Real Analysis, Folland Theorem 2.26 Integration of Complex Functions Background information:

Theorem 2.10 - Let $(X,M)$ be a measurable space.
a.) If $f:X\rightarrow [0,\infty]$ is measurable, there is a sequence $\{\phi_n\}$ of simple functions such that $0 \leq \phi_1 \leq \phi_2 \leq \ldots \leq f$, $\phi_n \rightarrow f$ pointwise, and $\phi_n\rightarrow f$ uniformly on any set on which $f$ is bounded.
b.) If $f:X\rightarrow \mathbb{C}$ is measurable, there is a sequence $\{\phi_n \}$ of simple functions such that $0 \leq |\phi_1| \leq |\phi_2| \leq \ldots \leq |f|$, $\phi_n\rightarrow f$ pointwise, and $\phi_n\rightarrow f$ uniformly on any set on which $f$ is bounded.  
Proposition 1.20 - If $E\in  M_\mu$ and $\mu(E) < \infty$, then for every $\epsilon > 0$ there is a set $A$ that is a finite union of open intervals such that $\mu(E \ \triangle \ A) < \epsilon$.

Question:

Theorem 2.26 - If $f\in L^1(\mu)$ and $\epsilon > 0$, then 
a.) there is an integrable simple function $\phi = \sum_{1}^{n}a_j\chi_{E_j}$ such that $\int |f - \phi|d\mu < \epsilon$.
b.) If $\mu$ is a Lebesgue-Stieltjes measure on $\mathbb{R}$, the sets $E_j$ in the definition of $\phi$ can be taken to be finite unions of open intervals.
c.) Moreover, in situation b.), there is a continuous function $g$ that vanished outside a bounded interval such that $\int |f - g|d\mu < \epsilon$.

Attempted Proof a.) - By theorem 2.10 we can find a sequence of simple functions $\{\phi_j\}$ with $\phi_j\rightarrow f$ pointwise and $|\phi_1|\leq |\phi_2|\leq \ldots \leq |f|$. Now, $|\phi_j - f|\rightarrow 0$ pointwise and $$|\phi_j - f|\leq |\phi_j| + |f| \leq 2|f|$$ Applying the Dominated Convergence Theorem, $$\lim_{j\rightarrow \infty}\int |\phi_j - f| = \int 0 = 0$$
Attempted Proof b.) - From a.) we can find a simple function $\phi = \sum_{1}^{n}a_j\chi_{E_j}$ within distance of $\epsilon/2$ from $f$. Since the simple sum is integrable, all of the $E_j$'s have finite measure. So by Theorem 1.20, for each $E_j$ there is a finite union of open intervals with $\mu(E \ \triangle \ F_j) < \frac{\epsilon}{2|a_j|n}$. Now, \begin{align*}\int_n\left|\sum_{1}^{n}a_j\chi_{E_j} - \sum_{1}^{n}a_j \chi_{F_j}\right| &\leq \sum_{1}^{n}|a_j|\int |\chi_{E_j} - \chi_{F_j}|\\ &= \sum_{1}^{n}|a_j|\mu(E_j \ \triangle \ F_j)\\ &\leq \sum_{1}^{n}\frac{\epsilon}{2n}\\ &= \frac{\epsilon}{2}
\end{align*}
I am not sure if this correct and I can't figure out c.) yet, I will post it once I figure it out.
 A: Your proof of item a is correct. I just copy it here for sake of completeness.
Your proof of item needs a little adjustment, mainly for clearness.

Theorem 2.26 - If $f\in L^1(\mu)$ and $\epsilon > 0$, then


a.) there is an integrable simple function $\phi = \sum_{1}^{n}a_j\chi_{E_j}$ such that $\int |f - \phi|d\mu < \epsilon$.


b.) If $\mu$ is a Lebesgue-Stieltjes measure on $\mathbb{R}$, the sets $E_j$ in the definition of $\phi$ can be taken to be finite unions of open intervals.


c.) Moreover, in situation b.), there is a continuous function $g$ that vanished outside a bounded interval such that $\int |f - g|d\mu < \epsilon$.

