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?

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

I thought about it and here is a solution along the above lines.

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 $(\sum_{k=1}^{i} \varepsilon_i)$.


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