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I was wondering if a Borel set $A \in \mathscr B (\mathbb R)$ has strictly positive Lebesgue measure ($\lambda(A) > 0$) iff it contains some interval $(a,b)$, $a<b$. I tried proving this, before I found a post with a few counterexamples, showing my 'proof' is obviously flawed. Could someone point out the error(s)?

My plan was to define a different measure for which it would be much easier to prove the claim (making it almost obvious), and then prove its equality to the Lebesgue measure for all Borel sets.

Define an equivalence relation on $A \in \mathscr B (\mathbb R)$ by:

$$ x \sim y \iff \text{for all $z$ between $x$ and $y$, we have } z \in A.$$

Then the equivalence classes of $A$ will be intervals (possibly of infinite length) or singletons. Name the set of all intervals, excluding the singletons, $\mathcal C_A$. All the intervals in $\mathcal C_A$ are disjoint, and there are only countably many of them, because each interval contains a rational number. Define:

$$ \mu (A) := \sum_{C \in \mathcal C_A} \lambda( C).$$ Then $\mu$ is a measure coinciding with $\lambda$ on all half-open intervals $[a,b)$, which are known to form a semi-ring $\mathscr S$. This $\mathscr S$ is stable under finite intersections and has an exhausting sequence (on which both $\mu$ and $\lambda$ are finite), so $\mu$ and $\lambda$ must also coincide on $\sigma(\mathscr S) = \mathscr B (\mathbb R)$.

One possibly erroneous statement is the claim that $\mu$ is a measure on $\mathscr B (\mathbb R)$. Any thoughts?

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  • $\begingroup$ You can't just throw away the singletons if there are uncountably many of them. For example, $\mathbb{R}\setminus \mathbb{Q}$ contains no intervals, and is therefore entirely singletons in the context of your usage, but has full measure in any interval $I\subset \mathbb{R}$. Of course, you can't produce $\mathbb{R}\setminus \mathbb{Q}$ as a finite intersection of half-intervals, but it is still a Borel set. $\endgroup$
    – Michael L.
    Aug 28, 2017 at 9:33

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You are correct that $\mu$ is not a measure since it is not even finitely additive. Note that $\mu(\mathbb{Q}) = 0 = \mu(\mathbb{R} - \mathbb{Q})$ since neither set contains any interval of positive length but $\mu(\mathbb{R}) = \infty$

Edit: This answer is essentially the same idea as in the comment given by Michael Lee. There it is noted that $\mathbb{R}-\mathbb{Q}$ has "the wrong measure". I exploit this to break additivity.

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If $A$ is the Cantor set, then its equivalence classes are all singletons. There are uncountably many of them; note a singleton is an interval and need not contain a rational. You have an uncountable sum $\sum\lambda(C)$.

You may object that this is irrelevant; in this example one is adding uncountably many zeroes to get zero. But there are "fat" Cantor sets: these are homeomorphic to the standard Cantor set, yet have positive measure. In this case, can you add uncountably many zeroes to get exactly the right nonzero number?

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  • $\begingroup$ I see that this is where it differs from the Lebesgue measure, yet they still coincide on all the half-open intervals. So is my use of that theorem incorrect? $\endgroup$
    – SvanN
    Aug 28, 2017 at 9:33
  • $\begingroup$ Your $\mu$ is not a measure. It gives $0$ on a fat Cantor set $A$, yet the complement of $A$ within a bounded open interval $I$ is a union of open intervals, so $\mu(I\setminus A)=\lambda(I\setminus A)$. But $\mu(A)=0<\lambda (A)$ and $\mu(I)=\lambda(I)$ so we don't have finite additivity. $\endgroup$ Aug 28, 2017 at 9:38
  • $\begingroup$ I was just about to realise that myself. Thank you for your help! $\endgroup$
    – SvanN
    Aug 28, 2017 at 9:39

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