(EDIT:) Definition of Atom considered for this problem: A set $A$ in $(Ω, \mathscr{A})$ is Atom if $A \in \mathscr{A}$, but no non-empty proper subset of $A$ belongs to $\mathscr{A}$.
(a) Show that if $f : (Ω, \mathscr{A}) \to \Bbb R \cup \{\pm \infty \}$ is a measurable function, then $f$ must be constant on each atom $A$ of $\mathscr{A}$.(b) Is the converse true ?
My attempt:
(a) For the sake of contradiction let us assume that $\exists A \in \mathscr{A}$ such that $f$ is not constant on $A$ i.e. $\exists A_1 \subset A$ such that $f(A_1) =c $ and $f(A- A_1) \ne c$. But then, $A_1 = f^{-1}(c) \cap A \in \mathscr{A}$ , contradicting the fact that $A$ is an atom of $\mathscr{A}$.
(b) If I am able to show that pre-image of a generating class of the Borel sigma algebra under $f$ is in $\mathscr{A}$, then the converse is also true. But how to proceed?
Thanks in advance for help!
I think you are interpreting the converse wrongly.The converse simply says that if every measurable function is constant on $A$ then $A$ is necessarily an atom. This is quite obvious: if $B$ is a proper non-empty subset of $A$ belonging to $\mathcal A$ then $I_B$ is a measurable function which takes both the values $0$ and $1$ on $A$. So we have a contradiction. [Def.: $I_B(x)=1$ if $x \in B$, $0$ otherwise]. Incidentally, the answer to the question in title is no; the atoms of the Borel sigma algebra on $\mathbb R$ are the singletons and every map is constant on singletons. Not every map is measurable. This makes it clear that you interpretation of the converse is wrong.