Suppose $f$ is a real-valued function on $\mathbb R$ such that $f^{−1}(c)$ is measurable for each number $c$. Is $f$ necessarily measurable?
Must $f$ be measurable if each $f^{-1}(c)$ is?
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$\begingroup$
real-analysis
measure-theory
lebesgue-integral
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0What does the notation 1(c) mean? – 2012-12-24
3 Answers
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Hint: Every injective function satisfies the hypothesis. You can take any nonmeasurable set, map it injectively into $(0,\infty)$, and map its complement injectively into $(-\infty,0)$ (for example using $e^x$ for a simple formula).
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0sorry i can not understand Can you explain more – 2012-12-24
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0user54144: Sorry I don't know what you don't understand. Could you please ask a more specific question? – 2012-12-24
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0i read royden all talk about if f is measurable =>........ – 2012-12-24
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0but don not talk about if my question so i can not find any thm or def from my book – 2012-12-24
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0user54144: The condition in your question is much weaker than the definition of measurable. And it turns out that there are examples showing that the condition can hold without the function being measurable. This will not come from a particular theorem in Royden's book, but it can come from you writing down an example (and the reasons it is an example). Asaf and I have each offered suggestions for an approach to showing counterexamples exist. Let me ask, do you see why an injective function satisfies the hypothesis? – 2012-12-24
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0so your mean is(The exponential function exp : R → R defined by exp(x) = ex is injective (but not surjective as no real value maps to a negative number) so can not satisfies measurable 4 Equivalent Property – 2012-12-24
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0user54144: I do not know what you mean by "your mean is". No, the exponential function itself is not an example, but it can be used in finding an example. I think it would be a good exercise for you to think about why the exponential function is measurable, because it is much more basic than the problem you brought here, and may help you have a better understanding of the ideas involved. Also let me ask again, do you see why an injective function satisfies the hypothesis? – 2012-12-24
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0i mean the question answer is"No" but i can not find Counterexample – 2012-12-24
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1@user54144: Could you please respond directly to specific questions? The *hypothesis* in your problem is the condition that $f^{-1}(c)$ is measurable for each $c\in\mathbb R$. **Question for you:** Do you see why an injective function satisfies this hypothesis? – 2012-12-24
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0In fact there are nonmeasurable one-to-one functions. Since singletons and the empty set are measurable, {x:f(x)=a} is always measurable. For example, let {xα:α∈A} be a Hamel basis of R over the rationals Q, and let f be the function that is linear over Q, interchanges two basis elements x1 and x2, and leaves the other basis elements fixed. This is easily seen to be one-to-one. For convenience we may choose x1 and x2 so that x1−x2 is rational. Then {x:f(x)=x} is a Vitali set (i.e. a set that contains exactly one element from each equivalence class of R/Q), and therefore is not measurable. – 2012-12-24
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0i see the answer but i can not reaily understand sorry i study by myself i am so fool – 2012-12-24
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1user54144: Is your comment with a solution taken from some source? Where is it from? I am not asking you for the full solution, I have been trying to ask if you understand just the small part about why an injective function satisfies the condition that $f^{-1}(c)$ is measurable. Do you? (From there, nothing so fancy as Hamel bases are needed for this problem, although that is an interesting approach. Incidentally, this might be my last comment for a while.) – 2012-12-24
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0thank very much i have to study hard – 2012-12-24
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0@user54144: Not every non-measurable set is a Vitali set. There is a plethora of non-measurable sets. – 2012-12-24
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Hint: Note that the Vitali set has size $2^{\aleph_0}$ and therefore there is a bijection between $\mathbb R$ and the Vitali set.
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0Why is it not sufficient to map the vitali set to itself? – 2015-10-02
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0Because that might not be true. – 2015-10-02
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0What do you mean? – 2015-10-02
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0Why would a general, arbitrary function $f$ will map a Vitali set to itself? (I don't know if that would be enough to show measurability, since there are other non-measurable sets too). – 2015-10-02
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0I mean why not use the identity map on the vitali set? then $f^{-1}(c)$ is the closed singleton, yet the inverse image of the whole set is not measurable. – 2015-10-02
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0But the identity function is measurable. Measurable means that the preimage of a measurable set, is again a measurable set. Using a bijection between a Vitali set and $[1,2]$, and the complement of a Vitali set has a bijection with $[4,5]$. Such function has fiber which are empty or singleton; however the preimage of $[1,2]$ is not measurable. So the function itself is not measurable. – 2015-10-02
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0I know this is obviously wrong, but whats wrong with the identity function $\iota: [0,1] \to [0,1]$. If $E$ is the vitali set, then $f^{-1}([0,1]) = f^{-1}([0,1] \setminus E) \cup f^{-1}(E)$. Then the latter set is not measurable – 2015-10-03
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0Clearly that example is wrong because $\iota(x)=x$ is measurable, but could you briefly say why? – 2015-10-03
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0Because a measurable set (of positive measure) can always be written as the union of two non measurable sets. That's not a big deal. But $f^{-1}([0,1])=[0,1]$, and $[0,1]$ is clearly a measurable set. – 2015-10-03
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Consider for example a set $E \subset [0,1]$; then the function $$ f(x)=\begin{cases} x & x \in E\\ -x & x \in [0,1]\setminus E \end{cases} $$ is measurable if and only if $E$ is measurable. On the other hand, $f$ is injective, hence $f^{-1}(c)$ is either empty, either a singleton (in both cases, anyway, $f^{-1}(c)$ is measurable).
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0Your means question answer is"NO" because your example???? – 2012-12-24
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0Yes, you are right: the answer to your question is "No". – 2012-12-24
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0can you explain why the f is not measurable???? thx – 2012-12-25
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0You have that $f$ is nonmeasurable if $E$ is nonmeasurable beacuse $f^{-1}((0,1))=E$. – 2012-12-25
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0your example means f is injective implus f−1(c) is measurable can you explain more?? thx – 2012-12-25
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0No, I'm sorry, but I can't explain more. There's nothing I can add. You simply have to study definitions. – 2012-12-25