How to prove $\int_{\Omega} \sum_{i=1}^{N} f_i(x)dx$ is equivilant with $\sum_{i=1}^{N} \int_{\Omega} f_i(x)u_i(x)dx$
Starting from your $E_2$ and exchanging the finite sum with the integral we get this:

$$E_2=\sum_{i=1}^N \int_\Omega f_i(x) u_i(x) dx = \int_\Omega \sum_{i=1}^N f_i(x) u_i(x) dx$$

Next we can use the disjoint cover $\{\Omega_i\}$ of $\Omega$ to split up the integral:

$$E_2 = \sum_{k=1}^N \int_{\Omega_k} \sum_{i=1}^N f_i(x) u_i(x) dx$$

The $x$ in the integral are now from $\Omega_k$ and we can apply your identity for the sum over $f_i(x) u_i(x)$:

$$E_2 = \sum_{k=1}^N \int_{\Omega_k} f_k(x) dx$$

This is almost your $E_1$, but not quite. We cannot exchange the integral with the sum because the integral depends on the summation index. Extending the integration region to $\Omega$ is also not possible without reintroducing a selection function. So your $E_1$ and $E_2$ are not equal. The closest thing to $E_1$ is what I have derived above.