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I'm reading a proof and in it they construct a set, then just assert they can choose a Borel Set contained in the set with a particular "size". Why can I do this? How do I know something is in there?

The explicit construction is ($(r_n) \in \mathbb{Q}$):

$$V_n = (r_n - \frac{1}{3^{n+1}},r_n + \frac{1}{3^{n+1}}) \\\ W_n = V_n - \bigcup_{k=n+1}^{\infty}V_k$$

They assert there's a Borel set, $A_n$ in $W_n$ with the property $0 < m(A_n) < m(W_n)$.

(The relevant post is here: Construction of a Borel set with positive but not full measure in each interval)

yoshi
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1 Answers1

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The (inner) regularity of Lebesgue measure implies that a closed set is contained in $W_n$ having measure arbitrarily less than the measure of $W_{n}.$ In fact, "closed" can be strengthened to "perfect and nowhere dense". See also Inner regularity of Lebesgue measurable sets.

Dave L. Renfro
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