Approximate partition of unity by characteristic functions Consider a measure space $(X, \Sigma, \mu)$ which is assumed to be complete, finite and without atoms. I would like to show the following:

Let $v_1,  \ldots, v_n \in L^\infty(\mu)$ be given, such that $v_i \ge 0$ for all $i = 1, \ldots, n$, and $\sum_{i=1}^n v_i \equiv 1$. Then, there are measurable sets $E_{ij}$, $i = 1,\ldots, n$, $j \in \mathbb{N}$, such that:
  
  
*
  
*$\Omega = E_{1j} \cup \ldots \cup E_{nj}$ is a disjoint partition
  
*$ \chi_{E_{ij}} \stackrel\star\rightharpoonup v_i$ in $L^\infty(\mu)$ as $j \to \infty$ for all $i = 1,\ldots, n$.
  

Roughly speaking, we would like to approximate the partition of unity $1 = \sum v_i$ by the partition into characteristic functions $1 = \sum \chi_{E_{ij}}$.
I am pretty sure that this is possible and I think I have a messy proof in the case that $X$ is a measurable subset of $\mathbb{R}^n$ equipped with the Lebesgue measure. Hence, my question is whether there is a nice and elegant proof of the above statement.
 A: Say $\mu$ is a non-atomic probability measure. We can assume $L^1(\mu)$ is separable: Let $B$ be the $\sigma$-algebra generated by the $\nu_j$, and let $L^1_B(\mu)$ be the $B$-measurable integrable functions. Then $L^1_B$ is separable, so if we prove the result for separable $L^1$ then we get sets converging weak* to $\nu_j$ in $L^1_B(\mu)^*$. Now if $f\in L^1(\mu)$ let $g$ be the conditional expectation $g=\Bbb E [f|B]$; then $\int_{E_i,j}f=\int_{E_{i.j}}g\to\int g\nu_i=\int f\nu_i$.
Edit: It's been pointed out that there's a problem here: $\mu|_B$ need not be non-atomic. We only need the following lemma:
Lemma  Suppose $(X,A,\mu)$ is a non-atomic measure space and $S\subset A$ is countable. There exists a $\sigma$-algebra $B$ with $S\subset B\subset A$ such that $(X,B,\mu)$ is non-atomic and separable.
The lemma "must" be true, and "can't" be all that hard. A possible proof is at the bottom of this post; for now we proceed assuming the lemma holds:
So assume $\mu$ is a non-atomic probability measure such that $L^1(\mu)$ is separable:
There exists a countable collection of simple functions dense in $L^1$. So there exist countably many measurable sets $A_j$ such that the span of the characteristic functions of the $A_j$ is dense. So it's enough to show there exist sets $E_{i,j}$ such that $$\lim_{j\to\infty}\mu(A_k\cap E_{i,j})=\int_{A_k}\nu_i$$for every $k$.
Let $B_N$ be the $\sigma$-algebra generated by $A_1,\dots,A_N$. Since $B_N$ is a finite $\sigma$-algebra there exists a finite partition $F_N$ of $X$ such that the elements of $B_N$ are precisely the unions of subsets of $F_N$. Choose $E_{i,N}$ in such a way that for every $S\in F_N$ we have $$\mu(S\cap E_{j,N})=\int_S\nu_j.$$Then for every $k$ we actually have
$$\mu(A_k\cap E_{i,j})=\int_{A_k}\nu_i\quad(j>k)$$since if $j>k$ then $A_k$ is a union of the elements of $F_j$.

The following is either a proof of the lemma or not.
Say a collection of sets $T$ is divisible if for every $E\in T$ there exist $E_0,E_1\in T$ such that $E_j\subset E$, $E_0\cup E_1=E$, $E_0\cap E_1=\emptyset$, and $\mu(E_j)=\mu(E)/2$. For every $E\in A$ choose a partition $(E_L,E_R)$ into two sets of measure $\mu(E)/2$.
In general let $a(T)$ be the algebra generated by $T$ and let $$d(T)=T\cup\{E_L,E_R:E\in T\}.$$Say $S=S_0$ is as in the lemma, and define $$S_{n+1}=a(d(S_n)).$$ Let $$S_\omega=\bigcup_{n=0}^\infty S_n.$$Then $S_\omega$ is a countable divisible algebra.
Let $B$ be the collection of all $E\in A$ such that for every $\epsilon>0$ there exists $F\in S_\omega$ with $$\mu(E\Delta F)<\epsilon.$$Then $B\subset A$ is a separable $\sigma$-algebra containing $S$, and for every $E\in B$ there exists $E'\in  B$ with $E'\subset E$ and $$|\mu(E)/2-\mu(E')|<2\epsilon,$$so $(X,B,\mu)$ is non-atomic.
