Let $(X,\Sigma, \mu)$ be a measurable space and let $f: X \mapsto \mathbb{R}$ be an integrable function. I want to show that for $\epsilon > 0$, I can find a simple function g such that $\int_{X} |f - g| d\mu < \epsilon$.
Since f is integrable, we know that $\int f = \text{sup}\{\int g :\text{g is a simple function and } g \leq_{a.e} f\}$ so if $g \leq_{a.e} f$ we have the inequality $\int f d\mu > \int g d\mu = \int \sum_\limits{i=1}^{n} a_{i} \chi(E_{i})d\mu$ where each $E_{i}$ are measurable sets.
But now I'm not sure how to proceed.
Start by writing $f=f_+-f_-$ where $f_+(x):=\max(f(x),0)$ and $f_-(x):=\max(-f(x),0)$.
Then $f_+,f_-$ are nonnegative functions that inherit measurability and integrability from $f$.
By definition: $$\int f_+=\sup\left\{\int s\mid s\text{ is a simple function with }0\leq s(x)\leq f_+(x)\text{ for every }x\in X\right\}$$
So for any $\epsilon>0$ we can find a simple function $g_1$ with $0\leq g_1(x)\leq f_+(x)$ for every $x\in X$, and: $$\int g_1\leq\int f_+<\frac12\epsilon+\int g_1$$
Likewise we can find a simple function $g_2$ with $0\leq g_2(x)\leq f_-(x)$ for every $x\in X$, and: $$\int g_2\leq\int f_-<\frac12\epsilon+\int g_2$$
Then for $A:=\{x\mid f(x)\geq0\}$ we find: $$\int|f-g|=\int_A |f-g|+\int_{A^c}|f-g|-=\int (f_+-g_1)+\int (f_--g_2)<\frac12\epsilon+\frac12\epsilon=\epsilon$$