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I want to prove that $[0,1]$ does not have measure zero but the book says $``$Explain why the following observation is not a solution to the problem: Every open interval that contains $[a,b]$ has length $> b-a",$ which was exactly identical to my initial reasoning.

Why is this bad reasoning?

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    $\begingroup$ The same specious reasoning would imply that the set $\{a, b\}$ (consisting of only two points) has positive measure. $\endgroup$
    – user169852
    Commented Jan 26, 2019 at 8:21
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    $\begingroup$ For others, I think this question merits reopening. Perhaps David could provide a specific reference for "the book", but even without such a reference, the question seems to be something that most anyone might naturally consider when first learning about Lebesgue measure at this level, and thus (to me, at least) the question is "relevant to you and our community". (Besides, I wrote what I thought was a nice detailed answer to complement the answer by twnly before I realized further answering is disabled!) $\endgroup$
    – Dave L. Renfro
    Commented Jan 26, 2019 at 10:18
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    $\begingroup$ In a nutshell, you have to consider not only coverings of the set by a single interval, but coverings by finitely or countably infinitely many intervals. As the answer by twnly demonstrates, sometimes you can achieve a more efficient covering (as measured by the sum of the lengths of the intervals in the covering) by using infinitely many intervals. For this problem, you'd have to show that even if you use countably many intervals $\{I_n\}_{n=1}^{\infty}$ to cover $[a,b]$, there's no way to do it without having $\sum_{n=1}^{\infty}m(I_n) \geq b-a$ $\endgroup$
    – user169852
    Commented Jan 26, 2019 at 23:57
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    $\begingroup$ As a hint to simplify the problem, you can use the compactness of $[a,b]$ to reduce any infinite covering of $[a,b]$ by open intervals to a finite subcover. $\endgroup$
    – user169852
    Commented Jan 27, 2019 at 0:01

2 Answers 2

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To see why this reasoning doesn't work, this following statement is true but does not imply that the set is not of measure 0.

Every open interval that contains all the rationals in $[0,1]$, or $\mathbb{Q} \cap [0,1]$, is of length $> 1 - \epsilon$, but the rationals are of measure 0.

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  • $\begingroup$ I still cannot figure why the statement can imply the set is not of measure 0. Isnt the example you gave was about set instead of intervals? $\endgroup$
    – David
    Commented Jan 26, 2019 at 1:39
  • $\begingroup$ I am just saying that following your reasoning in the question, we can also show that rationals are of measure greater than zero, just by replacing $[a,b]$ by the rationals in $[a,b]$, which is not true. And thus the reasoning is not valid. $\endgroup$
    – twnly
    Commented Jan 26, 2019 at 1:56
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twnly has given an example that shows acceptance of the reasoning can lead to a false result. This might be sufficient to answer the book’s question (it will depend on how your teacher interprets “Explain why”, which should be made clear in advance if the answer is to be graded for credit), but here’s an explanation that is a bit more constructive (in the mathematical sense) because I’ll point out where and how the argument breaks down. (Later: In a later comment @Bungo has given essentially the same explanation.)

(Technical comment for others) The reason I say “constructive” is that I’m thinking analogously of situations where one proves the existence of a mathematical object by a constructive existence proof rather than by a non-constructive existence proof. The answer twnly gave is analogous to a non-constructive existence proof.

To show that $[0,1]$ doesn’t have measure zero, you have to show there exists $\delta > 0$ such that no matter how you cover the set with COUNTABLY MANY open intervals, the sum of the lengths of the intervals in your covering is $\geq \delta$ (or equivalently, $> \delta).$ The argument you’re presenting only shows $\delta$ exists no matter how you cover $[0,1]$ with a SINGLE open interval. Specifically, any positive real number $\leq 1$ can be a choice for $\delta.$ Thus, there still remains to be proved that no such $\delta$ exists for coverings of $[0,1]$ by two open intervals, and no such $\delta$ exists for coverings of $[0,1]$ by three open intervals, and …, and no such $\delta$ exists for coverings of $[0,1]$ by countably many open intervals. Thus, not all of the cases that need to be considered have been considered (whether by a correct argument or by an incorrect argument).

Technical Note: There are 4 versions of what “countable cover” could mean, depending on whether (a) “countable” means “countable or finite” OR “countably infinite”, and (b) “cover” means a set of open intervals OR a sequence of open intervals. All 4 versions are equivalent for the purpose of defining Lebesgue measure. For example, any finite sequence of open intervals can be replaced with a countably infinite sequence of open intervals having the same union and the same sum of lengths by including countably many occurrences of the empty set. Also, if the coverings have to be sets of open intervals and we have appropriate sequences of open intervals available, then finite or infinite repetitions of open intervals in a sequence of open intervals can be replaced with one occurrence for each case of repetition, and the duplicates that originally appeared can be replaced with countably many open intervals, all different from each other and different from any open intervals already appearing, such that the sum of the lengths of the replacement intervals is arbitrarily small. An actual complete proof that all 4 versions are equivalent can be done by proving $\;2 \cdot {4 \choose 2} = 12\;$ implications (Version 1 implies Version 2, Version 2 implies Version 1, Version 1 implies Version 3, Version 3 implies Version 1, Version 2 implies Version 3, etc.), but in practice one usually tries to arrange things so that only 4 implications are needed --- Version 1 implies Version 2 implies Version 3 implies Version 4. Occasionally more than 4 implications might be proved in such situations, say Version 1 implies Version 2 implies Version 3 AND Version 2 implies Version 4 AND Version 4 implies Version 2, such as when some of the implications have proofs that are of special interest.

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