Lebesgue integral confusion

Question

I am trying to understand Lebesgue integration of simple functions.

If $(X,A,\mu)$ is a measure space, $E \in A$ and $\mu(E) = 0$ and $f(x) = \begin{cases} 0 & \text{if } x \in E \\ 1 & \text{if } x \not\in E \end{cases}$

then is $\int_X f(x)\mu(du) = \sum_i^k Xi\mu(Ei) = (1)\mu(E) + (0)\mu(E^c) = 1(0) + 0 = 0? $

Answer 1

$ \int_Xf(x) dx = \int_Ef(x)dx+\int_{E^c}f(x)dx = \int_E0\;dx+\int_{E^c}1\;dx = \mu (E^c) $. Since $\mu$ is a measure then it is countably additive so $ \mu (X) = \mu (E) + \mu(E^c) = \mu (E^c) = \int_Xf(x) dx $