I am just curious about why a function $f:\mathbb R \rightarrow \mathbb R $ is said to be Lebesgue measurable if $f^{-1}(U):=V$ is measurable, whenever $U$ is open in the reals. This seems to be counter to the more general definition between sigma algebras that says that the inverse image of a measurable set is measurable. Why don't we use this last definition instead of the open set one?
Definition of Measurability: Why use Open Sets?
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measure-theory
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0Actually, this last definition is more common than the first one, for exactly the reason you mention. They are equivalent, of course, but this needs to be proven. – 2011-12-16
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3This is a special case of the general definition for the Borel $\sigma$-algebra on the codomain (and the Lebesgue $\sigma$-algebra on the domain). – 2011-12-16
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3A related [MO thread](http://mathoverflow.net/questions/31603/why-do-probabilists-take-random-variables-to-be-borel-and-not-lebesgue-measurab). – 2011-12-16
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0when dealing with topological spaces, one often uses the sigma algebra generated by the topology. the lebesgue $\sigma$-algebra is the completion of the $\sigma$-algebra generated by the usual topology on $\mathbb{R}$ – 2011-12-16
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1Actually, we want to define measurable function from a measure space to the reals, where the reals is considered to be the space of scalars, and *not* as a measurable space. If you want, you can use this definition: $\{x:f(x)>t\}$ is measurable for all real $t$. No mention of your objectionable "open sets". – 2011-12-16
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0@Stefan: Those two definitions are not, in general equivalent; the link that t.b. posted has some details. – 2011-12-16
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0@Carl: Thanks for the correction. I overlooked the word "Lebesgue". – 2011-12-16