I am looking for a theorem that states the following:
Let $\Omega\in\mathbb{R}^n$, $n\in\mathbb{N}^*$, $D\subset\Omega$ arbitrarily chosen and $f$ is a continous function defined over $\Omega$.\ Then, $$ \int_D f(x) dx=0 \ \Longrightarrow f(x)=0,\ \forall x\in\Omega $$
On
The other answer has a nice and short proof. However, if you want a fancy name for what you seek, it is a variant of the fundamental lemma of calculus of variations, which says that if $\Omega$ is open and $f$ is continuous, then $$\int_\Omega f(x) h(x) dx = 0 \forall h\in C_c^\infty(\Omega) \Rightarrow f(x)=0 \forall x\in \Omega.$$
Your condition is equivalent to the one above:
On one hand, if $\int_D f(x) dx = 0$ for all measurable $D\subset \Omega$, then $\int_\Omega f(x) h(x) = 0$ for all step functions $h$ and then by approximation also for all $h \in C_c^\infty(\Omega)$. On the other hand, you can approximate a characteristic function $\chi_D$ by a sequence of smooth functions, so $\int_\Omega f(x) h(x) d x = 0$ for all $h\in C_c^\infty(\Omega)$ also implies $\int_D f(x) dx = \int_\Omega f(x) \chi_D(x) dx = 0$.
Note however that in general, if $\Omega$ is not open or $f$ not continuous, the theorem is no longer necessarily true.
If $f≠0$, then there is a point $x$ where $f(x) ≠ 0$, say without loss $f(x) > c > 0$. By continuity, $f(x) > c/2$ on an open neighbourhood $U$, and then $$ \int_U f(x) dx ≥ \frac{c}{2} |U| > 0.$$