what is the cardinality of set of all smooth functions in $L^1$? What is the cardinality of set of all smooth functions belonging to $L^1$ or $L^2$ ? What is that of set of all integrable or square integrable functions ? 
 A: The second part of the question asks for the cardinality of the set of integrable or square-integrable functions.  I will assume you actually mean "functions" rather than "equivalence classes of functions under the relation of equality almost everywhere". 
Let $\beta$ be the cardinality of the set of all functions from $\mathbb{R}$ to $\mathbb{R}$; by standard set theory this is the same as the cardinality of the powerset of the real numbers: $\beta = 2^{|\mathbb{R}|} = 2^{2^{\aleph_0}}$. 
Certainly the set of integrable functions, and the set of square integrable functions, can have cardinality no more than $\beta$. It turns out this is exactly the cardinality.
Let $E$ be a Cantor set; the key properties are that $|E| = |\mathbb{R}|$ and the measure of $E$ is $0$. Consider the set of all functions that are $0$ for every $x$ that is not in $E$. All of these functions are both integrable and square integrable, because the measure of $E$ is zero. The cardinality of this set is the cardinality of the set of functions from $E$ to $\mathbb{R}$, which is exactly $\beta$ because $|E| = |\mathbb{R}|$. 
A: A continuous function is determined by its value on the rational points, so there are at most $\aleph^{\aleph_0} = \aleph$ of them.
Conversely, it's not difficult to find $\aleph$ smooth (integrable) functions in $L_1 \cap L_2$, just take any such non-zero function $f$ and consider $\{rf : r \in \mathbb{R}\}$.
EDIT: For definiteness, $x \mapsto e^{-x}$ is $C^\infty$ and in $\bigcap_{p>0} L_p$.
A: You can also find the cardinality of smooth functions by first considering the set of constant functions, having cardnality c, and then determine the cardnality of all of the other smooth functions by knowing that, because it is not constant (by definition) that at each rational number, (rational only because they are dense in the reals and thus define any continuous function) there must be either positive or negative infinitesimal change in the function (not zero which would cause a non-differentiable "sharp corner"),the set of all such "decisions" defining each function having cardinality 2^(ℵ0) (since the rationals are countable).
This gives the set of smooth functions to have cardinality
c+2^(ℵ0)  =  c+c  =c 
