there exists a sequence of simple functions $\{g_n\}$ bounded in $L^1$, such that $f_n-g_n\to 0 \text{ a.e and in } L^1.$ Let $(E,\mathcal{A},\mu)$ be a finite measure space, and $\{f_n\}$ be a sequence bounded in $L^1$.
Why does there exist a sequence of simple functions $\{g_n\}$ bounded in $L^1$, such that:
$$
f_n-g_n\to 0 \text{ a.e and in } L^1.
$$
 A: For a fixed $n$, a standard construction found in most introductory measure theory textbooks shows that there are sequences of simple functions $g_n^{k,+}$ and $g_n^{k,-}$ indexed by $k$ such that $$0 \leq g_n^{k,+} \leq g_n^{k+1,+} \leq f_n^+ \qquad 0 \leq g_n^{k,-} \leq g_n^{k+1,-} \leq f_n^-$$
and such that $g_n^{k,+} \to f_n^+$ and $g_n^k := g_n^{k,-} \to f_n^-$ as $k \to \infty$ a.e. and in $L^1$. Hence $g_n^{k,+} - g_n^{k,-} \to f_n$ a.e. and in $L^1$. 
Additionally, the inequalities above show that $\|g_n^{k, \pm}\|_{L^1} \leq \|f_n^\pm\|_{L^1} \leq \|f_n\|_{L^1}$. Since $(f_n)_{n \geq 1}$ is bounded in $L^1$, this shows that $\{g_n^k: n, k \geq 1\}$ is bounded in $L^1$ also. 
From the above, $g_n^k \to f_n$ in $L^1$ and hence for each $n$, there is a $k_1(n)$ such that $\|f_n - g_n^{j}\|_{L^1} \leq 2^{-n}$ for $j \geq k_1(n)$.
Also, by Egorov's Theorem, for each $n$ we can find a measurable set $B_n$ such that $\mu(E \setminus B_n) < 2^{-n}$ and $g_n^k \to f_n$ uniformly on $B_n$. Thus there exists a $k_2(n)$ such that for $j \geq k_2(n)$ and $x \in B_n$, $|f_n(x) - g_n^j(x)| \leq 2^{-n}$. 
Hence, if we define $k(n) = \max\{k_1(n), k_2(n)\}$ and $g_n = g_n^{k(n)}$ then it is immediate that 
$$\|f_n - g_n\|_{L^1} \leq 2^{-n} \to 0$$
so that $f_n - g_n \to 0$ in $L^1$.
Also, for $x \in \bigcup_{k \geq 0} \bigcap_{j \geq k} B_j$ and $n$ sufficiently large, we have that 
$$|f_n(x) - g_n(x)| \leq 2^{-n}$$
so that the a.e. convergence follows if we can show that $\mu\left(E \setminus \bigcup_{k \geq 0} \bigcap_{j \geq k} B_j \right) = 0$. Now \begin{align*}\mu\left(E \setminus \bigcup_{k \geq 0} \bigcap_{j \geq k} B_j \right) =& \mu \left(\bigcap_{k \geq 0} \bigcup_{j \geq k} E \setminus B_j \right) \\=& \lim_{k \to \infty} \mu \left(\bigcup_{j \geq k} E\setminus B_j \right) \\ \leq& \lim_{k \to \infty} \sum_{j \geq k} 2^{-j} \\ =& \lim_{k \to \infty} 2^{-k+1} = 0\end{align*}
as desired
