Characteristic functions dense in simple functions in $L^1$? Consider $L^1(X,Y)$ where $X=Y=[0,1]$ and $ \langle f, g\rangle =\int fg $.  Is the set of characteristic functions $\{\chi_{A}  \}$   dense in the set of simple functions $\{ s\}$, in the sense there is a sequence $\{\chi_{A_n}\}$
$$\langle \chi_{A_n}, g\rangle \to \langle s, g\rangle$$
for any continuous $g\in L^{\infty}(X,Y)$?
I read this answers here (http://mathforum.org/kb/message.jspa?messageID=6119055) but I didn't fully understand. It seems the answer is yes? Hope anyone offer some further explanation.

This does seem to work. This seems to be the idea of the construction:
Start by fixing a subinterval $I \subset [0,1]$. Recursively cut $I$ into $n$ pieces and define a function which is $1$ on $m$ of the pieces and $0$ on the rest. The resulting sequence, which I'll call $f_{q,k}$ where $q=\frac{m}{n}$, converges in the sense you've described to $\frac{m}{n} \chi_I$, as $k \to \infty$. This is essentially the content of your link. Roughly speaking, a sequence of characteristic functions which are somehow both "uniformly" spread across $I$ and have integral $\frac{m}{n} |I|$ will converge to $\frac{m}{n} \chi_I$.
Now for arbitrary $\alpha \in [0,1]$, get a rational sequence $q_n \to \alpha$. Consider the sequences $f_{q_n,k}$ as before. Take the diagonal sequence: $f_{q_n,n}$ will converge as $n \to \infty$ to $\alpha \chi_I$. Call the sequence that we just extracted $g_{\alpha,I,n}$.
Now we handle $\alpha \chi_A$ for a general measurable $A$. The idea is another diagonalization argument. Take a sequence of finite unions of intervals that approximate $A$, say $\{ \{ I_{n,k} \}_{n=1}^{N_k} \}_{k=1}^\infty$. Then consider $\sum_{n=1}^{N_k} g_{\alpha,I_{n,k},m}$. Diagonalize again, by considering $\sum_{n=1}^{N_k} g_{\alpha,I_{n,k},k}$. Then we've approximated $\alpha \chi_A$. There may be some technical issues with overlaps here, but I think those go away if we go far enough out into the sequence. At any rate, call the resulting approximation $h_{\alpha,A,k}$.
Now we finally approximate $\sum_{m=1}^N \alpha_m \chi_{A_m}$ by $\sum_{m=1}^N h_{\alpha,A_m,k}$. Again there may be some technicality about overlapping, but it should go away for large $k$.
