What does it mean to say $f\in L^{\infty}(\mathbb{R}^N)$? I'm trying to understand $L^p$ spaces. I know that a function is $L^1(\Omega)$ if $\int_{\Omega}|f|<\infty$ and $L^2(\Omega)$ if $\int_{\Omega}|f|^2>\infty$. However, what does it mean to say  $f\in L^{\infty}(\mathbb{R}^N)$?. 
According to wikipedia, it is the space of essentially bounded functions. It means that a function of the set $L^{\infty}$ (of $\Omega$?) looks like this:
$$||f||_{\infty} = \inf\{C\ge 0: |f(x)|<C\mbox{ for almos every $x$}\}$$
I don't see this as a good way to understand what are $L^{\infty}$ funcions (and in which set?). Could someone provide me a better understanding of wht is happening?
 A: For $f :\Omega \to \mathbb R $ measurable, you can define its essential supremum by
$$ E(f) := \inf_{\substack{U:   \mu(U) = 0}} \sup_{x\in \Omega \setminus U} |f(x)|$$ 
Suppose that $f=0$ a.e. Let the null set where $f\neq 0$ be $U_0$. Then
$$ 0 \le E(f) \le \sup_{x\in \Omega\setminus U_0} |f(x)| = 0$$
so $E(f) = 0$. 
Conversely, if $E(f)=0$, Then for any $\epsilon>0$ there is a null set $U$ such that $$\sup_{x\in\Omega\setminus U}|f(x)| < \epsilon  $$
Thus, $|f|<\epsilon$ a.e. for every $\epsilon>0$, and
$$\mu(|f|>0) = \mu \left(\bigcup_{n=1}^\infty \{ |f|>1/n \}\right) \le \sum_{n=1}^\infty \mu(|f|>1/n) = \sum_{n=1}^\infty 0 = 0$$
so $f=0$ a.e. This proves that $E(f)=0 \iff f=0$ a.e.
Now, if $f=g$ a.e. then $E(f-g)=0$, and so taking sups and then infs of $|f(x) \le   |g(x)|+|f(x) -g(x)|$ gives
$$ E(f) \le E(g) + E(f-g) = E(g)$$
and similarly $E(g)\le E(f)$, so $E(f) = E(g)$.
Hence, if you take the set
$$ \mathcal L^\infty(\Omega):= \{ f:\Omega \to \mathbb R \text{ measurable} : E(f) <\infty \}$$
and quotient out by $f\sim g \iff f=g$ a.e., $E([f]):=E(f)$ is well defined on the equivalence classes $[f]\in L^\infty(\Omega):=\mathcal L^\infty(\Omega)/\sim  $. One can then go through the standard arguments to check that $E$ is a norm. The above takes care of the hardest step, namely  $E([f])=0$ iff $E(f) = 0$, iff $f=0$ a.e., iff $[f]=[0]$.
To match with the definition in Wikipedia, note that for each $\epsilon$, there is a null set $U$ such that
$|f|<E(f)+\epsilon$ outside $U$, so that $\|f\|_\infty \le E(f)$ (with $\|f\|_{\infty}$ as defined in Wikipedia). Conversely, if $|f(x)|<C$ a.e., Then $E(f) \le C$. Taking infimums over allowable $C$s gives $E(f) \le \|f\|_{\infty}$, so $E(f) = \|f\|_{\infty}$.
A: $f$ in $\mathcal{L}^{\infty}(\mathbb{R}^n)$ if and only if there exists a constant $C \geq 0$ such that $|f(x)| \leq C$ almost everywhere. Then we say that $f \sim g$ if and only if $f = g$ almost everywhere, and define (as for the $L^p$ spaces) $L^{\infty} = \mathcal{L}^{\infty} / \sim$.$
The "best constant" $C$, i.e.,
$$\inf \{C : |f(x)| \leq C \text{ almost everywhere}\}$$ is defined to be the $L^{\infty}$-norm of $f$.
