If I define a measurable function $f:E \to \Bbb R \cup \pm\{\infty\}$ where $E \subset \Bbb R^d$ is a measurable set as if $\forall a \in \Bbb R$, the set
$$f^{-1}([-\infty,a))=\{x \in E :f(x) < a\}$$ is measurable. Under this assumption is it true that inverse image of a borel set under a measurable function is borel set ? I think not we can only say it is measurable.
Best Answer
It depends on the topology of $E$. Here is a simple example showing $f$ might not be Borel.
By definition, a function is Borel-measurable if the preimage of every open set is Borel. However, it can be shown that it suffices to show the the preimage of every half ray $[-\infty, a )$ is Borel.
Now the exampple. Set $E = \{0,1,2\}$ where the probability (measure) of each point is $1/3$. Now let $f(x)=x$. It is measurable since every subset of $E$ is measurable (this applies also to your definition).
Equip $E$ with the indiscrete topology. That is, $E$ and $\emptyset$ are the only open sets. This implies that the Borel $\sigma$-algebra is $\{\emptyset,E\}$.
Consider $f^{-1}([-\infty,2))=\{0,1\}$. It is the inverse image of a open set but $\{0,1\}$ is not a borel set of $E$. Hence $f$ is not a Borel-measurable function.
Lebesgue gave a characterization of all Borel functions in terms of pointwise convergence and continuity. But it requires a basic knowledge of the Borel hierarchy. Roughly speaking, functions that can be obtained : (1) initially, as the pointwise limit of continous functions (call this class 1) and (2) as the pointwise limit of functions of the previous class (there are called Baire classes). have a tight relationship to each level of the Borel hierarchy. Their preimages lie at each level.
If you are interested in these topic, look at A. Kechris's book Classical Descriptive Set Theory, which is the traditional discipline where Borel sets are studied in depth.