Proof of the simple approximation lemma a) For the proof of the simple approximation lemma, our textbook says,

Let (c,d) be an open, bounded interval that contains the image of E, f(E), and $c=y_0 < y_1 < ... < y_n = d$ be a partition of the closed bounded interval [c,d] such that $y_k - y_{k-1} < \epsilon$ for $1 \leq k \leq n$. Define $I_k = [y_{k-1}, y_k)$ and $E_k = f^{-1}(I_k)$ for $ 1 \leq k \leq n$.
Since each $I_k$ is an interval and the function f is measurable, each set $E_k$ is measaurable.

I was a bit confused about this last sentence. I'm not sure what theorem they are using to say that $E_k$ is measurable because $I_k$ is an interval and f is measurable...

A function $f$ defined on a measurable set is measurable iff inverse image of every open set is measurable.
This is an easy consequence of the fact that any open set can be divided into disjoint open intervals and each one of them is an intersection of set of kind $(-\infty,b) \cup (a,\infty)$.
