Say we take the measure of a countable set, we obtain that $\mu=0$. Now if this is the case, does this automatically imply that it is Lebesgue integrable as well?
The reason I bring up the set is because that set is used to build the characteristic function, which gives us $1$ for elements in the set, but $0$ for elements not in the set. If this set has measure $0$ (or better yet is measurable), then I want to say that if we take the characteristic function, it must be Lebesgue integrable.
On
A function, $f:X \to \mathbb{R}$ is measurable if for all $r\in \mathbb R$, $f^{-1}((-\infty, r]) = \{ x\in X \ |\ f(x)\leq r\}$ is a measurable set. From this, you should be able to conclude that $\chi_A$ is measurable for all measurable $A$.
Integrability is harder, but the idea is that a function is integrable if it is measurable and has a finite integral over any set (so no cancelling positive bits with negative bits). Your question comes down to "is the measure of a countable set finite?" Well, you can rewrite a countable set as $\cup_{i\in\mathbb N} \{a_i\}$, and from there you can write its measure as $\sum_i \mu(\{a_i\}).$ So, what is the measure of a singleton, and what do you get if you sum that a countable number of times?
Say you want yo integrate your characteristic function of a measurable set $A$ with $\mu(A)=0$ contained in a measurable space $X$. By the definition of the Lebesgue integral:
$$ \int_X \chi_A(x)\, dx = 1\cdot \mu (A)+ 0\cdot \mu(X\backslash A)=1\cdot 0 + 0 = 0. $$
Also, your set need not be countable, but is only required to have measure zero. Think of the Cantor set, for example, which is uncountable and has measure zero.