Tips to show that the Random Variable $X$ is $mathcal{B}(mathbb R)-mathcal{B}(mathbb R)-$measurable












0














Let $X: mathbb R to mathbb R$, where $forall omega in mathbb R -mathbb Q: X(omega)=0$.
Show $X$ is $mathcal{B}(mathbb R)-mathcal{B}(mathbb R)-$measurable function:



My ideas:
Using the definition of measurability I could take any $C in mathcal{B}(mathbb R)$ and I then want to show $X^{-1}(C) in mathcal{B}(mathbb R)$, but from the definition of the $X$ , all I can take away is that $mathbb R - mathbb Q subseteq X^{-1}({0})$ but this in no way shows measurability.



Another approach could be to take a generator $mathcal{E}$, and prove $X^{-1}(mathcal{E}) subseteq mathcal{B}(mathbb R)$. I think the biggest problem is that I have no definition of $X(omega)$ where $omega in mathbb Q$.



Any tips?






probability random-variables borel-sets borel-measures







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    0














    Let $X: mathbb R to mathbb R$, where $forall omega in mathbb R -mathbb Q: X(omega)=0$.
    Show $X$ is $mathcal{B}(mathbb R)-mathcal{B}(mathbb R)-$measurable function:



    My ideas:
    Using the definition of measurability I could take any $C in mathcal{B}(mathbb R)$ and I then want to show $X^{-1}(C) in mathcal{B}(mathbb R)$, but from the definition of the $X$ , all I can take away is that $mathbb R - mathbb Q subseteq X^{-1}({0})$ but this in no way shows measurability.



    Another approach could be to take a generator $mathcal{E}$, and prove $X^{-1}(mathcal{E}) subseteq mathcal{B}(mathbb R)$. I think the biggest problem is that I have no definition of $X(omega)$ where $omega in mathbb Q$.



    Any tips?






    probability random-variables borel-sets borel-measures







    share|cite|improve this question

























      0












      0








      0







      Let $X: mathbb R to mathbb R$, where $forall omega in mathbb R -mathbb Q: X(omega)=0$.
      Show $X$ is $mathcal{B}(mathbb R)-mathcal{B}(mathbb R)-$measurable function:



      My ideas:
      Using the definition of measurability I could take any $C in mathcal{B}(mathbb R)$ and I then want to show $X^{-1}(C) in mathcal{B}(mathbb R)$, but from the definition of the $X$ , all I can take away is that $mathbb R - mathbb Q subseteq X^{-1}({0})$ but this in no way shows measurability.



      Another approach could be to take a generator $mathcal{E}$, and prove $X^{-1}(mathcal{E}) subseteq mathcal{B}(mathbb R)$. I think the biggest problem is that I have no definition of $X(omega)$ where $omega in mathbb Q$.



      Any tips?






      probability random-variables borel-sets borel-measures







      share|cite|improve this question













      Let $X: mathbb R to mathbb R$, where $forall omega in mathbb R -mathbb Q: X(omega)=0$.
      Show $X$ is $mathcal{B}(mathbb R)-mathcal{B}(mathbb R)-$measurable function:



      My ideas:
      Using the definition of measurability I could take any $C in mathcal{B}(mathbb R)$ and I then want to show $X^{-1}(C) in mathcal{B}(mathbb R)$, but from the definition of the $X$ , all I can take away is that $mathbb R - mathbb Q subseteq X^{-1}({0})$ but this in no way shows measurability.



      Another approach could be to take a generator $mathcal{E}$, and prove $X^{-1}(mathcal{E}) subseteq mathcal{B}(mathbb R)$. I think the biggest problem is that I have no definition of $X(omega)$ where $omega in mathbb Q$.



      Any tips?






      probability random-variables borel-sets borel-measures




      probability random-variables borel-sets borel-measures






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 21 '18 at 21:14









      SABOY

      568311




      568311






















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          Hint using your first approach:




          Let $A in mathcal{B}(mathbb{R})$, then consider the two cases: $0 in A$ and $0 notin A$. Now try using that all closed and all open subsets of $mathbb{R}$ are contained in $mathcal{B}(mathbb{R})$ and the definition of a $sigma$-algebra (in particular, that it is closed under countable unions).







          share|cite|improve this answer























          • I am still lost. I mean if I take the case $0 in A$: all I am able to say is that $X^{-1}(A)=X^{-1}({0} cup (A - {0}))$. This yields $X^{-1}({0})cup X^{-1}(A - {0})= (mathbb R - mathbb Q) cup X^{-1}(A - {0})$, and I do not know what to do with $X^{-1}(A - {0})$
            – SABOY
            Nov 21 '18 at 22:00












          • Think of what values by definition of $X$ map to $(Asetminus {0}) subset mathbb{R}setminus {0}$ and then try to connect this with the hint above.
            – Jonathan
            Nov 21 '18 at 22:02












          • $X^{-1}(A-{0})= mathbb Q$ , so then $X^{-1}(A)= mathbb R$, which is $in mathcal{B}(mathbb R)$, and then measurability immediately follows. Is this correct?
            – SABOY
            Nov 21 '18 at 22:07










