I have a pretty dumb doubt.
For a function $f$ with $f: \mathbb{R}^d \rightarrow \mathbb{R}$, we say
$f$ is measurable if $f^{-1}((\infty, a))$ is (Lebesgue) measurable for every $a \in \mathbb{R}$.
Define $\mathcal{P}$ to be the set of all subsets of $\mathbb{R}$ such that for $P \in \mathcal{P}$, $f^{-1}(P)$ is measurable. Now note that $f^{-1}((\infty, b)) \cap f^{-1}((\infty,a))^c = (a,b)$, so all the open intervals lie in $\mathcal{P}$.
Also since we know that each open set in $\mathbb{R}$ can be written as countable union of disjoint open intervals, by countable additivity of the Lebesgue measure, we have
All the open sets in $\mathbb{R}$ is in $\mathcal{P}$ ($\dagger$)
Also by the properties of Lebesgue measure, we have countable union, intersects and complements of elements lie in $\mathcal{P}$ too, so we have
$\mathcal{P}$ is a sigma algebra ($\ddagger$)
- It follows from $\dagger$ and $\ddagger$ that $\mathcal{P}$ contains the Borel $\sigma$ algebra of $\mathbb{R}$ (since Borel $\sigma$ algebra is pretty much defined as the smallest $\sigma$ algebra containing all the open sets, so since $\mathcal{P}$ contains all the open sets the Borel set must be a subset of that), right ?
- It doesn't follow from the definition that inverse image of a Lebesgue measurable set should be a Borel set, right ?
- The why the definition of a measurable function is not strengthened to show that $f^{-1}(M)$ is Lebesgue measurable whenever $M$ is a measurable subset of $\mathbb{R}$ ? If you define $f: [0,1] \rightarrow [0,2]$ by $f(x) = c(x) + x$ (where $c(x)$ is the Cantor Function) then $f(x)$ is measurable by the original criterion but not by the modified one. What more do we gain by relaxing the definition to only have $f^{-1}(B)$ is Lebesgue measurable whenver $B$ is a Borel subset $\mathbb{R}$ ?