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Let $(X, \mathcal{X})$ and $(Y,\mathcal{Y})$ be measurable spaces, and $f: X \to Y$ a measurable function. Let $A \in \mathcal{X}$ be a measurable subset of $X$.

Is it guaranteed that $f_{\mid A}: A \to Y$ is measurable?

The measurable space on $A$ is $(X, \mathcal{X})$ restricted to $A$. Formally, the $\sigma$-algebra on $A$ is $\mathcal{A}=\{ S \cap A \mid S \in \mathcal{X} \}$.


Proof suggestion:

My guess is that it is, because for a measurable $B \in \mathcal{Y}$, $f_{\mid A^{-1}}(B)=f^{-1}(B) \cap A$, and $f^{-1}(B)$ and $A$ are both measurable.


On different notions of measurability: Does anything change if we use one of the following definitions for measurable functions?

  • The preimage of every measurable set is measurable (this is the standard definition in my opinion)
  • The preimage of every open set is measurable
  • The preimage of every open set is open (this is a sub-case of the last case)
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  • $\begingroup$ Can you write explicitly the measure and the sigma-algebra that make $A$ a measurable space? $\endgroup$ – SiD Oct 27 '17 at 10:07
  • $\begingroup$ Like this? I did not add a measure because that seems irrelevant to the question. $\endgroup$ – Peter Oct 27 '17 at 10:16
  • $\begingroup$ It is quite relevant, as the "restriction" (that is, the pull-back along the inclusion function) of the measure on $X$ is null if $A$ has null measure - and this is exactly what happens for any piecewise smooth curve in math.stackexchange.com/questions/2489673/…. If you do not agree, please define your notion of restriction $\endgroup$ – SiD Oct 27 '17 at 10:24
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The proof, as pointed out, is correct.

Regarding the notions of measurability: no, they are not the same. First of all, the last notion, preimage of open set is open, is the notion of continuity. Continuity is in general much stronger than measurability. Secondly, observe that you haven't defined any topology nor on $X$ or $Y$. So you don't even know that open sets are measurable, in order to understand whether the notions are equivalent you would need to assume it. However, if you take the sigma algebra on $Y$ to be the borel sigma algebra, i,e. the sigma algebra generated by open sets of $Y$, then the first two notions are equivalent.

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That’s true. If $f$ is measurable then $E_a=f^{-1}((a,+\infty))$ is a measurable set, for every $-\infty<a<+\infty$. Hence, the set $E_a \cap A$ is measurable. Noticing that $f_{|A}(a,+\infty)=E_a \cap A$, we are done. You are correct!

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