I need help correcting my real analysis proof












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Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
$m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
I’m going to create a closed set for F so the function f will be bounded on it.
$$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.



Nothing in your work has included f. I do not see why suddenly f should be bounded.






real-analysis







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    $begingroup$


    Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
    Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
    I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
    $m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
    Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
    I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
    I’m going to create a closed set for F so the function f will be bounded on it.
    $$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
    Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
    Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
    Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.



    Nothing in your work has included f. I do not see why suddenly f should be bounded.






    real-analysis







    share|cite|improve this question











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      1








      1





      $begingroup$


      Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
      Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
      I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
      $m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
      Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
      I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
      I’m going to create a closed set for F so the function f will be bounded on it.
      $$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
      Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
      Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
      Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.



      Nothing in your work has included f. I do not see why suddenly f should be bounded.






      real-analysis







      share|cite|improve this question











      $endgroup$




      Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < infty$. For each $epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (Esetminus F) < epsilon$. What is $F$? I know $F subset E$ do I have to say that in my proof?
      Let $Esetminus F = E cap F^c$ which implies $m^*(Esetminus F) = m^*(E cap F^c)$.
      I know $m^*(F) = m^*( Fcap E) + m^*(Fsetminus E^c)$ but I have
      $m^*(F) = m^*(Ecap F^c ) + m^*(Fcap E)$. I don’t see where this came from.
      Since $m^*(E) < infty$ that implies $m^*( Ecap F^c) = m^*(F) – m^*( Fcap E)$.
      I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
      I’m going to create a closed set for F so the function f will be bounded on it.
      $$ F_n=[a_n,b_n- epsilon /2^n ] $$ so let $$ F = bigcup F_n. $$
      Using the definition of outer measure I get $$ m^*(F) = sum l(F_n )=sum (b)_n- a_n-epsilon/2^n )= sum(b_n-a_n-epsilon).$$
      Since $Fsubset E$, it is easy to see that $m^*(Ecap F)=m^*(F)$ so $$m^*(Ecap F) = m^*(F) leq m^*(I_n )=sum (b_n-a_n)$$ which implies $$m^*(Ecap F) leq sum l(I_n )- sum l(F_n )=sum (b_n-a_n )-sum (b_n-a_n-epsilon)=epsilon.$$
      Thus there is an $ F$ such that $Fsubset E$ and $F$ is closed so $f$ is bounded on it and $m^*(Esetminus F) leq epsilon$.



      Nothing in your work has included f. I do not see why suddenly f should be bounded.






      real-analysis




      real-analysis






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      edited Dec 13 '18 at 5:58









      Fiori_Modena

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      asked Oct 11 '15 at 18:08









      TessTess

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          $begingroup$

          You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.



          $F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.






          share|cite|improve this answer









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          • $begingroup$
            I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
            $endgroup$
            – user146269
            Nov 4 '15 at 19:10












          • $begingroup$
            and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
            $endgroup$
            – user146269
            Nov 4 '15 at 19:15














          Your Answer





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          $begingroup$

          You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.



          $F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.






          share|cite|improve this answer









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          • $begingroup$
            I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
            $endgroup$
            – user146269
            Nov 4 '15 at 19:10












          • $begingroup$
            and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
            $endgroup$
            – user146269
            Nov 4 '15 at 19:15


















          3












          $begingroup$

          You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.



          $F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
            $endgroup$
            – user146269
            Nov 4 '15 at 19:10












          • $begingroup$
            and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
            $endgroup$
            – user146269
            Nov 4 '15 at 19:15
















          3












          3








          3





          $begingroup$

          You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.



          $F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.






          share|cite|improve this answer









          $endgroup$



          You need to construct such an $F$ with the properties outlined. The set $A_n={xin E:|f(x)|>n}$ is an open set with $A_nsubseteq E$. Since f finite a.e. and $m^*(E)<infty$, there exists a $s$ such that $m^*(A_s)<varepsilon$. Take $F=Esetminus A_s$.



          $F$ is closed and $Fsubseteq E$ with $m^*(Esetminus F)< varepsilon$. Also, $|f(x)|leq s$ on $F$ by definition of $A_s$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 11 '15 at 18:29









          Konrad WrobelKonrad Wrobel

          419212




          419212












          • $begingroup$
            I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
            $endgroup$
            – user146269
            Nov 4 '15 at 19:10












          • $begingroup$
            and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
            $endgroup$
            – user146269
            Nov 4 '15 at 19:15




















          • $begingroup$
            I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
            $endgroup$
            – user146269
            Nov 4 '15 at 19:10












          • $begingroup$
            and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
            $endgroup$
            – user146269
            Nov 4 '15 at 19:15


















          $begingroup$
          I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
          $endgroup$
          – user146269
          Nov 4 '15 at 19:10






          $begingroup$
          I have two question, the first question how is $F$ is closed. and the second question how does $m^*(E - F) < epsilon$ i can not seem to follow
          $endgroup$
          – user146269
          Nov 4 '15 at 19:10














          $begingroup$
          and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
          $endgroup$
          – user146269
          Nov 4 '15 at 19:15






          $begingroup$
          and one more question how did you deduce that $|f(x)| leq s$ which show boundeness
          $endgroup$
          – user146269
          Nov 4 '15 at 19:15




















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