inverse image of measurable functions of all borel sets are measurable

[[concept]]

Topics

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Theorem

If $E$ is measurable and $f:E\to \mathbb{R}$ a measurable function, then for all $F\in \mathcal{B}$ (the borel sigma algebra) we have $f^{-1}(F)$ is measurable.

Proof

If $f$ is measurable, then for all intervals $(a,b)$ we have

f^{-1}((a,b)) &= f^{-1}([-\infty,b)\cap(a, \infty]) \\ &= f^{-1} ([-\infty, b)) \cap f^{-1}((a, \infty]) \end{align}$$ Both sets of RHS are [[Concept Wiki/lebesgue measurable\|measurable]], and thus so is their intersection. So each open interval is measurable. Thus $f^{-1}(U)$ is measurable for all open $U \subset \mathbb{R}$. But then $${\cal A} = \{ F \subset \mathbb{R}: f^{-1} (F) \text{ is measurable} \}$$ is a [[Concept Wiki/sigma-algebra]] that contains all open sets, and thus ${\cal B}$ bust be a subset of ${\cal A}$. $$\tag*{$\blacksquare$}$$

Combined with the result that the inverse inverse image of infinity of measurable functions is measurable, this means that the inverse image of the borel sets of the extended reals are measurable for a measurable function!

References

References

See Also

Mentions

Mentions

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