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I am trying to show that the family of sets whose inverse images are measurable is a $\sigma$-algebra, and the only part I am really having difficulty with is showing that the empty set is contained in the family.

Might be a stupid question, but how do I show that $f^{-1}(\emptyset)$ is measurable?

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    $\begingroup$ For a real (or more generally complex) valued function $f$, can $f(\text{something}) \in \emptyset$? $\endgroup$ – Cameron Williams Oct 28 '15 at 23:59
  • $\begingroup$ $\emptyset$ itself, I suppose. Is that all there is? $\endgroup$ – ALannister Oct 29 '15 at 0:52
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    $\begingroup$ The inverse image by $f$ of $B$ is denoted by $f^{-1}(B)$ and it is defined as $$ f^{-1}(B)= \{ x \:|\: f(x)\in B\}$$ So $$ f^{-1}(\emptyset)= \{ x \:|\: f(x)\in \emptyset\}= \emptyset$$ So the inverve image of $\emptyset$ is just $\emptyset$ itself and it is measurable. $\endgroup$ – Ramiro Oct 29 '15 at 3:00
  • $\begingroup$ However, it is worth to note that a family of sets whose inverse images by a function $f$ are measurable may not be a $\sigma$-algebra, if $f$ is not surjective. $\endgroup$ – Ramiro Oct 29 '15 at 3:09
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    $\begingroup$ @JessyCat You're on the right track. You can't have an element sent to $\emptyset$ (since $f$ only takes real (or complex) values.. it doesn't take the empty set as a value), thus the inverse image of $\emptyset$ is $\emptyset$. $\endgroup$ – Cameron Williams Oct 29 '15 at 3:24

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