Real Analysis, Folland Theorem 2.41/Exercise 53 The $n$ dimensional Lebesgue integral 
Theorem 2.41/Exercise 53 - If $f\in L^1(m)$ and $\epsilon > 0$, there is a simple function $\phi = \sum_{1}^{n}a_j\chi_{R_j}$ where each $R_j$ is a product of intervals, such that $\int |f - \phi| < \epsilon$, and there is a continuous function $g$ that vanishes outside a bounded set such that $\int |f - g| < \epsilon$.

Attempted proof - From theorem 2.10 (Folland) we can find a sequence of simple functions $\{\phi_j\}$ with $\phi_j\rightarrow f$ pointwise and $|\phi_1| \leq |\phi_2| \leq \ldots \leq |f|$. Note that $|\phi_j - f|\rightarrow 0$ pointwise and $$|\phi_j - f|\leq |\phi_j| + |f| \leq 2|f|$$ By Dominated convergence theorem $$\lim_{j\rightarrow \infty}\int |\phi_j - f| = \int 0 = 0 $$
Folland then suggests to use 2.40 Theorem c.) which is found here to approximate the latter functions $\phi$ of the desired form. Then he says that you should approximate such $\phi$'s by continuous functons by applying an "obvious" generalization of Theorem 2.26. I am not sure how to apply this, any suggestions is greatly appreciated.
 A: Note that in the statement of the theorem, I renamed $\phi$ by $\psi$. It will help to highlight the relationship between the proof of this theorem and the proof of Theorem 2.26.

Theorem 2.41/Exercise 53 - If $f\in L^1(m)$ and $\epsilon > 0$, there is a simple function $\psi = \sum_{1}^{n}a_j\chi_{R_j}$ where each $R_j$ is a product of intervals, such that $\int |f - \psi| < \epsilon$, and there is a continuous function $g$ that vanishes outside a bounded set such that $\int |f - g| < \epsilon$.

Proof:
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, $$\lim_{j\rightarrow \infty}\int |\phi_j - f| = \int 0 = 0$$
So, 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 2.40 item c, we have that, for each $E_j$, there is a finite union of rectangles $F_j$ whose sides are intervals 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
$$\int_n\left|f - \sum_{1}^{n}a_j \chi_{F_j}\right| \leq \int |f- \phi| + \int\left|\phi - \sum_{1}^{n}a_j \chi_{F_j}\right|< \epsilon/2 + \epsilon/2 =\epsilon $$
Since, for each $j$, $F_j$ is a finite union of rectangles  whose sides are intervals, we have that $$\sum_{1}^{n}a_j \chi_{F_j}=\sum_{1}^{m}c_k \chi_{R_k}$$ where, for each $k$, $R_k$ is a rectangle whose sides are intervals, $c_k \neq 0$. Take $\psi= \sum_{1}^{m}c_k \chi_{R_k}$.
So we proved that, for any $\epsilon >0$, there is a simple function $\psi = \sum_{1}^{m}c_k \chi_{R_k}$ such that, for each $k$, $R_k$ is a rectangle whose sides are intervals, $c_k \neq 0$ and
$$ \int |\psi - f|<\epsilon$$
For the last part, note that, from what we have just proved we can find  a simple function $\psi = \sum_{1}^{m}c_k \chi_{R_k}$ such that, for each $k$, $R_k$ is a rectangle whose sides are intervals, $c_k \neq 0$ and
$$ \int |\psi - f|<\epsilon/2$$
For each $k$, let $\varphi_k $ be a continuous function such that
$$\left | \chi_{R_k}- \varphi_k \right |<  \epsilon/(2|c_k|m)$$
(such function always exist)
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_{R_k}-\sum_{k=1}^m c_k \varphi_k \right | \leq \\
& \leq  \epsilon/2 +  \sum_{1}^{m}|c_k| \int \left | \chi_{R_k}- \varphi_k \right | \leq \\ 
& \leq \epsilon/2 +  \sum_{1}^{m}|c_k| (\epsilon/(2|c_k|m))= \\
& = \epsilon/2 + \epsilon/2 = \epsilon
\end{align*}
Remark: Note that this proves is "parallel" to the proof of Theorem 2.26. See, for instance, my answer in
Real Analysis, Folland Theorem 2.26 Integration of Complex Functions
