# finding a finite-measure open cover of a finite-measure set of reals

This is probably false, but I'll ask anyway, just in case.

Conjecture 1. Suppose $$E\subset[0,\infty)$$ has finite Lebesgue measure. Then there is a subset $$E_0$$ of measure zero such that $$E\setminus E_0$$ admits a countable and order-respecting (in the sense that $$I_k) cover of disjoint intervals $$\{I_k\}_{k=1}^\infty$$, where $$\bigcup_{k\in\mathbb{N}}I_k$$ also has finite measure.

Since $$\mathbb{R}$$ is Lindelof under the Lebesgue topology, it should be enough to find an open cover $$\{U_\alpha\}_{\alpha\in I}$$ of $$E$$ such that $$\bigcup_{\alpha\in I}U_\alpha$$ has finite measure. Then we convert to intervals and pass to a countable subcover. Order the intervals $$(I_k)_{k=1}^\infty$$ according to their infimum, and throw out any $$I_k$$ satisfying $$\sup I_k<\sup I_j$$ for some $$j. Then relabel $$I_k$$ to denote $$I_k\setminus\bigcup_{j. And that should do the trick.

So, really, Conjecture 1 is equivalent to:

Conjecture 2. If $$E\subset[0,\infty)$$ has finite Lebesgue measure, then there is a subset $$E_0$$ of measure zero such that $$E\setminus E_0$$ admits an open cover $$\{U_\alpha\}_{\alpha\in I}$$, where $$\bigcup_{\alpha\in I}U_\alpha$$ also has finite measure.

To disprove Conjecture 1, it is enough to prove this anti-conjecture:

Conjecture 3. For any $$\epsilon\in(0,1)$$, there is a measurable subset $$K_\epsilon$$ of $$(0,1)$$ such that $$\mu(K_\epsilon\cap(a,b))>0$$ for any $$0, and $$\mu(K_\epsilon)<\epsilon$$. (Here, $$\mu$$ is the Lebesgue measure.)

If Conjecture 3 is true, then let $$E=\bigcup_{j=0}^\infty(j+K_{2^{-j}})$$. That will break Conjecture 1.

Conjecture 1 is false and in fact there is a counterexample $$E$$ which is an open set. Just take $$E$$ to be any dense open set of finite measure (to construct such a dense open set, enumerate the rationals and take an interval of length $$1/2^n$$ around the $$n$$th rational). Then $$E\setminus E_0$$ is still dense for any null set $$E_0$$. If intervals $$I_k$$ existed as in the conjecture, then the right endpoint of $$I_k$$ would have to be the same as the left endpoint of $$I_{k+1}$$ for each $$k$$ (since if they were different, $$E\setminus E_0$$ would fail to be dense in the interval between them), and so their union contains all but countably many points and has infinite measure.
Conjecture 2 is true, though. Indeed, for at least one definition of Lebesgue measure it is true immediately from the definition. Namely, the Lebesgue (outer) measure of a set $$E$$ is defined as the infimum of all sums of lengths of countable collections of open intervals that cover it. So, if $$E$$ has finite measure, there must exist a cover by open intervals whose union has finite measure (indeed, measure arbitrarily close to that of $$E$$).
Your argument deducing Conjecture 1 from Conjecture 2 is wrong; in particular, you cannot necessarily order the intervals as $$(I_k)$$ by their infima, since the set of the infima of the intervals may not be order-isomorphic to $$\mathbb{N}$$. (For instance, think about the intervals that form the complement of a Cantor set.)
Conjecture 3 is also true. Just fix an enumeration $$(I_n)$$ of all the intervals in $$(0,1)$$ with rational endpoints, and pick a subinterval $$J_n\subseteq I_n$$ of length at most $$\epsilon/2^{n+1}$$ for each $$n$$. Then $$K_\epsilon=\bigcup J_n$$ will intersect every subinterval of $$(0,1)$$ on a set of positive measure but $$K_\epsilon$$ has measure at most $$\epsilon$$.