Tips to show that the Random Variable $X$ is $mathcal{B}(mathbb R)-mathcal{B}(mathbb R)-$measurable
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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
asked Nov 21 at 21:14
SABOY
498311
add a comment |
up vote
0
down vote
favorite
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
asked Nov 21 at 21:14
SABOY
498311
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
asked Nov 21 at 21:14
SABOY
498311
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
asked Nov 21 at 21:14
SABOY
498311
asked Nov 21 at 21:14
SABOY
498311
asked Nov 21 at 21:14
SABOY
498311
asked Nov 21 at 21:14
SABOY
498311
asked Nov 21 at 21:14
SABOY
498311
498311
add a comment |
add a comment |
1 Answer
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up vote
1
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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).
edited Nov 21 at 22:45
LaserPineapple
403
answered Nov 21 at 21:33
Jonathan
13812
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 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 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 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 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 at 22:23
|
show 4 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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).
edited Nov 21 at 22:45
LaserPineapple
403
answered Nov 21 at 21:33
Jonathan
13812
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 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 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 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 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 at 22:23
|
show 4 more comments
up vote
1
down vote
accepted
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).
edited Nov 21 at 22:45
LaserPineapple
403
answered Nov 21 at 21:33
Jonathan
13812
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 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 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 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 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 at 22:23
|
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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).
edited Nov 21 at 22:45
LaserPineapple
403
answered Nov 21 at 21:33
Jonathan
13812
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).
edited Nov 21 at 22:45
LaserPineapple
403
answered Nov 21 at 21:33
Jonathan
13812
edited Nov 21 at 22:45
LaserPineapple
403
edited Nov 21 at 22:45
LaserPineapple
403
edited Nov 21 at 22:45
LaserPineapple
403
403
answered Nov 21 at 21:33
Jonathan
13812
answered Nov 21 at 21:33
Jonathan
13812
answered Nov 21 at 21:33
Jonathan
13812
13812
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 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 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 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 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 at 22:23
|
show 4 more comments
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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 at 22:23
|
show 4 more comments
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