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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?

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Actually, 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. –  Stefan Walter Dec 16 '11 at 14:39
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This is a special case of the general definition for the Borel $\sigma$-algebra on the codomain (and the Lebesgue $\sigma$-algebra on the domain). –  Qiaochu Yuan Dec 16 '11 at 14:40
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A related MO thread. –  t.b. Dec 16 '11 at 14:40
    
when 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 $\mathbb{R}$ –  yoyo Dec 16 '11 at 14:43
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Actually, 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". –  GEdgar Dec 16 '11 at 15:47
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1 Answer 1

The definition you stated is equivalent to:

The inverse image of any Borel set is measurable

but when the definition is stated for open sets it can be easier to verify in particular instances. Actually it can be narrowed further, to just say that the inverse images of some sufficient collection of intervals are all measurable.

There are two reasons not to look at functions with the property that the inverse image of any Lebesgue measurable set is Lebesgue measurable:

1) For the purposes of Lebesgue integration, the subsets of $\mathbb{R}$ that we need to take inverse images of are Borel sets, so for this purpose there is no need for further generality.

2) It is not even true that every continuous function $\mathbb{R} \to \mathbb{R}$ has the property that the pre-image of every Lebesgue measurable set is Lebesgue measurable. This was pointed out by Nate Eldridge in this MO answer.

All that we need to construct an example is a continuous bijection $q$ between a set of positive measure $A$ and a set of zero measure $B$. The set $A$ will have a non-Lebesgue-measurable subset $M$, but $q(M)$ will be a subset of a measure zero set and hence $q(M)$ will be measurable. Being a bijection, $q$ will pull back the measurable set $q(M)$ to the non-measurable set $M$.

To take one example from Folland's Real Analysis, exercise 2.2.9, let $g(x) \colon [0,1] \to [0,2]$ be $x + f(x)$ on $[0,1]$, where $f(x)$ is the so-called Cantor function $$ f(x) = P(C \cap [0,x]) $$

where $P$ is the standard fair-coin probability measure on the Cantor set $C$. Then $g$ is continuous (because $P$ is non-atomic) and $g$ is strictly increasing, so it is bijective, and $g(C)$ has positive Lebesgue measure. Since $[0,1]$ is compact, $g$ is an open map, by standard topology, so $q = g^{-1}$ is a continuous bijection with the desired property: $M = g(C)$ has positive measure and $q(M) = C$ has measure $0$.

Going further, this MO question has a hint about how to make an example $q$ that is $C^\infty$.

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