The following citation is from Folland's Real Analysis.

Let $m$ denote the Lebesgue measure on $\mathbb{R}$ and $\{ r_j \}$ be an enumeration of the rational numbers in $[0,1]$, and given $E > 0$, let $I_j$ be the interval centered at $r_j$ of length $\epsilon \, 2^{-j}$. Then the set $U = (0,1)\cap \left( \bigcup_{1}^{\infty} I_j \right)$ is open and dense in $[0,1]$, but $m(U) < \sum_{1}^{\infty}\epsilon\, 2^{-j} = \epsilon$;

Since $U$ is a non-empty open set, $m(U)>0$. But why does the above argument show that $m(U)=0$?


It seems that you're arguing that $m(U) < \varepsilon$ for all $\varepsilon > 0$, so $m(U) = 0$.

But $U$ depends on the choice of $\varepsilon$, so the above argument doesn't work!

All this argument shows is that there are open and dense sets of arbitrarily small positive Lebesgue measure.

In other words, this is the difference between

$\qquad\exists U$ open and dense $\forall \varepsilon > 0 : m(U) < \varepsilon$ (which is not true)


$\qquad \forall \varepsilon > 0 \exists U$ open and dense $: m(U) < \varepsilon$ (which is what Folland shows).

  • $\begingroup$ That makes sense. I appreciate your answer. $\endgroup$ Apr 14 '15 at 13:20
  • $\begingroup$ Isn't the next point that $(0,1)\backslash U$ is a nowhere-dense set of positive measure, and by changing $\varepsilon$ that measure can be as close to $1$ as desired? $\endgroup$
    – Henry
    Apr 14 '15 at 20:12

The argument in the yellow box does not show that $m(U)=0$. It does show that for any $\epsilon>0$ there is a set $U\supset(\Bbb Q\cap [0,1])$ such that $m(U)<\epsilon$. The conclusion that can be immediately derived from all this is

$$m(\Bbb Q\cap [0,1])=0$$

This argument can be easily extended to show that the Lebesgue measure of any countable infinite set is zero. Isn't that the continuation in Folland's Real Analysis?

  • $\begingroup$ Thanks for your answer. Until seeing your comment, I mistakenly assumed that Lebesgue measure satisfies $\sigma$-additivity and concluded that the fact that one point set is measure zero implies any countable set is measure zero. This is clearly because I haven't perused the section 1.5 yet, so I should do as soon as possible. $\endgroup$ Apr 14 '15 at 19:03
  • 1
    $\begingroup$ @AranKomatsuzaki: The Lebesgue measure does satisfy $\sigma$-additivity, but perhaps that has not been shown in your book yet. The argument you showed is a quick and early way to show that any countable set has measure zero. At least, I've seen that argument used that way in at least one other place, and it also showed that for the rationals before showing it for all countable sets. $\endgroup$ Apr 14 '15 at 19:13
  • $\begingroup$ Thanks for solving my misunderstanding. Now the Lebesgue measure looks quite reasonable to me. $\endgroup$ Apr 14 '15 at 21:09

U is dependent on $\epsilon$ .

  • 4
    $\begingroup$ As already mentioned in the other, more complete, answer. $\endgroup$ Apr 14 '15 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.