4
$\begingroup$

Let $f: \mathbb{R} \to \mathbb{R} $ be a continuous function. I need to show that is a measurable function.

I tried working with the definition: Let $f: X \to \mathbb{R}$ be a function. If $f^{-1}(O)$ is a measurable set for every open subset $O$ of $\mathbb{R}$, then $f$ is called a measurable function.

Since $f^{-1}(O)$ also lies in $\mathbb{R}$, I think it is sufficient to show that every subset of $\mathbb{R}$ is measurable. But is this possible?

So far I concluded that $\mathbb{R}$ itself is measurable, since $$\mu(A) = \mu(A \cap \mathbb{R}) + \mu(A \cap \mathbb{R}^c) = \mu(A) + \mu(\emptyset) = \mu(A).$$ How do I need to approach?

$\endgroup$
  • 1
    $\begingroup$ No, in general and for the Lebesgue measure in particular not every subset of $\mathbb{R}$ is measurable. Consider approximating measurable sets with open or closed sets in measure. $\endgroup$ – Thomas Mar 2 '14 at 12:19
  • 3
    $\begingroup$ @TonyK: My point is: if I say in an analysis seminar "let $K$ be a nonmeasurable set", nobody will think this is unusual; the existence of nonmeasurable sets is a standard theorem in analysis texts. If I say "assume all subsets of $\mathbb{R}$ are measurable", I had better be ready to take some questions! To present these as easily interchangeable, rather than being the mainstream option and a specialized topic of set theory research, risks misleading students about actual mathematical practice. I feel many students think AC is more controversial than it truly is from seeing such comments. $\endgroup$ – Carl Mummert Mar 2 '14 at 21:02
  • 1
    $\begingroup$ @TonyK: of course, like many things in modern mathematics, analysis is inextricably linked to AC. It is also inextricably linked to the axiom of infinity and to classical logic. But it would be misleading to prefix every standard theorem with "if we accept classical logic and the axiom of infinity and AC"... Measure theorists who are not trying to study set theory simply assume the axiom of choice and move on. It might be less misleading to write "because of AC, nonmeasurable sets exist". Writing "if we assume AC" seems to suggest the other option is common, when it is not. $\endgroup$ – Carl Mummert Mar 2 '14 at 21:23
  • 1
    $\begingroup$ @TonyK: should I also put "if we assume the axiom of choice" in front of "a countable union of measure zero sets has measure zero"? Because, without AC, it is consistent that the continuum is a countable union of countable sets, each of which has measure zero. Am I am sinking further into denial if I claim that "a countable union of measure zero sets has measure zero" is true? Perhaps the last quoted statement is a common topic of debate in analysis seminars! :) cf. math.stackexchange.com/questions/389543/… $\endgroup$ – Carl Mummert Mar 2 '14 at 21:39
  • 1
    $\begingroup$ @TonyK: for the continuum being a countable union of countable sets, we can take the Feferman-Levy model of ZF, where measure theory goes horribly wrong; see math.stackexchange.com/a/188881/630 for more info. For each countable set being of measure zero, this is provable in ZF (without the axiom of choice) by inspection of the usual proof - since the set is already countable, we already have an enumeration of its points. (Also, please know the previous post was meant in good spirits.) $\endgroup$ – Carl Mummert Mar 2 '14 at 21:51
4
$\begingroup$

Since $f$ is continuous $f^{-1}(O)$ is open if $O$ is open. Open sets are measurable (if the space is equipped with the $\sigma$-algebra of the Borel-sets) so you are ready.

addendum:

It would indeed be sufficient if every subset of $\mathbb R$ was Borel-measurable, but that is not the case. For that see the comments on your question.

$\endgroup$
  • $\begingroup$ @Thomas Yes, you are right. I don't answer the question that was really posted here. I will try to find an answer or reference to that. If I don't succeed in it then I will delete my answer. $\endgroup$ – drhab Mar 2 '14 at 12:35
  • 2
    $\begingroup$ @Thomas: let $\mathcal{A}$ be the algebra of subsets of $\mathbb{R}$ whose preimages are measurable. The answer here shows $\mathcal{A}$ contains all open sets; so, by a standard theorem, it contains all Borel sets. It is not true, in general, that the inverse image of a Lebesgue measurable (but not Borel) set under a continuous function must be Lebesgue measurable. The definition of a measurable function in general is that the preimage of every Borel set is measurable. $\endgroup$ – Carl Mummert Mar 2 '14 at 12:39
  • $\begingroup$ @CarlMummert this is not shown, either. $\endgroup$ – Thomas Mar 2 '14 at 12:41
  • $\begingroup$ @Thomas: what exactly is not shown? I pointed out that it is an utterly standard fact that as soon as a $\sigma$-algebra contains the open sets, it contains the Borel sets. Are you asking for the answer to copy a proof of that result as well? I'm not trying to be too difficult - I just don't see what your objection is. $\endgroup$ – Carl Mummert Mar 2 '14 at 12:42
  • 2
    $\begingroup$ @Thomas as the OP says himself: $f$ is measurable if the preimages of open sets are measurable. He is not asking for a proof of that. If the Original question would be: 'How to prove that a continuous function is measurable?' then my answer is sufficient. $\endgroup$ – drhab Mar 2 '14 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.