Is Lebesgue integral over interior equal to the integral over the whole set?

Question

I have a measurable set $S\subset\mathbb{R}$ and a measurable function $f\colon\mathbb{R}\rightarrow \mathbb{R}$.

Is it true that $$\int\limits_Sf(x)\, dx=\int\limits_{\operatorname{int}(S)}f(x)\,d x?$$

Answer 1

No, the interior can even be empty. Consider $S = \Bbb{R} \setminus \Bbb{Q}$.

Answer 2

Another example: $S$ is a fat Cantor set, so $S$ is closed and bounded, but $\int_S 1\,dx \ne 0$, but $\mathrm{int}(S) = \varnothing$.