-x \in B(\frac{1}{2}, \delta)$, we get $\max(|c(1-x)|, |c(x)-1|) \ge \max(|\gamma|-\epsilon, |\gamma-1|-\epsilon) \ge \frac{1}{2}-\epsilon$. Hence for $x\in (\frac{1}{2},\frac{1}{2}+\delta)$, we have $\max(|c(1-x)-1_{[\frac{1}{2},1]}(1-x)|, |c(x)-1_{[\frac{1}{2},1]}(x)|) \ge \frac{1}{2}-\epsilon$.

In particular, $m \{x |\ |c(x)-1_{[\frac{1}{2},1]}(x)| \ge \frac{1}{2}-\epsilon \} \ge \delta >0$. The above observation shows that this is true for all $c'\in [c], f'\in [1_{[\frac{1}{2},1]}]$, so it follows from the definition of essential supremum that $\|[c]-[1_{[\frac{1}{2},1]}]\| \ge \frac{1}{2}$.

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