          • Correct thinking and close, but $X^{-1}(A setminus {0}) = mathbb{Q}$ is not quite correct. Remember, $A$ is a general subset of $mathcal{B}(mathbb{R})$.
            – Jonathan
            Nov 21 '18 at 22:09






          • 1




            yes of course! And of course any subset of $mathbb Q$ is measurable as a union of countable singletons. Correct?
            – SABOY
            Nov 21 '18 at 22:23











          Your Answer





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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          1














          Hint using your first approach:




          Let $A in mathcal{B}(mathbb{R})$, then consider the two cases: $0 in A$ and $0 notin A$. Now try using that all closed and all open subsets of $mathbb{R}$ are contained in $mathcal{B}(mathbb{R})$ and the definition of a $sigma$-algebra (in particular, that it is closed under countable unions).







          share|cite|improve this answer























          • I am still lost. I mean if I take the case $0 in A$: all I am able to say is that $X^{-1}(A)=X^{-1}({0} cup (A - {0}))$. This yields $X^{-1}({0})cup X^{-1}(A - {0})= (mathbb R - mathbb Q) cup X^{-1}(A - {0})$, and I do not know what to do with $X^{-1}(A - {0})$
            – SABOY
            Nov 21 '18 at 22:00












          • Think of what values by definition of $X$ map to $(Asetminus {0}) subset mathbb{R}setminus {0}$ and then try to connect this with the hint above.
            – Jonathan
            Nov 21 '18 at 22:02












          • $X^{-1}(A-{0})= mathbb Q$ , so then $X^{-1}(A)= mathbb R$, which is $in mathcal{B}(mathbb R)$, and then measurability immediately follows. Is this correct?
            – SABOY
            Nov 21 '18 at 22:07










          • Correct thinking and close, but $X^{-1}(A setminus {0}) = mathbb{Q}$ is not quite correct. Remember, $A$ is a general subset of $mathcal{B}(mathbb{R})$.
            – Jonathan
            Nov 21 '18 at 22:09






          • 1




            yes of course! And of course any subset of $mathbb Q$ is measurable as a union of countable singletons. Correct?
            – SABOY
            Nov 21 '18 at 22:23
















          1














          Hint using your first approach:




          Let $A in mathcal{B}(mathbb{R})$, then consider the two cases: $0 in A$ and $0 notin A$. Now try using that all closed and all open subsets of $mathbb{R}$ are contained in $mathcal{B}(mathbb{R})$ and the definition of a $sigma$-algebra (in particular, that it is closed under countable unions).







          share|cite|improve this answer























          • I am still lost. I mean if I take the case $0 in A$: all I am able to say is that $X^{-1}(A)=X^{-1}({0} cup (A - {0}))$. This yields $X^{-1}({0})cup X^{-1}(A - {0})= (mathbb R - mathbb Q) cup X^{-1}(A - {0})$, and I do not know what to do with $X^{-1}(A - {0})$
            – SABOY
            Nov 21 '18 at 22:00












          • Think of what values by definition of $X$ map to $(Asetminus {0}) subset mathbb{R}setminus {0}$ and then try to connect this with the hint above.
            – Jonathan
            Nov 21 '18 at 22:02












          • $X^{-1}(A-{0})= mathbb Q$ , so then $X^{-1}(A)= mathbb R$, which is $in mathcal{B}(mathbb R)$, and then measurability immediately follows. Is this correct?
            – SABOY
            Nov 21 '18 at 22:07










          • Correct thinking and close, but $X^{-1}(A setminus {0}) = mathbb{Q}$ is not quite correct. Remember, $A$ is a general subset of $mathcal{B}(mathbb{R})$.
            – Jonathan
            Nov 21 '18 at 22:09






          • 1




            yes of course! And of course any subset of $mathbb Q$ is measurable as a union of countable singletons. Correct?
            – SABOY
            Nov 21 '18 at 22:23














          1












          1








          1






          Hint using your first approach:




          Let $A in mathcal{B}(mathbb{R})$, then consider the two cases: $0 in A$ and $0 notin A$. Now try using that all closed and all open subsets of $mathbb{R}$ are contained in $mathcal{B}(mathbb{R})$ and the definition of a $sigma$-algebra (in particular, that it is closed under countable unions).







          share|cite|improve this answer














          Hint using your first approach:




          Let $A in mathcal{B}(mathbb{R})$, then consider the two cases: $0 in A$ and $0 notin A$. Now try using that all closed and all open subsets of $mathbb{R}$ are contained in $mathcal{B}(mathbb{R})$ and the definition of a $sigma$-algebra (in particular, that it is closed under countable unions).








