Prove there exists a measurable subset $E' subset E$ such that $f$ is almost everywhere of one sign on $E'$.












2












$begingroup$


Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.




My (weak) attempt:



Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.




I think all I need is a hint to start it.






measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.




    My (weak) attempt:



    Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.




    I think all I need is a hint to start it.






    measure-theory lebesgue-measure







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.




      My (weak) attempt:



      Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.




      I think all I need is a hint to start it.






      measure-theory lebesgue-measure







      share|cite|improve this question









      $endgroup$




      Let $f in Lleft(mathbb{R}right)$ where $L$ denotes Lebesgue integrable space of functions. Suppose there exists a subset $E subset mathbb{R}$ such that $m(E) > 0$ and $f(x) neq 0$ for all $x in E$. Prove there exists a measurable subset $E' subset E$ such $m(E') > 0$ and $f$ is almost everywhere of one sign on $E'$.




      My (weak) attempt:



      Let $E_alpha = {E setminus A: m(E) = m(E setminus A)}$.... I don't know. Rewrote this proof like 10 times with no insight. So now I'm just throwing it out there. Thank you.




      I think all I need is a hint to start it.






      measure-theory lebesgue-measure




      measure-theory lebesgue-measure






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 3 '18 at 21:11









      ZduffZduff

      1,663920




      1,663920






















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$

          Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,



          $$
          E = (E cap P) sqcup (E cap N).
          $$



          Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, thank you. But is it easy to show these two sets are measurable?
            $endgroup$
            – Zduff
            Dec 4 '18 at 2:51






          • 1




            $begingroup$
            Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 13:08










          • $begingroup$
            So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
            $endgroup$
            – Zduff
            Dec 4 '18 at 14:51










          • $begingroup$
            You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:09










          • $begingroup$
            (cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:10











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          1






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          active

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          2












          $begingroup$

          Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,



          $$
          E = (E cap P) sqcup (E cap N).
          $$



          Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, thank you. But is it easy to show these two sets are measurable?
            $endgroup$
            – Zduff
            Dec 4 '18 at 2:51






          • 1




            $begingroup$
            Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 13:08










          • $begingroup$
            So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
            $endgroup$
            – Zduff
            Dec 4 '18 at 14:51










          • $begingroup$
            You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:09










          • $begingroup$
            (cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:10
















          2












          $begingroup$

          Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,



          $$
          E = (E cap P) sqcup (E cap N).
          $$



          Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, thank you. But is it easy to show these two sets are measurable?
            $endgroup$
            – Zduff
            Dec 4 '18 at 2:51






          • 1




            $begingroup$
            Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 13:08










          • $begingroup$
            So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
            $endgroup$
            – Zduff
            Dec 4 '18 at 14:51










          • $begingroup$
            You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:09










          • $begingroup$
            (cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:10














          2












          2








          2





          $begingroup$

          Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,



          $$
          E = (E cap P) sqcup (E cap N).
          $$



          Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?






          share|cite|improve this answer









          $endgroup$



          Hint: take $P = {f > 0}$ and $N= {f < 0}$. Now,



          $$
          E = (E cap P) sqcup (E cap N).
          $$



          Certainly $f$ maintains its sign in each of these. Can they both be of measure zero?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 21:25









          Guido A.Guido A.

          7,5261730




          7,5261730












          • $begingroup$
            Ah, thank you. But is it easy to show these two sets are measurable?
            $endgroup$
            – Zduff
            Dec 4 '18 at 2:51






          • 1




            $begingroup$
            Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 13:08










          • $begingroup$
            So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
            $endgroup$
            – Zduff
            Dec 4 '18 at 14:51










          • $begingroup$
            You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:09










          • $begingroup$
            (cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:10


















          • $begingroup$
            Ah, thank you. But is it easy to show these two sets are measurable?
            $endgroup$
            – Zduff
            Dec 4 '18 at 2:51






          • 1




            $begingroup$
            Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 13:08










          • $begingroup$
            So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
            $endgroup$
            – Zduff
            Dec 4 '18 at 14:51










          • $begingroup$
            You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:09










          • $begingroup$
            (cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
            $endgroup$
            – Guido A.
            Dec 4 '18 at 20:10
















          $begingroup$
          Ah, thank you. But is it easy to show these two sets are measurable?
          $endgroup$
          – Zduff
          Dec 4 '18 at 2:51




          $begingroup$
          Ah, thank you. But is it easy to show these two sets are measurable?
          $endgroup$
          – Zduff
          Dec 4 '18 at 2:51




          1




          1




          $begingroup$
          Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
          $endgroup$
          – Guido A.
          Dec 4 '18 at 13:08




          $begingroup$
          Yes, intersection of measurable sets is measurable and $f in L(mathbb{R})$ implies measurability. Hence both level sets are measurable.
          $endgroup$
          – Guido A.
          Dec 4 '18 at 13:08












          $begingroup$
          So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
          $endgroup$
          – Zduff
          Dec 4 '18 at 14:51




          $begingroup$
          So if $f$ is Lebesgue integrable, then its level sets are always measurable? What is that theorem called, or do you possibly have a link?
          $endgroup$
          – Zduff
          Dec 4 '18 at 14:51












          $begingroup$
          You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
          $endgroup$
          – Guido A.
          Dec 4 '18 at 20:09




          $begingroup$
          You can find surely find it on Zygmund's Measure and Integral. I can give you a sketch of the proof: in the book it is proven (in the chapter on Fubini's theorem) that if $E subset mathbb{R}^{n_1 + n_2}$ is measurable, then $E^y := {x : (x,y) in E }$ is measurable a.e. $y in mathbb{R}^{n_2}$. Since the integral can be define as the measure of the set under the graphic of $f$, the fact that $f$ is integrable tells us that this set $R(f,E)$ is measurable and so $R(f,E)^y = { f > y}$ is measurable a.e. $y in mathbb{R}$.
          $endgroup$
          – Guido A.
          Dec 4 '18 at 20:09












          $begingroup$
          (cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
          $endgroup$
          – Guido A.
          Dec 4 '18 at 20:10




          $begingroup$
          (cont.) Since the set where the level sets are not measureable is of measure zero, the set where they are measurable must be dense, and so $f$ is measurable (this last result is also proved in the same book).
          $endgroup$
          – Guido A.
          Dec 4 '18 at 20:10


















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