I have a rather vague question, but, my analysis teacher left as an exercise to show that the continuous functions are dense in L1. There are other way to do this, but she specified that this must be done by: 1) Showing that simple functions are dense in L1, which is easy peasy lemony squeezy, just use the definition of the intergral!
2) Show that every characteristic function may be approximated by a continuous function by the means of showing that the set of all characteristic functions that can be approximated by continuous functions is a $\sigma$-algebra.
So question is, how to do 2) and how does that help me prove what she sets out to do? She left is a "fill in the gap" part of a proof, so no homework or anything. I tried to do 2) but feel stumped, I tried to do some sort of extension argument but am not sure where to start.
Since you didn't specify the measure space we are working with, I'll assume $\Omega = (a,b) \subset \mathbb R,$ equipped with the Lebesgue measure.
To show (2) implies the result: An clean way to see this is to set, $$X = \overline{C^0(\Omega)}^{L^1} \subset L^1(\Omega),$$ which is the closure of the space of continuous functions in $L^1(\Omega),$ where we identify $C^0(\Omega) \subset L^1(\Omega)$ in the natural way. This is a closed subspace of $L^1(\Omega)$ which contains $C^0(\Omega)$ as a dense subspace. Hence it suffices to show that in fact $X = L^1(\Omega).$
Suppose that every characteristic function can be approximated by continuous functions. Then we get $1_A \in X$ for all $A$ measurable with finite Lebesgue measure. Since $X$ is a subspace, it hence contains all simple functions (taking linear combinations). But then it's easy to show that $X$ is dense in $L^1(\Omega)$ from here, which implies they are equal.
Showing (2) is a classical measure theory argument. You first show that the result holds for intervals in $(a,b),$ by exhibiting explicit approximations using piecewise linear functions. Then as set of (open, closed or half-open) intervals forms a $\pi$-system which generates the Borel $\sigma$-algebra, you can consider the set, $$\mathcal F = \{A \in \mathcal{B}(\Omega) : 1_A \text{ can be approximated by continuous functions in } L^1(\Omega) \},$$ and show it contains all Borel sets. This can be proved using for example Dynkin's $\lambda$-$\pi$ system, after checking all the conditions are satisfied.
Finally note that for general $\Omega \subset \mathbb R^n$ the same argument more or less goes through, but you need to restrict to measurable subsets of finite measure. The added details aren't too hard to fill in, but depending on what theorem you invoke the proof of step 2 may need some more care.