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Is it true that every non-empty open set has Lebesgue measure greater than zero?

I could think of a proof along the following lines but not sure if that would be right:

Since every non-empty open set is a finite or countable union of open intervals where at least one open interval is nonempty, and since every nonempty open interval has measure greater than zero implies every nonempty open set has measure greater than zero.

Is my reasoning sound?

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Yes, take a point $x \in A$, with $A$ open. Then exists $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon) \subset A$. So: $${\frak m}(x-\epsilon,x+\epsilon) \leq {\frak m}A \implies 0 < 2\epsilon \leq {\frak m}A \implies {\frak m}A > 0.$$

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Every non-empty open set contains at least one open interval, and open intervals have a positive measure.

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