all borel sets are measurable

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Theorem

Every open subset of $R$ is measurable (ie $B \subset M$ - the borel sigma algebra is contained in the collection of all measurable sets)

Proof

Since intervals of the form $(a, \infty)$ are measurable for all $a \in R$ (open intervals with upper bound infinity are measurable), so is

(- \infty, b) = ⋃_{n = 1}^{\infty} (- \infty, b - \frac{1}{n}] = ⋃_{n = 1}^{\infty} {(b - \frac{1}{n}, \infty)}^{c}

This is because the measurable sets form a sigma algebra and they are therefore closed under complements, countable unions, and finite intersections. Thus any finite open interval is also measurable since

(a, b) = (- \infty, b) \cap (a, \infty)

Finally, every open subset of $R$ is a countable union of open intervals. Thus all open intervals are measurable.

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References

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all borel sets are measurable

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