Let $f$ be a Lebesgue measurable function and $B$ be a Borel set. Show that $f^{-1}(B)$ is also measurable.

Attempt at the proof:

Suppose $f$ is Lebesgue measurable. Then $f^{-1}((\alpha,\infty))$ is measurable as well (by definition of a measurable function), $\forall\alpha\in\mathbb{R}$. We note that $(\alpha,\infty)\in\mathcal{B}$, where $\mathcal{B}$ is the Borel $\sigma$-algebra, since:

  • $\emptyset\in (\alpha,\infty)$
  • $(\alpha,\infty)^c=\mathbb{R}\setminus(\alpha,\infty)\in\mathcal{B}$
  • the infinite union of open sets is once again an open set

So, $(\alpha,\infty)$ is a Borel set. So let $B=(\alpha,\infty)$. Since $f$ is measurable, $f^{-1}((\alpha,\infty))$ is measurable and so $f^{-1}(B)$ is measurable.

However I'm doubtful that this is correct, since I didn't choose an arbitrary Borel set $B$. Can anyone nudge me in the right direction? Thanks.


1 Answer 1


$\{A| f^{-1}(A) \text{is Borel}\}$ is a $\sigma$-algebra containing all open intervals. What can you say now?

  • $\begingroup$ I think that means that $\{A\mid f^{-1}(A)\text{ is Borel }\}$ is in fact a Borel $\sigma$-algebra, which is relevant because $\mathcal{B}\subset \mathcal{M}$, so this means that $f^{-1}(A)$ is Borel and hence measurable? $\endgroup$ Commented Feb 18, 2015 at 2:20
  • $\begingroup$ Isn't $f^{-1}(A)$ on the right side supposed to be Lebesgue mesurable instead of Borel? $\endgroup$
    – Adayah
    Commented Apr 12, 2015 at 22:03

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