all borel sets are measurable

[[concept]]

Topics

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Theorem

Every open subset of $\mathbb{R}$ is measurable (ie $\mathcal{B}\subset \mathcal{M}$ - 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 $a\in \mathbb{R}$ (open intervals with upper bound infinity are measurable), so is $(-\infty ,b)={\bigcup }_{n=1}^{\infty }\left(-\infty ,b-\frac{1}{n}\right]={\bigcup }_{n=1}^{\infty }{\left(b-\frac{1}{n},\infty \right)}^c$ 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 $\mathbb{R}$ is a countable union of open intervals. Thus all open intervals are measurable. \tag*{$\blacksquare$}

References

References

See Also

Mentions

Mentions

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