# An “elementary” proof that “small” subsets can not exhaust the interval?

Suppose $$S_1\subseteq S_2\subseteq \ldots \subset I=[0,1]$$ is a such that each $$I\setminus S_j$$ contains a finite union of segments of total length $$\varepsilon>0$$. Then $$\cup S_j\neq I$$.

Here is a proof: By continuity of outer Lebesgue measure from below $$S=\cup S_j$$ has outer measure at most $$1-\varepsilon$$, and, in particular $$S\neq I$$.

I am interested in finding a proof that avoids measure theory and is as simple as possible.

Perhaps one can start by replacing $$S_j$$s by their closures $$\Sigma_j$$. Then one could apply the classification of closed sets in $$I$$ to say that each $$I\setminus \Sigma_j$$ is a countable union of intervals then do something with those intervals. Anything simpler?

• in this case, mimicking the inner measure of $I \setminus S_i$ seems more intuitive, but is in no way simpler – dcolazin May 12 at 14:29
• This statement implies Lebesgue dominated convergence theorem for step functions, which fairly straightforwardly implies Lebesgue dominated convergence theorem itself. – Max May 14 at 12:31

Let $$U_i=int (I\setminus S_j)=I\setminus \bar{S}_j$$.
These are monotone decreasing (countable) unions of open intervals $$I_{i,j}$$, of total length of $$l_i=l(U_i)=\sum_j l(I_{i,j}) \geq \varepsilon$$. We want to build a sequence of nonempty nested closed sets $$K_i\subset U_i$$. Then of course we would have $$\cap U_i \supseteq \cap K_i \neq \emptyset$$.
Fix a sequence of positive numbers $$\varepsilon_i$$, summing up to $$\varepsilon/2$$.
For each $$i$$, there is union of finite number of closed segments $$C_i=\cup J_{i,j}$$ inside $$U_i$$ of total length at least $$l(C_i)=\sum_j l(J_{i,j})> l(U_i)-\varepsilon_i$$. We let $$K_i=\cap_{k=1}^{i} C_k$$. Clearly, these are nested closed subsets. We claim that each $$K_i$$, which is a finite union of closed segments, has total length at least $$l(U_i)- (\sum_{k=1}^{i} \varepsilon_i)>0$$, hence in particular, non-empty, proving what we want. The claim about the length follows from the fact that $$U_i\setminus K_i \subseteq \cup_{j=1}^{i} U_j\setminus C_j$$ (if you are not in $$K_i$$, then you are not in one of the $$C_j$$'s), and so has length of at most $$(\sum_{k=1}^{i} \varepsilon_i)$$.