Theorem 4.3 approximating measurable functions with step functions

Question

I was reading Stein & Shakarchi's Real Analysis and got stuck understanding this proof:

Definition: A step function is $$ f = \sum_{k=1}^{N} a_k \chi_{R_k} $$ where each $R_k$ is a rectangle, and $a_k$ are constants. A (closed) rectangle in $\mathbb{R}^d$ is given by $$ R = [a_1, b_1] \times \cdots \times [a_d, b_d] .$$

Theorem 4.3 (p32): Suppose $f$ is measurable on $\mathbb{R}^d$. Then there exists a sequence of step functions $\{ \psi_{k} \}_{k=1}^{\infty}$ that converges pointwise to $f(x)$ a.e.

Proof: By the previous theorem, there are simple functions such that $\lim_{k \rightarrow \infty} \phi_k(x) = f(x)$ for all $x$. To approximate each $\phi_k$ by a step function. We recall part (iv) of Theorem 3.4, which states that if $E$ is a measurable set of finite measure, there exists cubes $Q_1, \ldots, Q_N$ such that $m(E \Delta \bigcup_{j=1}^{N} Q_j) \le \epsilon$. We may ssume these cubes are almost disjoint rectangles. By taking closed rectangles $\tilde{Q_j}$ contained in $Q_j$, we find a collection $\{ \tilde{Q_j} \}_{j=1}^N $ that satisfy $m(E \Delta \bigcup _{j=1}^{N} \tilde{Q_j}) \le 2 \epsilon.$

It follows from this observation and the definition of a simple function that for each $k$, ther exists a step function $\psi _k $ and a measurable function $F_k$ so that $m(F_k) < 2^{-k}$ and $\phi_k(x) = \psi_k(x)$ for all $x \notin F_k.$

I do not understand why the italicized part holds.

Firstly, what exactly is this $F_k$, and secondly, how do we define $\psi_k$ ?

Thanks in advance.

Answer 1

From this observation: For each $k\ge 1$, there exists a collection of closed disjoint rectangles $\{\tilde{Q}_j^{(k)}\}_{j=1}^{N_k}$ such that $m(E\Delta\bigcup_{j=1}^{N_k}\tilde{Q}_j^{(k)})\le 2^{-k}$. Let $\psi_k(x)=\sum_{j=1}^{N_k}\chi_{\tilde{Q}_j^{(k)}(x)}$. Since $$E_k=\{x:f(x)\ne\psi_k(x)\}=\{x:\chi_E(x)\ne\psi_k(x)\}\subset E\Delta\Bigl(\bigcup_{j=1}^{N_k}\tilde{Q}_j^{(k)}\Bigr),$$ Hence $m(E_k)\le 2^{-k}$. Now let $F_k=\bigcup_{j=k+1}^\infty E_j$, then $m(F_k)\le 2^{-k}$ and if $x\in F_k^c=\bigcap_{j=k+1}^\infty E_j^c$, we have $$ f(x)=\psi_j(x),\qquad \forall j\ge k+1,\qquad\text{and}\qquad\lim_{j\to\infty}\psi_j(x)=f(x).$$ Therefore, let $F=\bigcap_{k=1}^\infty F_k=\lim\limits_{k\to\infty}\!\!\downarrow\!\! F_k$, then $m(F)=0$ and $$\lim_{j\to\infty}\psi_j(x)=f(x)\qquad \forall x\in F^c=\bigcup_{k=1}^\infty F_k^c. $$