Diffuse-like decomposition of the segment $[0,1]$ in accordance with Lebesgue measure Consider the segment $[0,1]\subset\mathbb{R}$ and the standard Lebesgue measure $\mu$ on $\mathbb{R}$. I wonder 

if we can find such decomposition $A\sqcup B=[0,1]$, that for any subsegment $[a,b]\subset[0,1]$ we'd have $\mu(A\cap [a,b])=\mu(B\cap [a,b])$?

More particular case is that for any $x\in[0,1]$ we have $\mu(A\cap [0,x])=\mu(B\cap [0,x])$ and hence $\mu(A\cap [0,x])=\frac{x}{2}$.
Metaphor. Such decomposition can be assosiated with mixture of liquids. Suppose we have two liquids of equal amount. We bottle them, shake them up and then pour some into the glass. No matter how much we pour, the glass will contain equal amount of both liquids.
Ideas. We can decompose $[0,1]$ as $X\sqcup Y$ where $X=[0,1]\setminus\mathbb{Q}$ and $Y=[0,1]\cap\mathbb{Q}$. Here we have $\mu(X)=1$ and $\mu(Y)=0$. Maybe it's possible to describe a procedure of moving points from $X$ to $Y$ (i.e. excluding them from $X$ and including in $Y$) that will lead to desired decomposition. Another thought is to set $A_n=\bigsqcup_{k=0}^{2^{n-1}-1}[2k\cdot2^{-n},\ (2k+1)\cdot2^{-n})$ and $B_n=[0,1]\setminus A_n$ for all $n\in\mathbb{N}$. Then for any $n\in\mathbb{N}$ we'll have


*

*$A_n\sqcup B_n=[0,1]$

*$\mu(A_n)=\mu(B_n)$

*$|\mu(A\cap [a,b])-\mu(B\cap [a,b])|\leq 2^{-n}$ for any $[a,b]\subset[0,1]$
I wonder if we can take some kind of limit here raising $n\longrightarrow\infty$.
Origin. This question arose after I read this post. It's interesting whether the condition of Riemann integrability is crucial there or we could weaken it with Lebesgue integrability. In attempts to find counterexample I thought of above mentioned decomposition. If it exists then we could define $f(x)=1$ if $x\in A$ and $f(x)=-1$ if $x\in B$. That would be our counterexample because $\int_I f\,d\mu$ would be zero for any interval $I\in[0,1]$.
Generalization. Suppose $W\subset\mathbb{R}^n$ is a Lebesgue measurable set, and $\{r_i\}_{i=1}^k\subset\mathbb{R}_+$ such that $r_1+\ldots+r_k=\mu(W)$. Is it possible to choose such decomposition $A_1\sqcup\ldots\sqcup A_k=W$ that for any measurable $V\subset W$ we'd have $\mu(A_i \cap V)=r_i\cdot\frac{\mu(V)}{\mu(W)}$ for any $1\leq i\leq k$.
In other words, can we uniformly "blend" measurable subsets of a measurable set in any proportions? It seems to me like a very interesting result, provided it's true.
 A: The answer is that we cannot find such subsets $A$ and $B$. Of course, $A$ and $B$ need to be Lebesgue measurable so that $\mu (A)$ and $\mu(B)$ are well-defined.
Let us defined $\nu (C) = \mu (A \cap C)$ for all Lebesgue measurable subsets $C$. Then:


*

*$\nu$ is a measure on Lebesgue-measurable sets, as it satisfies all axioms;

*$\nu(C) = \mu(C)/2$ for all open intervals.


But open intervals generate the Borel sigma-algebra, so $\nu = \mu/2$ on Borel sets. Then their completion also satisfy this relation, so $\nu = \mu/2$ on Lebesgue measurable sets.
But then $\mu(A)/2 = \nu (A) = \mu (A \cap A) = \mu(A)$, so $\mu(A) = 0$, which contradicts our hypotheses.

If you want to describe a mixture of liquids exactly in proportion $1-1$, the most natural way is to define a function $f$ as the (local) proportion of the first liquid in the mixture. For a homogeneous mixing in equal parts, one would have $f \equiv 1/2$ almost everywhere, and indeed $\int_a^b f(t) dt = (b-a)/2$ is the quantity of the first liquid in the mix between points $a$ and $b$.
This point of view also behaves very well with respect to your "limiting process". Let $f_n = 1_{A_n}$ be the repartition of the first liquid. Then, for any measurable and bounded function $g$,
$$\lim_{n \to + \infty} \int_0^1 f_n (x) \cdot g(x) dx = \int_0^1 \frac{1}{2} \cdot g(t) dt,$$
which is to say, the sequence of functions $(f_n)$ converges weakly in $\mathbb{L}^1 ([0,1],\mu)$ to the function $f \equiv 1/2$. So, in this sense at least, the function $f$ is indeed what you get as a limit of your mixing process.
N.B.: If you want to prove the convergence above, and elementary way if to prove it when $g = 1_{[a,b]}$ for some $a<b$, then when $g$ is simple, and finally use and approximation argument - for instance in $\mathbb{L}^2$ - to extend it to all measurable and bounded $g$.
