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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?

Bach
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user54297
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    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. – Thomas Mar 02 '14 at 12:19
  • You can't show that every subset of $\mathbb R$ is measurable, because it's not true if you assume the Axiom of Choice. (If you don't assume the Axiom of Choice, then it is consistent that every subset of $\mathbb R$ is measurable; but that doesn't make it provable.) – TonyK Mar 02 '14 at 12:20
  • Yes, it would be sufficient to prove that every subset of $\mathbb{R}$ is measurable, but that is not true, so you will have to prove that just the sets you need to be measurable are measurable. This is typical whenever you are trying to prove a function is measurable. – Carl Mummert Mar 02 '14 at 12:23
  • @Carl: "but that is not true": see my earlier comment for a more nuanced take. – TonyK Mar 02 '14 at 20:13
  • @TonyK: apart from specialized research in set theory, the axiom of choice (AC) has become a standard part of modern mathematics, so I'm not convinced (personally) that it is beneficial when speaking at an elementary level to treat AC as if it's still a major area of disagreement. The state of affairs is different in 2014 than it was in 1914 or 1940. I think it may be more misleading to emphasize the dependence on AC (as if this was still a major area of dispute) than to simply state what modern texts do: there is a nonmeasurable set. This question is about analysis, after all, not set theory. – Carl Mummert Mar 02 '14 at 20:42
  • @Carl: If you say "that is not true", then at an elementary level the obvious response is "please show me a non-measurable set". But you can't, can you? Not without the Axiom of Choice, which is precisely what you are trying to ignore. – TonyK Mar 02 '14 at 20:54
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    @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. – Carl Mummert Mar 02 '14 at 21:02
  • @Carl: But measure theory is inextricably linked with AC. You seem to be in denial about this! – TonyK Mar 02 '14 at 21:15
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    @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. – Carl Mummert Mar 02 '14 at 21:23
  • Carl: So you're in denial then. – TonyK Mar 02 '14 at 21:25
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    @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. http://math.stackexchange.com/questions/389543/is-the-countably-union-of-measure-zero-sets-zero – Carl Mummert Mar 02 '14 at 21:39
  • @Carl: "without AC, it is consistent that the continuum is a countable union of countable sets, each of which has measure zero": I am surprised. Can you please supply a link to this? – TonyK Mar 02 '14 at 21:43
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    @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 http://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.) – Carl Mummert Mar 02 '14 at 21:51
  • @Carl: Thank you for that! The more we debate this, the more surprised I am that you objected to my "more nuanced" comment :-) – TonyK Mar 02 '14 at 21:56

1 Answers1

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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.

drhab
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  • @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. – drhab Mar 02 '14 at 12:35
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    @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. – Carl Mummert Mar 02 '14 at 12:39
  • @CarlMummert this is not shown, either. – Thomas Mar 02 '14 at 12:41
  • @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. – Carl Mummert Mar 02 '14 at 12:42
  • @CarlMummert The original question was not answered, only the fact that the preimage of on open set is measurable. At the level of the question, I do expect some more details, or indication that it is only a hint. – Thomas Mar 02 '14 at 12:43
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    @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. – drhab Mar 02 '14 at 12:45
  • ok, withdrawing my comments. My apologies. – Thomas Mar 02 '14 at 12:54
  • @Thomas Ego te absolvo. – drhab Mar 02 '14 at 16:03