Why is $L^2$ not a RKHS? Let's take $L^2[0,1]$ as an example. Obviously $L^2[0,1]$ is not a Hilbert space because you can have a function $f(0)=1$ but equal to $0$ everywhere else, so you don't have an unique element with norm $0$, violating the definition of the inner product. So it only makes sense to talk about $L^2[0,1]$ as a Hilbert space in the sense of equivalence classes of functions that differ on at most a (Lebesgue) measure $0$ set. So why is  $L^2[0,1]$ not a RKHS, when talking in terms of equivalence classes of functions? Is it because there is no unique representative for each equivalence class so the evaluation functional $\delta_x:f \rightarrow f(x)$ is undefined?
If so, is there a way to choose $f$ for each equivalence class of functions in $L^2[0,1]$ such that the evaluation functional becomes well defined and continuous, and thus making $L^2[0,1]$ a RKHS?
 A: As you said, $L^2([0, 1])$ is NOT a function space - and it can't be made one easily (or even at all). So the notion of $\delta_x$ does not make sense at all.
You can impose weak differentiability, i.e. go to $H^1([0, 1]) \subseteq L^2([0, 1])$. Due to integral representation, we know that every $f \in H^1([0, 1])$ has a continuous representative and there is even $C>0$ such that
$$
\lVert f \rVert_{C([0, 1])} \leq C\lVert f \rVert_{H^1([0, 1])}
$$
for all $f \in H^1([0, 1])$. So evaluation makes sense if we define it for the continuous representative. Now, for $x \in [0, 1]$ and $H^1([0, 1])$:
$$
\lvert f(x) \rvert \leq \lVert f \rVert_{C([0, 1])} \leq C \lVert f \rVert_{H^1([0, 1])}
$$
So this is a RKHS.
But we also know that an embedding $C([0, 1]) \hookrightarrow L^2([0, 1])$ does not exist - this makes it hard to figure out a representative that has some regularity. There might be such thing but usually $L^2([0, 1])$ equivalence classes can contain highly irregular functions like $\chi_{\mathbb{Q}}$.
A: We can actually prove mathematically, that there is no possibility to choose representatives that make the evaluation functionals well-defined and continuous.
Suppose there is a way to choose a function for each equivalence class such
that the evaluation functionals $\delta_x$ become well defined and continuous.
For each number $r\in[0,1]$, consider the function
$$
g_r(x)
= \begin{cases} 0 & :\; x < r \\ 1&:\; x \ge r.  \end{cases}
$$
Let us denote by $f_r$ the chosen representative of $g_r$.
For each $r$, let $A_r\subset [0,1]$ be a set of measure zero
such that $f_r=g_r$ outside of $A_r$.
We define
$$
A_{\Bbb Q} := \bigcup_{q\in\Bbb Q\cap [0,1]} A_q.
$$
The set $A_{\Bbb Q}$ is still a set of measure zero (due to the countability of the rationals),
and for all rational $q\in[0,1]$ we have $f_q=g_q$ outside of $A_{\Bbb Q}$.
Let $x\in(0,1)$ be a point outside of $A_{\Bbb Q}$ (such a point must exist
because $A_{\Bbb Q}$ has measure zero).
Let us also choose two rational monotone approximations of $x$,
$q_n\nearrow x$ from below and $p_n\searrow x$ from above.
Note that we have $f_{q_n}(x) = g_{q_n}(x)=1$
and $f_{p_n}(x)=g_{p_n}(x)=0$ for all $n\in \Bbb N$.
One can also easily show that $f_{p_n} \to f_x$
and $f_{q_n}\to f_x$ in the $L^2[0,1]$ norm.
If $\delta_x$ was continuous, this would imply
$$
0 = f_{p_n}(x) = \delta_x(f_{p_n}) \to \delta_x(f_x) = f_x(x)
$$
and
$$
1 = f_{q_n}(x) = \delta_x(f_{q_n}) \to \delta_x(f_x) = f_x(x).
$$
This, however, is a contradiction.
some remarks/intuitions/ideas:
As mentioned in the comment below, picking a fixed point $x$ like $x=1/2$
a-priori and then showing that $\delta_x$ is discontinuous cannot work,
because the representatives can be chosen such that they all have the same value at $x$.
Because you do not know a-priori, which point $x$ to look at,
we need to study a larger class of functions.
This class should have two properties: First, it should be only countable,
(to allow for the trick with the countable union of sets of measure zero).
Second, there should be enough functions in this class so that we can
potentially "create problems" at many points in $[0,1]$.
The class of functions of $g_q$ for rational $q$ satisfies these properties.