Proof
a.) - By theorem 2.10 we can find a sequence of simple functions $\{\phi_j\}$ with $\phi_j\rightarrow f$ pointwise and $|\phi_1|\leq |\phi_2|\leq \ldots \leq |f|$. Now, $|\phi_j - f|\rightarrow 0$ pointwise and $$|\phi_j - f|\leq |\phi_j| + |f| \leq 2|f|$$ Applying the Dominated Convergence Theorem, $$\lim_{j\rightarrow \infty}\int |\phi_j - f| = \int 0 = 0$$
b.) - From a.) we can find a simple function $\phi = \sum_{1}^{n}a_j\chi_{E_j}$ within $L^1$-distance of $\epsilon/2$ from $f$, that means
$$ \int |\phi - f|<\epsilon/2$$
We can assume, without loss of generality, that the $E_j$'s are disjoint and for all $j$, $a_j \neq 0$. So, since
$$\sum_{j=1}^n |a_j| \mu(E_j) = \int |\phi| \leq \int |f| + \int |\phi - f|<  \int |f| + \epsilon/2 <+\infty $$
we have that all of the $E_j$'s have finite measure. So by Theorem 1.20, for each $E_j$, there is a finite union of open intervals $F_j$ such that $\mu(E_j \ \triangle \ F_j) < \frac{\epsilon}{2|a_j|n}$. Now,
\begin{align*}\int\left|\sum_{1}^{n}a_j\chi_{E_j} - \sum_{1}^{n}a_j \chi_{F_j}\right| &\leq \sum_{1}^{n}|a_j|\int |\chi_{E_j} - \chi_{F_j}|\\ &= \sum_{1}^{n}|a_j|\mu(E_j \ \triangle \ F_j)\\ &\leq \sum_{1}^{n}\frac{\epsilon}{2n}\\ &= \frac{\epsilon}{2}
\end{align*}
It follows that
$$\int_n\left|f - \sum_{1}^{n}a_j \chi_{F_j}\right| \leq \int |f- \phi| + \int\left|\phi - \sum_{1}^{n}a_j \chi_{F_j}\right|< \epsilon/2 + \epsilon/2 =\epsilon $$
c.) From item b.) we can find  a simple function $\psi = \sum_{1}^{m}c_k \chi_{I_k}$ such that, for each $k$, $I_k$ is an open interval, $c_k \neq 0$ and
$$ \int |\psi - f|<\epsilon/2$$
For each $k$, suppose $I_k = (a_k,b_k)$ and let
$\delta = \min \{ \epsilon/(4|c_k|m) , (b_k-a_k)/3\}$. Let $\varphi_k$ be a  continuous function such that, for all $x$, $\varphi_k(x) \in [0,1]$ and $\varphi_k(x)=0$ if $x \notin I_k$ and $\varphi_k(x)=1$ if $x \in [a_k+\delta, b_k-\delta]$. Let $ g= \sum_{k=1}^m c_k \varphi_k $.
It is clear that $g$ is continuous and we have
\begin{align*}\int |f-g| &\leq  \int |f-\psi|+\int |\psi - g| < \\
& < \epsilon/2 + \int \left | \sum_{1}^{m}c_k \chi_{I_k}-\sum_{k=1}^m c_k \varphi_k \right | \leq \\
& \leq  \epsilon/2 +  \sum_{1}^{m}|c_k| \int \left | \chi_{I_k}- \varphi_k \right | \leq \\ 
& \leq \epsilon/2 +  \sum_{1}^{m}|c_k| (2\epsilon/(4|c_k|m))= \\
& = \epsilon/2 + \epsilon/2 = \epsilon
\end{align*}
Remark: We used the fact that
$$ | \chi_{I_k}- \varphi_k  | \leqslant | \chi_{I_k} -\chi_{[a_k+\delta, b_k-\delta]}| = \chi_{(a_k,a_k+\delta) \cup (b_k-\delta, b_k)} $$
Since $\delta \leq  (b_k-a_k)/3$, we have that $(a_k,a_k+\delta) \cup (b_k-\delta, b_k)$ is a disjoint union. So
$$\int \left | \chi_{I_k}- \varphi_k \right | \leq 2 \delta$$
Since $\delta \leq   \epsilon/(4|c_k|m)$, we have
$$\int \left | \chi_{I_k}- \varphi_k \right | \leq 2 \delta \leq 2 \epsilon/(4|c_k|m)$$
