I showed that $L^{p}$ is seprable when $X$ is a topological space with countable base and $\mu$ a radon measure on $T$.
Do you think it's the right assumptions ? I mean does it exist a counter exemple when the basis of $T$ is not countable ?
I wish you a very good day.
On
It seems like a reasonable assumption.
The converse can't be literally true. For a trivial counterexample, say $X$ is not second-countable but $\mu$ is supported on a second-countable subset of $X$.
On
If $X$ is compact, this is Exercise 43 in the book by Reed and Simon vol 1 on p. 123 (1980 edition). They offer the following hint: Let $(A_n)$ be countable basis for each pairs $(m,n)$ such that $\overline{A_m}\cap\overline{A_n}=\varnothing$ construct a $[0,1]$-valued continuous function $f_{m,n}$ equal to 1 on $A_m$ and to 0 on $A_n$. Then the $\mathbb{Q}$-algebra generated by these functions is dense in $C(X)$ and thus in $L^p(X)$ for $1\le p<\infty$. They also mention the extension to $X$ locally compact using $\sigma$-compactness.
BTW, Exercise 44 right after that is: "Do any fifty problems in Kelley's book."
See Theorem 4.13 in Brezis's functional analysis. If the measure space $(\Omega, \Sigma,\mu)$ is separable, meaning that
then $L^p(\Omega)$ is separable (as a metric space) for $1\leq p<\infty$.
Now suppose $\Omega=X$ is a second-countable topological space, $\Sigma=\mathcal{B}(X)$ is the Borel $\sigma$-algebra, and $\mu$ is a Borel measure. Then there exists a countable collection $\mathcal{F}\subset \mathcal{B}(X)$ of open sets such that any open set in $X$ is the countable (since $\mathcal{F}$ itself is countable) union of sets in $\mathcal{F}$. Hence in the Borel $\sigma$-algebra, $\mathcal{F}$ generates all open sets, and therefore the whole $\mathcal{B}(X)$ (by definition of Borel $\sigma$-algebra). So, $(\Omega,\Sigma,\mu)$ is separable, and the theorem can be applied.
In fact this argument does not depend on the properties of $\mu$ - it can be any Borel measure.