$E_{1}=\{ x \in E : f(x) < g(x)\}$ is also measurable and
$E_{2} = \{ x \in E : f(x) = g(x)\}$ is measurable
2. Let h:=f-g. Then your second problem is equivalent to showing $\{ x\in E : h(x) = 0\}$ is measurable. But this is $h^{-1}(0)$ and since h is measurable, and {0} is measurable, $E_2$ is too. Use similar ideas for $E_1$.