$f \in L^p(]0,1[)$ almost everywhere.

Question

Why are $L^p$ functions defined almost everywhere? As if we are defining an $L^p$ function on a set $\Omega=[0,1]$ , so we say $f \in L^p(]0,1[)$ almost everywhere. And not on $[0,1]$ !

Answer 1

That's because of the way we construct the $L^p$ spaces. First you look at the vector space (let $\Omega \subseteq \mathbb{R}$) $$ \mathcal{L}^p = \left\{ f\colon \Omega \to \mathbb{R} : f \text{ measurable and} \int_\Omega |f(x)|^p dx < \infty\right\} $$ and the mapping $$ q\colon \mathcal{L^p} \to [0,\infty[,\ f\mapsto \Bigg(\int_\Omega |f(x)|^p dx\Bigg)^{\frac{1}{p}} $$ and notice that it defines a seminorm (use Minkowski's inequality). Next you look want to "convert" this into a norm. When you have a seminorm you can look at the set $$ N = \left\{f \in \mathcal{L}^p | q(f)=0\right\} = \left\{f \in \mathcal{L}^p : \int_\Omega |f(x)| dx = 0\right\}. $$ Now you define $L^p := \mathcal{L}^p / N$ as the quotient vector space.

What this essentially does is the following: We have $$ f=g \text{ a.e.} \Leftrightarrow \int_\Omega |(f-g)(x)| dx= 0\\ \Leftrightarrow q(f-g)= \Bigg(\int_\Omega |(f-g)(x)|^p dx\Bigg)^{\frac{1}{p}} = 0. $$

so every time $f=g$ almost everywhere, they land in the same equivalence class. This means, while they may be different functions (they can differ on a null set), they are "the same thing" in $L^p$. When we integrate, we can't tell the difference.

For example you could look at $L^p([0,1])$ and the functions $$ f\colon [0,1]\to \mathbb{R},\ f(x) = \begin{cases}0,\ \text{if } x\in \mathbb{Q},\\ 1,\ \text{otherwise.}\end{cases}\\ g\colon [0,1]\to \mathbb{R},\ g(x) = 1. $$ Now $g$ and $f$ only differ on a set of measure null (they are equal almost everywhere). But this means they lie in the same equivalence class. Therefore they are the same element in $L^p$. So although we might have different functions $f$ and $g$, if we can't tell them apart by our integral ($q(f)=q(g)$), we say they are the same element in $L^p$. That's why it's okay to define them almost everywhere. We can always set them to $0$ or basically anything we want on the null set and get a function that represents the same thing and is fully defined.

In your case $\{0,1\}$ is null set, therefore it doesn't matter if we talk about $[0,1]$ or $[0,1]\setminus \{0,1\} = ]0,1[$ and define the function only on the smaller set.

[I'm working with the Lebesgue measure in this post]

edit regarding your comment: Well, we know that for $x\in [0,\infty[$ and $r\in ]0,\infty[$ $$ x^r = 0 \Leftrightarrow x=0, $$ therefore $$ q(f) = 0 \Leftrightarrow \int_\Omega |f|^p dx = 0. $$ By the same argument we know $|f(x)|=0 \Leftrightarrow |f(x)|^p = 0\ (\forall x\in\Omega)$ and thus $$ \int_\Omega |f|^p dx = 0 \Leftrightarrow \int_\Omega |f| dx = 0. $$