I have a question about a claim in a proof given by Stein and Shakarchi of the following theorem: Suppose $f$ is measurable and finite valued on $E$ with $E$ of finite measure. Then for every $\epsilon>0$ there exists a closed set $F_\epsilon$ such that $f|_{F_\epsilon}$ is continuous.
The proof starts by letting $f_n$ be a sequence of step functions such that $f_n\rightarrow f$ a.e.
The authors then claim that "we may find sets $E_n$ so that $m(E_n)<1/2^n$ and $f$ is continuous outside $E_n$."
What allows us to be able to find sets $E_n$ as such? Is this a property due to step functions?
This is a direct result from the definition.
We call $f$ a step function if it is a finite sum $$f=\sum_{k=1}^N a_k\chi_{R_k},$$ where each $R_k$ is a rectangle, and the $a_k$ are constants.
What you cited is Theorem 4.5. If you feel not comfortable with the choice of $E_n$, we can consider in the setting of $\mathbb R^1$, in which case $f$ a step function if$$f=\sum_{k=1}^N a_k\chi_{R_k},$$ where $R_k$ is a interval, and the $a_k$ are constants.
To make things more clear, we consider an example: $f=\sum_{k=1}^N \chi_{(a_k,b_k)}$, where $a_1< b_1< a_2< b_2< \cdots< a_N< b_N$. In this example, the discontinuous points of $f$ are precisely $\{a_i,b_i: 1\leq i\leq N\}$, which is finite and thus of measure zero. So for every $\delta>0$ we can take $E$ with $m(E)<\delta$ such that $f$ is continous outside $E$.