I was trying to prove that piecewise-constant functions on $\mathbb{R}$ are not dense in $L^\infty(\mathbb{R})$ and I was looking for a $L^\infty$ function that cannot be "approximated" with piecewise-constant functions.
At first I tried with the Dirichlet function $\chi_{\mathbb{Q}}$, but then I realized that $\chi_{\mathbb{Q}}=0$ in $L^\infty$ because $\mathbb{Q}$ has zero measure.
So I thought about this:
If $\mathbb{R}=D_1\sqcup D_2$ where $D_1,D_2$ are two dense disjoint measurable (with non-zero measure) subsets, then $\chi_{D_1}$ is an example of function in $L^\infty$ that cannot be approximated with piecewise constant functions.
But I cannot find such a decomposition. Does it exist?
Giulio Sassatelli made me notice that such a decomposition exists but it doesn't necessarily work as a counterexample. So I'd like to add the condition that the sets are "locally non-zero measure" (look in the comments for a definition).
Best Answer
Yes, such a decomposition exists.
(*update: As Nate Eldredge points out in the comments, your motivating question on $L^\infty$ denseness of piecewise constant functions is already contradicted by a single fat cantor set - no need for both the set and complement to be dense. Nonetheless here is such a decomposition.)
For each $\epsilon\in (0,1)$, let $C_{\epsilon}\subseteq [0,1]$ be a Cantor set of measure $\epsilon$, containing $0$ and $1$, whose complementary intervals in $[0,1]$ all have length at most $\frac{1}{2}$. If $I\subseteq \mathbb R$ is a compact interval, let $C_{\epsilon}^I\subseteq I$ be $C_\epsilon$ translated and scaled by an affine map from $[0,1]$ to $I$, so that $|C_\epsilon^I|=\epsilon |I|$.
For a closed set $S\subseteq \mathbb R$, let $\mathcal C(S)$ denote the family of complementary intervals to $S$.
Finally, let $\{\epsilon_k\}$ be a sequence with each $\epsilon_k>0$ such that $\sum_{k=0}^\infty\epsilon_k\leq \frac{1}{2}$.
Then construct an increasing sequence of closed nowhere dense sets $D_n$ for $n\in\mathbb N$ as follows:
Let $D_0=\mathbb Z+C_{\epsilon_0}$. Then having defined $D_k$, define $$D_{k+1}=D_k\cup\bigcup_{I\in\mathcal C(D_k)} C_{\epsilon_{k+1}}^I\text{.}$$
Finally let $D=\bigcup_{n=0}^\infty D_k$.
Observe first that $D$ is dense by construction, as the complementary intervals decline in length by a factor of at least $2$ at every stage of the construction. Moreover, as $D$ is a countable union of nowhere dense sets, the complement $D'=\mathbb R\backslash D$ is also dense, by the Baire Category Theorem.
Observe moreover that since $D'$ is dense, every open interval $J\subseteq \mathbb R$ contains a point $x\in D'$. Also, since the lengths of complementary intervals in the construction decay by a factor of at least two at each stage, and $x$ is in an intersection of such intervals, we have some $n$ for which the component $I$ of $x$ in $\mathbb R\backslash D_n$ is contained in $J$. But on that component $I$, we have $$0< |D\cap I|\leq \left(\sum_{k=n+1}^\infty\epsilon_k\right)|I|<\frac{|I|}{2}\text{.}$$ If follows that $|J\cap D|>0$ and $|J\cap D'|>0$.