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I'm working on a past exam to help me study for my finals and I came across the following question:

Let $\varepsilon>0$ and define $A=\bigcup_{j=1}^\infty(x_j-\varepsilon,x_j+\varepsilon)$ where $x_j\in\mathbb{R}$. Suppose that $A\cap[0,1]$ is dense in $[0,1]$. Then $|A\cap[0,1]|=1$ (Lebesgue measure).

A hint is provided, suggesting the use of Lebesgue's Differentiation Theorem. On one hand, I don't see how to use the theorem, but, is it necessary? My line of thinking is as follows:

Since we're in $\mathbb{R}$ and $A$ is open (as a union of open sets), that means $A$ can be written as a countable union of disjoint open intervals, say $\bigcup_{j=1}^\infty(a_j,b_j)$. Assume the measure of $A\cap[0,1]$ is strictly less than $1$, say $1-\delta$ for some $\delta>0$. By the minimum fixed length of $b_j-a_j$, this implies there is some nondegenerate interval in $[0,1]$, which in turn implies there are elements with open sets disjoint from $A$, which is a contradiction since $A$ is dense in $[0,1]$.

In fact, if I'm thinking about the construction correctly, it would seem to contain all but finitely many (based on the idea that $b_j-a_j\geq2\varepsilon$ for each $j$). I realize my proof doesn't generalize to any higher dimensions (although something similar might work) since an open set can't be written as a countable union of disjoint open sets in higher dimensions. Will anyone critique my proof and possibly suggest a line of proof using Lebesgue's Differentiation Theorem? Thanks!

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Your reasoning is not all correct. It is possible for an open set to be dense in $[0,1]$ while having measure less than $1$. E.g., let $\{r_1,r_2,\ldots\}$ be the rational numbers in $[0,1]$, and take $B=\cup_{n=1}^\infty(r_n-4^{-n},r_n+4^{-n})$. Or take the complement of a fat Cantor set. –  Jonas Meyer Dec 13 '12 at 5:18
    
In your example, though, the "$\varepsilon$" changes as $n$ changes. I see your point (and I'm thankful for the insight), but I feel like my argument is still valid. –  Clayton Dec 13 '12 at 5:23
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The set $A \cap [0,1]$ is the union of open intervals of length at least $1/\varepsilon$ each (if $\varepsilon < 1$); except possibly for two intervals, which might be shorter. Thus there are finitely many intervals in the union $\bigcup_j (a_j,b_j)$. Now it follows that $|A \cap [0,1]| = 1$. –  Yury Dec 13 '12 at 5:24
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@Clayton: In that case, perhaps the important part of your argument was implicit rather than communicated to us. We can only judge what was written. –  Jonas Meyer Dec 13 '12 at 5:30
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I don't see how to meaningfully use LDT here. It is much easier to prove directly that $[0,1]\setminus A$ is finite (it contains at most $1 + 1/(2\varepsilon)$ points) and thus $\mu([0,1]\setminus A) = 0$. –  Yury Dec 13 '12 at 6:01

1 Answer 1

up vote 1 down vote accepted

Let $\varepsilon>0$ and $\{x_j\}_{j=1}^\infty\subseteq\mathbb{R}$. Define $$A=\bigcup_{j=1}^\infty(x_j-\varepsilon,x_j+\varepsilon).$$ Then, since $A$ is an open set in $\mathbb{R}$, we can rewrite $A$ as a disjoint union of open sets, say $$A=\bigcup_{j=1}^\infty\:(a_j,b_j).$$ Observe that $b_j-a_j\geq2\varepsilon$. Now, we can order these intervals from left to right on $[0,1]$ (since $0$ and $\frac{\varepsilon}{2}$ can be in at most two different intervals, we have a "first" interval). After ordering the intervals, note that if $a_{j+1}-b_j>0$, then $A\cap[0,1]$ is not dense in $[0,1]$, a contradiction. Thus, $a_{j+1}-b_j=0$ for all $j$ where $(a_j,b_j)\cap[0,1]\neq\varnothing$. $^{(*)}$ Observe that since $2\varepsilon>0$, there exists a positive integer $N$ such that $N(2\varepsilon)>1$, thus if we take $N$ distinct points in $[0,1]$, there can be at most $N-1$ points not in $A$ (otherwise $m(A\cap [0,1])\geq N(2\varepsilon)>1$, a contradiction). Hence, $A$ contains all but finitely many points of $[0,1]$; therefore, clearly $m(A\cap[0,1])=1$, as was to be shown.

$^{(*)}$Here we could also argue that that there are at most countably many points not in the union and deduce the result from that, but we show something stronger.

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