Complement of set $\bigcup_{n\in \mathbb{N}} I_n$ has zero measure.

Question

Let $r_n$ be an enumeration of $\mathbb{Q}$ at $I$ where $I=[0,1]$ and $I_n=[r_n-4^{-n},r_n+4^{-n}]\cap I$ for each $n$.

Prove that the set $$D=(\bigcup I_n)^c$$ has zero measure.

I was trying to prove that given $f(x)=-1$ when $x\in\bigcup I_n$ and $0$ otherwise, if we had some step function $s(x)$ which satisfies $s(x)\leq f(x)$ for all $x\in I$, then $s(x)\leq -1$ almost everywhere on $I$, and the problem reduced to prove that $(\bigcup I_n)^c$ had zero measure.

How can I do that?

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To give a more profound insight about the original problem, I was trying to prove that there is a function $f\in U(I)$ such that $-f\notin U(I)$, where $U(I)$ is the set of all functions which are the limit of an increasing sequence of step functions almost everywhere on $I$, such that the limit of the integral of each element of the sequence also exists.

The "$f$" in this question is $-f$ in the original problem.

Answer 1

It's not that the complement has zero measure. Even more, since (as charMD says) $$\mu\left(\bigcup_{n\in\mathbb{N}}I_n\right)\leq \sum_{n=1}^\infty \mu(I_n)=\sum_{n=1}^\infty 2\cdot 4^{-n}=2\sum_{i=1}^\infty 4^{-n}=\frac{2}{1-\frac{1}{4}}-2=\frac{2}{3}<1=\mu(I),$$ that's definitely not the case.

But since $s$ is a step function, there is a partition $P$ of $I$ such that $s$ is constant in all the subintervals of $P$. So if $P=\{x_0,\dots,x_n\}$ and $x\in (x_{k-1},x_k)$ is such that $s(x)>-1$, then $s(x)>-1$ for all $x\in (x_{k-1},x_k)$. But we can find $N\in\mathbb{N}$ such that $I_N\subseteq (x_{k-1},x_k)$ and then we'd have that $f(x)=-1$ in $I_N$, and $s(x)> f(x)$ in $I_N$, a contradiction. Then if $s_n(x)>-1$, it must be for $x\in P$, a finite set, of measure zero, of course.