inverse image of measurable functions of all borel sets are measurable

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Theorem

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

Proof

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

\begin{aligned} f^{- 1} ((a, b)) & = f^{- 1} ([- \infty, b) \cap (a, \infty]) \\ = f^{- 1} ([- \infty, b)) \cap f^{- 1} ((a, \infty]) \end{aligned}

Both sets of RHS are measurable, and thus so is their intersection. So each open interval is measurable.

Thus $f^{- 1} (U)$ is measurable for all open $U \subset R$ . But then

A = {F \subset R : f^{- 1} (F) is measurable}

is a sigma-algebra that contains all open sets, and thus $B$ bust be a subset of $A$ .

\begin{matrix} ◼ \end{matrix}

References

See Also

Mentions

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