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 '18 at 22:45









          LaserPineapple

          403




          403










          answered Nov 21 '18 at 21:33









          Jonathan

          16412




          16412












          • I am still lost. I mean if I take the case $0 in A$: all I am able to say is that $X^{-1}(A)=X^{-1}({0} cup (A - {0}))$. This yields $X^{-1}({0})cup X^{-1}(A - {0})= (mathbb R - mathbb Q) cup X^{-1}(A - {0})$, and I do not know what to do with $X^{-1}(A - {0})$
            – SABOY
            Nov 21 '18 at 22:00












          • Think of what values by definition of $X$ map to $(Asetminus {0}) subset mathbb{R}setminus {0}$ and then try to connect this with the hint above.
            – Jonathan
            Nov 21 '18 at 22:02












          • $X^{-1}(A-{0})= mathbb Q$ , so then $X^{-1}(A)= mathbb R$, which is $in mathcal{B}(mathbb R)$, and then measurability immediately follows. Is this correct?
            – SABOY
            Nov 21 '18 at 22:07










          • Correct thinking and close, but $X^{-1}(A setminus {0}) = mathbb{Q}$ is not quite correct. Remember, $A$ is a general subset of $mathcal{B}(mathbb{R})$.
            – Jonathan
            Nov 21 '18 at 22:09






          • 1




            yes of course! And of course any subset of $mathbb Q$ is measurable as a union of countable singletons. Correct?
            – SABOY
            Nov 21 '18 at 22:23


















          • I am still lost. I mean if I take the case $0 in A$: all I am able to say is that $X^{-1}(A)=X^{-1}({0} cup (A - {0}))$. This yields $X^{-1}({0})cup X^{-1}(A - {0})= (mathbb R - mathbb Q) cup X^{-1}(A - {0})$, and I do not know what to do with $X^{-1}(A - {0})$
            – SABOY
            Nov 21 '18 at 22:00












          • Think of what values by definition of $X$ map to $(Asetminus {0}) subset mathbb{R}setminus {0}$ and then try to connect this with the hint above.
            – Jonathan
            Nov 21 '18 at 22:02












          • $X^{-1}(A-{0})= mathbb Q$ , so then $X^{-1}(A)= mathbb R$, which is $in mathcal{B}(mathbb R)$, and then measurability immediately follows. Is this correct?
            – SABOY
            Nov 21 '18 at 22:07










          • Correct thinking and close, but $X^{-1}(A setminus {0}) = mathbb{Q}$ is not quite correct. Remember, $A$ is a general subset of $mathcal{B}(mathbb{R})$.
            – Jonathan
            Nov 21 '18 at 22:09






          • 1




            yes of course! And of course any subset of $mathbb Q$ is measurable as a union of countable singletons. Correct?
            – SABOY
            Nov 21 '18 at 22:23
















          I am still lost. I mean if I take the case $0 in A$: all I am able to say is that $X^{-1}(A)=X^{-1}({0} cup (A - {0}))$. This yields $X^{-1}({0})cup X^{-1}(A - {0})= (mathbb R - mathbb Q) cup X^{-1}(A - {0})$, and I do not know what to do with $X^{-1}(A - {0})$
          – SABOY
          Nov 21 '18 at 22:00






          I am still lost. I mean if I take the case $0 in A$: all I am able to say is that $X^{-1}(A)=X^{-1}({0} cup (A - {0}))$. This yields $X^{-1}({0})cup X^{-1}(A - {0})= (mathbb R - mathbb Q) cup X^{-1}(A - {0})$, and I do not know what to do with $X^{-1}(A - {0})$
          – SABOY
          Nov 21 '18 at 22:00














          Think of what values by definition of $X$ map to $(Asetminus {0}) subset mathbb{R}setminus {0}$ and then try to connect this with the hint above.
          – Jonathan
          Nov 21 '18 at 22:02






          Think of what values by definition of $X$ map to $(Asetminus {0}) subset mathbb{R}setminus {0}$ and then try to connect this with the hint above.
          – Jonathan
          Nov 21 '18 at 22:02














          $X^{-1}(A-{0})= mathbb Q$ , so then $X^{-1}(A)= mathbb R$, which is $in mathcal{B}(mathbb R)$, and then measurability immediately follows. Is this correct?
          – SABOY
          Nov 21 '18 at 22:07




          $X^{-1}(A-{0})= mathbb Q$ , so then $X^{-1}(A)= mathbb R$, which is $in mathcal{B}(mathbb R)$, and then measurability immediately follows. Is this correct?
          – SABOY
          Nov 21 '18 at 22:07












          Correct thinking and close, but $X^{-1}(A setminus {0}) = mathbb{Q}$ is not quite correct. Remember, $A$ is a general subset of $mathcal{B}(mathbb{R})$.
          – Jonathan
          Nov 21 '18 at 22:09




          Correct thinking and close, but $X^{-1}(A setminus {0}) = mathbb{Q}$ is not quite correct. Remember, $A$ is a general subset of $mathcal{B}(mathbb{R})$.
          – Jonathan
          Nov 21 '18 at 22:09




          1




          1




          yes of course! And of course any subset of $mathbb Q$ is measurable as a union of countable singletons. Correct?
          – SABOY
          Nov 21 '18 at 22:23




          yes of course! And of course any subset of $mathbb Q$ is measurable as a union of countable singletons. Correct?
          – SABOY
          Nov 21 '18 at 22:23


















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