Confusion caused by the name $L^\infty$ The notation $L^\infty(\mu)$(the linear normed space made up of essentially bounded measurable functions) in Functional Analysis gives people the impression that the space  $L^\infty(\mu)$ is obtained by finding the "limit" of the sequence $L^1,L^2,...$, however I cannot explain how this works.
Let $(\Omega,\Sigma,\mu)$ be a measure space.  A function $f$ is defined to be in $L^\infty(\mu)$ if $f$ is almost everywhere equal to a bounded function. Therefore, with the norm
$$
\|f\|_\infty=\inf_{E\subset\Omega, \mu(E)=0}\sup_{t\in \Omega-E} |f(t)|
$$
the space $L^\infty(\mu)$ is a Linear normed space.
I wish to see whether or not $\lim_{p \to \infty} L^p=L^\infty$. I begin by a special case, $\ell^p$ spaces.
It is clear that $\ell^1 \subset \ell^p \subset \ell^q \subset \ell^\infty,q>p$. Let's consider the sequence $\ell^1,\ell^2,\ell^3,...$, so  $\lim \ell^p=\bigcup_{i=0}^\infty \ell^i$, which is a linear space. However, $\bigcup_{i=0}^\infty \ell^i$ doesn't include some elements in $\ell^\infty$. For example, the sequence $(1,1,1,...)$ lies in $\ell^\infty$, but not in $\bigcup_{i=0}^\infty \ell^i$, because $(1,1,1,...)$ doesn't belong to any $\ell^p, p<\infty$.
So How can I explain why the notation $L^\infty$ or $\ell^\infty$ are used?
 A: If $\Omega$ is any measure space, and $f\in L^p(\Omega)\cap L^{\infty}(\Omega)$ for some $p\in [1,\infty)$, then $f\in L^q(\Omega)$ for all $q\in [p,\infty]$ and
$$\lim_{q\to \infty}\|f\|_{L^q(\Omega)}=\|f\|_{L^{\infty}(\Omega)} $$
So the limit of the $L^q$ norms is the $L^{\infty}$ norms. In this sense, the space $L^{\infty}$ is the limit of the $L^q$ spaces as $q\to \infty$.
To prove this result one might start from the Riesz-Thorin interpolation inequality:

If $\Omega$ is a measure space and $f\in L^p(\Omega)\cap L^q(\Omega)$, with $1\leq p<q\leq\infty$, then $f\in L^r(\Omega)$ for all $r$ with $p\leq r\leq q$ and 
  $$ \|f\|_{L^r(\Omega)}\leq \|f\|_{L^p(\Omega)}^{\lambda}\|f\|_{L^q(\Omega)}^{1-\lambda}$$
  Where $\lambda\in [0,1]$ is such that 
  $$\frac{1}{r}=\frac{\lambda}{p}+\frac{1-\lambda}{q} $$

A: Lorenzo Quarisa's answer contains the essential point: Think of the limit of the norms, not of the spaces. Let me write out how this works in the simple example of the $\ell^p$ spaces, and let me begin with an even simpler (but incomplete) vector space included in all of the $\ell^p$ spaces, namely the space $F$ of sequences with only finitely many non-zero terms. 
Consider a sequence $x\in F$, and let $n$ be the number of its non-zero components. Also, let $m$ be the largest of the absolute values of the terms $x_i$ in $x$. Then the $p$-norm is bounded above and below in terms of $m$:
$$
m\leq \Vert x\Vert=\left(\sum_i|x_i|^p\right)^{1/p}\leq n^{1/p}m.
$$
(The first inequality is because $m^p$ is one of the summands; the second is because there are only $n$ summands, each majorized by $m^p$.)  If we let $p\to\infty$ while keeping $x$ and therefore $n$ fixed, we have $n^{1/p}\to1$ and therefore $\Vert x\Vert\to m$.  That is, the $p$-norm approaches the $\infty$-norm, for each fixed $x$.
It is in this sense that the $\infty$-norm is the limit of the $p$-norms. EDIT: The next sentence is wrong in the case of $\ell^\infty$; see correction at the end.  As for the $\ell^p$ spaces, notice that they are the completions of $F$ with respect to the $p$-norms. So the limiting behavior of the norms is ascribed (guilt by association?) to the spaces.
CORRECTION: The closure of $F$ in the $\infty$-norm is not $\ell^\infty$ but the subspace $c_0$ of sequences that converge to zero. So to get this approach to work, one would need to double-dualize, which has no effect on $\ell^p$ for $1<p<\infty$ but sends $c_0$ to $\ell^\infty$.
