Prove that if $f: X \rightarrow \mathbb{R}$ is measurable, and $g: \mathbb{R} \rightarrow \mathbb{R}^*$ is Borel measurable

then $g \circ f:X \rightarrow \mathbb{R}^*$ is measurable.

Could someone please explain to me first, what it means for a function to be Borel Measurable, I can't seem to find an exact definition.

So far the best one that I have is that a function is Borel Measurable if the pre-image of any open set is again a Borel set.

Now I try to use the definition that a for a measure space $(X, \mathcal{A}, \mu)$ and topological space $(Y,T)$ with $f: X \rightarrow Y$, $f$ is measurable if and only if

$f^{-1}(B) \in \mathcal{A}, \forall B \in B(Y,T)$

so I try to say that let $A \in \mathcal{A}$ be an open set.

Now $(g \circ f)^{-1}(A) = f^{-1}(g^{-1})(A)$ and since $g$ is Borel measurable would that mean that $g^{-1}(A) \in B(Y,T)$? so then we would have $f^{-1}(g^{-1}(A)) \in \mathcal{A}$ and hence $g \circ f $ is measurable?

  • $\begingroup$ Your definition of Borel measurable is correct. $\endgroup$
    – Batman
    May 20, 2014 at 2:01

1 Answer 1


I think I have a better solution to this question,

$f: X \rightarrow Y$ is Borel Measurable if $A \in \mathcal{B}(Y,T)$ then $f^{-1}(A) \in \mathcal{B}(X,T)$

Now with this new definition:

Let $B \in \mathcal{B}(Y,T)$

Now $(g \circ f)^{-1}(B) = f^{-1}(g^{-1})(B),$ and since $g$ is Borel Measurable we have that $g^{-1}(B) \in \mathcal{B}(Y,T)$

Finally, since $f$ is measurable, for the measure space mentioned above $(X, \mathcal{A}, \mu)$ we have the $f^{-1}(g^{-1}(B)) \in \mathcal{A}$ and hence $g \circ f$ is measurable.

  • $\begingroup$ Fixed it. Would that be a complete proof? $\endgroup$
    – Pablo
    May 20, 2014 at 2:16
  • $\begingroup$ Yes. Also, minor typo in the definition of Borel measurable. $\endgroup$
    – Ink
    May 20, 2014 at 2:18
  • $\begingroup$ could you point it out to me? $\endgroup$
    – Pablo
    May 20, 2014 at 2:20
  • $\begingroup$ You write $A \in \mathcal{B}(Y, T)$, then $f^{-1}(A) \in \mathcal{B}(Y,T)$. It should be $f^{-1}(A) \in \mathcal{B}(X, T)$, unless we have $f: Y \to Y$. $\endgroup$
    – Ink
    May 20, 2014 at 2:21
  • 1
    $\begingroup$ No, given $f: X \to Y$ if $A \in \mathcal{B}(Y, T)$, then it's possible that $f^{-1}(A) \in \mathcal{B}(X, T)$. But you have $f^{-1} (A) \in \mathcal{B}(Y, T)$. $X$ and $Y$ could be different. $\endgroup$
    – Ink
    May 20, 2014 at 2:33

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