Is it possible for an integral between $a$ and $a$ to have a value other than 0 Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ with a constant $a \in \mathbb{R}$ such that
$$\int_{a}^{a} f(x) \, dx  \neq 0 \quad$$
holds?

No, because $\{a\}$ is a null set.
To be more precise, for any  function $f \colon \mathbb{R} \to \mathbb{R}$ you have that $$ f \cdot 1_{\{a\}}  = f(a) \cdot 1_{\{a\}}  $$ is Lebesgue measurable. Thus the integral is always defined and zero because of
$$ \int_\mathbb{R}  f(x) \cdot 1_{\{a\}}(x) \, dx = \int_{\{a\}} f(a) \, dx = f(a) \cdot \lambda(\{a\}) = 0.$$
Remark 1: 
For any null set $A$, and any Lebesgue measurable function $f$ holds
$$ \int_A f(x) \, dx = 0 $$
and every Riemann integrable function is Lebesgue measurable.
Remark 2: 
For any null set $A$ and any function $f \colon \mathbb{R} \to \mathbb{R}$ you have that $1_A \cdot f$ is Lebesgue measurable and thus $$\int_A f(x) \, dx = 0.$$ 
