Equivalent conditions for a measurable function I am reading Stein and Shakarchi volume 3 and on page 28 they give the definition of a Lebesgue  measurable (real - valued) function $f: \Bbb{R}^d \to \Bbb{R}$ to be on in which for any $a \in \Bbb{R}$, $f^{-1}(-\infty,a)$ is Lebesgue measurable. Now I am wondering it this is equivalent to the following conditions:


*

*$f$ is a Lebesgue measurable function if for any Lebesgue measurable set $E$, $f^{-1}(E)$ is Lebesgue measurable

*For almost every $a \in \Bbb{R}$, $f^{-1}\left((-\infty,a)\right)$ is Lebesgue measurable.




My question is: Are these equivalent to the definition given in Stein and Shakarchi? 


 A: (2) is equivalent to the Stein-Shakarchi definition (which let's call (0)).  (1) is strictly weaker.
(0) is actually equivalent to the following statement: (3) there is a dense set $E \subset \mathbb{R}$ such that for every $a \in \mathbb{R}$, $f^{-1}((-\infty, a))$ is Lebesgue measurable.  To see (3) implies (0), note that for any $a \in \mathbb{R}$ we can find a sequence $a_n \in E$ with $a_n \uparrow a$.  Now observe that
$$f^{-1}((-\infty, a)) = f^{-1}\left(\bigcup_n (-\infty, a_n) \right) = \bigcup_n  f^{-1}((-\infty, a_n))$$
So $f^{-1}((-\infty, a))$ is a countable union of Lebesgue measurable sets, hence Lebesgue measurable.
Now (2) implies (3) because every set of full Lebesgue measure is dense.  (If it were not dense, its complement would contain an interval, but intervals have positive Lebesgue measure.)  Obviously (0) implies (2).
Clearly (1) implies (0).  However, the converse fails; for instance, every continuous function satisfies (0) (open sets are Lebesgue measurable) but there are continuous functions which do not satisfy (1).  For more information on this, see an answer I wrote on MathOverflow.
