Definition of measurable function - connection with being measurable f is measurable if for all Borel sets B, the preimage is in the sigma-algebra.
This is basically saying that for all elements in the sigma algebra, f of it is a borel set.
What part of this definition explains the measurability of f— is it that borel sets are measurable? Thus, taking the function f of anything in the sigma algebra will produce a measurable output, hence the function is “measurable”?

The definition I refer to is as follows.
$f$ is measurable with respect to a measurable space $(X, A)$ if for every borel set B, $$f^{-1}(B)=\{x\in X:f(x)\in B\}\in A$$
This basically means: $f^{-1}(B) \in A$, the $\sigma$-algebra $\implies$ $f$ is measurable
Thus, $f$ is not measurable if its preimage of any Borel set is not contained in the $\sigma$-algebra. My question is as follows.
What is the meaning of this definition?

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*We have a $\sigma$-algebra mapping to another set that contains Borel sets. If those Borel sets have preimages all of which lay in the $\sigma$-algebra then $f$ is measurable. Is it because of the Borel sets that are measurable, that makes $f$ measurable?


If we associate $f$ with the range $f(X):= R $ then for borel subset $B$ of $R$, we can assign a measure to it $\mu_X (f^{-1}(B))$ so the function (in terms of its range) can be measured.
