Consider $f\in L^1(I,I)$ where $I=[0,1]$ and $ \langle f, g\rangle =\int fg $. For any given $\frac{m}{n}\chi_{A}$ where $\frac{m}{n}$ rational and $A$ an subinterval in $I$, how would I show that a sequence of characteristic functions $\{\chi_{A_n}\}$
$$\langle \chi_{A_n}, g\rangle \to \langle \frac{m}{n}\chi_{A}, g\rangle$$
for all continuous $g\in L^{\infty}(I,I)$?
Suppose first we divide the interval $A$ into $n$ parts with $m$ parts with values equal to $1$ and do this recursively. Then how could I continue?
Recall that $g$ is uniformly continuous; let $\omega$ be its modulus of continuity.
Divide the interval $A$ into $kn$ equal subinterval, where $k$ is large. Let's say the subintervals are $I_1,\dots,I_{kn}$. For $j=1,\dots, n$ define $$ B_j = \bigcup_{r=0}^{k-1} I_{j+rn} $$ Simply put, $B_j$ is the union obtained by taking every $n$th interval starting with $I_j$. The key point is for $j,\ell\in \{1,\dots,n\}$ $$ \left|\int_{I_j+rn}g-\int_{I_\ell+rn}g\right| \le \frac{|A|}{kn}\omega(|A|/k) $$ which follows from the uniform continuity of $g$. (The number $|A|/k$ bounds the offset between the two intervals of integration.) Sum over $r$ to get $$ \left|\int_{B_{j}}g-\int_{B_{\ell}}g\right| \le \frac{|A|}{n}\omega(|A|/k) $$ Note that the right-hand side is small when $k$ is large. So, the integral over all sets $B_{j}$ are about the same. In particular, taking $$A_k=B_1\cup\dots\cup B_m$$ achieves the goal $$ \int_{A_k}g\to \frac{m}{n}\int_A g $$