$f\in L^1(\mathbb{R})$ there exists step functions $\{ g_n\}$ such that $\lim_{n\to \infty}\int_{-\infty}^{\infty}|f(x)-g_n(x)|dx=0$ Question: If $f\in L^1(\mathbb{R})$ then prove that there exists step functions $\{ g_n\}$ such that $$\displaystyle\lim_{n\to \infty}\int_{-\infty}^{\infty}|f(x)-g_n(x)|dx=0$$
I defined $g_n(x)=\displaystyle\sum_{i=1}^{2n^2}f\Big(-n+\frac{i-1}{2n}\Big) \chi_{[-n+\frac{i-1}{2n},-n+\frac{i}{2n}]}$ which I believe works but I cannot find a way to prove this. Any help is appreicated. Thank you!

To prove this for general integrable functions, not just for bounded functions, consider the following.
Suppose $f \in L^1( \mathbb{R})$.  We want to show that for any positive integer $n$ there exists a step function $g_n$ with support on a finite interval $[-c_n,c_n]$ such that
$$\tag{1}\int_{-\infty}^{+\infty} | f - g_n| = \int_{-c_n}^{c_n} | f - g_n| + \int_{-\infty}^{-c_n} |f| + \int_{c_n}^{+\infty} |f| < \frac{1}{n}$$
Since $f$ is integrable we can find $c_n$ such that 
$$\int_{-\infty}^{-c_n} |f| + \int_{c_n}^{+\infty} |f| < \frac{1}{2n}.$$
It remains to find the step function $g_n$ such that the first integral on the RHS of (1) is less than $1/(2n)$.
Using the simple approximation theorem, since $f$ is Lebesgue integrable on $[-c_n,c_n]$, there exists a simple function $\phi_n$ such that
$$\int_{-c_n}^{c_n} |f - \phi_n| < \frac{1}{4n}$$.
If we can find a step function $g_n$ such that
$$\int_{-c_n}^{c_n} | g_n - \phi_n| < \frac{1}{4n}$$ we are done since
$$\int_{-c_n}^{c_n} | f - g_n| \leqslant\int_{-c_n}^{c_n} | f - \phi_n|+ \int_{-c_n}^{c_n} | g_n - \phi_n| < \frac{1}{2n}.$$
To find $g_n$ it is enough to prove this for a characteristic function $\phi_n = \chi_E$ of a measurable set $E \subset [-c_n,c_n]$, since any simple function is a finite combination of characteristic functions.
There is an open set $G$ such that $E \subset G$ and $m(G\setminus E) < 1/(4n)$. Since $G$ is open it is the union of countably many disjoint open intervals $(a_j,b_j)$ and 
$$m(G) = \sum_{j=1}^\infty (b_j - a_j) < m(E) + 1/(4n)$$
Fix $N$ and let $g_n$ be the characteristic function of the set $\cup_{j=1}^N (a_j,b_j) \cap [-c_n,c_n]$. This also is a step function. Let $h_n$ be the characteristic function of $G \cap [-c_n,c_n]$.
We have
$$\tag{2}\int_{-c_n}^{c_n}|\chi_E - g_n| \leqslant \int_{-c_n}^{c_n}|\chi_E - h_n| + \int_{-c_n}^{c_n}|g_n - h_n| \\ \leqslant m \left(G \setminus E \right) + m \left(\bigcup_{j = N+1}^\infty (a_j,b_j) \right) 
$$
The first term on the RHS of (2) is less than $1/(4n)$ and the second term can be made less that $1/(4n)$ by choosing $N$ sufficiently large since $\cup_{j=1}^\infty (a_j,b_j)$ has finite measure.
Thus,
$$\int_{-c_n}^{c_n}|\chi_E - g_n| < \frac{1}{2n},$$
which completes the proof.
