With some simple theorems about integration over subsets we have that
$\qquad\displaystyle\int_E f \,d\mu = \int_X \chi_E\cdot f\,d\mu = \int_X 0 \,d\mu = 0 \cdot \mu(X) = 0$,
where we've used that $f \cdot \chi_E = 0$ a.e. and that the integral of two a.e. equal functions agree.
Of course this all depends on the order that one develops measure theory in and thus what definitions one is working with. One can also define integrals over subsets by restriction of measure in a suitable way and then prove that
$\qquad\displaystyle\int_E f\,d\mu = \int_X \chi_E\cdot f\,d\mu$.