Example of $0\leq f\leq g$ and $E$ non measurable set s.t. $\int\limits_E f>\int\limits_E g$

Question

It is known that if $0\leq f\leq g$ and $E$ is lebesgue measurable then $\int\limits_E f\leq\int\limits_E g$ but why is the measurability of $E$ important? I guess we cannot really find a counterexample where $\int\limits_E f>\int\limits_E g$ but the integral is just not well defined if the set $E$ is non measurable...

Answer 1

By definition, $\int_E$ for a simple function $f = \sum_{i=1}^n a_i 1_{A_i}$ becomes $$ \int 1_E f d\mu = \sum_{i=1}^n a_i \mu(A_i \cap E). $$ Measurability of $E$ makes sure that $\mu(A_i \cap E)$ is well-defined. So the measurability is not so much to make sure that the inequality you showed is true, but rather that whatever we mean with $\int_E$ makes sense.