Proving that the function $x^2$ is a Borel function [closed]
$begingroup$
How should I prove that the function $x^2$ where $x^+=max(x,0)$, $x^-=-min(x,0)$ and $|x|=x^+ + x^-$ is a Borel function?
I know that a Borel function is a random variable $mathbb{R} rightarrow mathbb{R}$. But how to mathematically prove it? I have no idea... May be I should use indicators? But how?
probability probability-theory borel-sets
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
$endgroup$
closed as off-topic by user21820, Lord Shark the Unknown, Lord_Farin, jgon, Cesareo Dec 25 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How should I prove that the function $x^2$ where $x^+=max(x,0)$, $x^-=-min(x,0)$ and $|x|=x^+ + x^-$ is a Borel function?
I know that a Borel function is a random variable $mathbb{R} rightarrow mathbb{R}$. But how to mathematically prove it? I have no idea... May be I should use indicators? But how?
probability probability-theory borel-sets
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
$endgroup$
closed as off-topic by user21820, Lord Shark the Unknown, Lord_Farin, jgon, Cesareo Dec 25 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Not answered but this question has useful tips in the comments.
$endgroup$
– Boshu
Dec 10 '18 at 13:53
$begingroup$
An easier thing to do might be to prove that all continuous functions are Borel, thus the results for all the functions you mention will follow
$endgroup$
– Gâteau-Gallois
Dec 10 '18 at 14:42
$begingroup$
@Gâteau-Gallois Isn't this kind of cyclic though ? To show continuity he must show that the pre image of any open set is an open set.
$endgroup$
– Digitalis
Dec 10 '18 at 14:52
add a comment |
$begingroup$
How should I prove that the function $x^2$ where $x^+=max(x,0)$, $x^-=-min(x,0)$ and $|x|=x^+ + x^-$ is a Borel function?
I know that a Borel function is a random variable $mathbb{R} rightarrow mathbb{R}$. But how to mathematically prove it? I have no idea... May be I should use indicators? But how?
probability probability-theory borel-sets
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
$endgroup$
How should I prove that the function $x^2$ where $x^+=max(x,0)$, $x^-=-min(x,0)$ and $|x|=x^+ + x^-$ is a Borel function?
I know that a Borel function is a random variable $mathbb{R} rightarrow mathbb{R}$. But how to mathematically prove it? I have no idea... May be I should use indicators? But how?
probability probability-theory borel-sets
probability probability-theory borel-sets
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
edited Dec 10 '18 at 13:57
Brahadeesh
6,21242361
6,21242361
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
asked Dec 10 '18 at 13:44
AtstovasAtstovas
1109
1109
closed as off-topic by user21820, Lord Shark the Unknown, Lord_Farin, jgon, Cesareo Dec 25 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Lord Shark the Unknown, Lord_Farin, jgon, Cesareo Dec 25 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Not answered but this question has useful tips in the comments.
$endgroup$
– Boshu
Dec 10 '18 at 13:53
$begingroup$
An easier thing to do might be to prove that all continuous functions are Borel, thus the results for all the functions you mention will follow
$endgroup$
– Gâteau-Gallois
Dec 10 '18 at 14:42
$begingroup$
@Gâteau-Gallois Isn't this kind of cyclic though ? To show continuity he must show that the pre image of any open set is an open set.
$endgroup$
– Digitalis
Dec 10 '18 at 14:52
add a comment |
$begingroup$
Not answered but this question has useful tips in the comments.
$endgroup$
– Boshu
Dec 10 '18 at 13:53
$begingroup$
An easier thing to do might be to prove that all continuous functions are Borel, thus the results for all the functions you mention will follow
$endgroup$
– Gâteau-Gallois
Dec 10 '18 at 14:42
$begingroup$
@Gâteau-Gallois Isn't this kind of cyclic though ? To show continuity he must show that the pre image of any open set is an open set.
$endgroup$
– Digitalis
Dec 10 '18 at 14:52
$begingroup$
Not answered but this question has useful tips in the comments.
$endgroup$
– Boshu
Dec 10 '18 at 13:53
$begingroup$
Not answered but this question has useful tips in the comments.
$endgroup$
– Boshu
Dec 10 '18 at 13:53
$begingroup$
An easier thing to do might be to prove that all continuous functions are Borel, thus the results for all the functions you mention will follow
$endgroup$
– Gâteau-Gallois
Dec 10 '18 at 14:42
$begingroup$
An easier thing to do might be to prove that all continuous functions are Borel, thus the results for all the functions you mention will follow
$endgroup$
– Gâteau-Gallois
Dec 10 '18 at 14:42
$begingroup$
@Gâteau-Gallois Isn't this kind of cyclic though ? To show continuity he must show that the pre image of any open set is an open set.
$endgroup$
– Digitalis
Dec 10 '18 at 14:52
$begingroup$
@Gâteau-Gallois Isn't this kind of cyclic though ? To show continuity he must show that the pre image of any open set is an open set.
$endgroup$
– Digitalis
Dec 10 '18 at 14:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$f: mathbb{R} rightarrow mathbb{R}$ is borel $iff ; forall B in mathcal{B}(mathbb{R}) : f^{-1}(B) in mathcal{B}(mathbb{R}) iff ; forall b in mathbb{R} : f^{-1}(-infty,b) in mathcal{B}(mathbb{R}).$
$$ f^{-1}((-infty,b)) = {x : x^2 in (-infty,b) } = { x : x^2 < b }$$
If $b < 0 :$ then ${ x : x^2 < b } = varnothing.$
If $b geq 0$ then
$$ x^2 < b iff - sqrt b < x < sqrt b.$$
So $$ f^{-1}((-infty,b))= begin{cases} varnothing &in mathcal{B}(mathbb{R}) & b < 0 \ (-sqrt b, sqrt b)& in mathcal{B}(mathbb{R}) & b geq 0end{cases}$$
and $f$ is borel.
Consider $f(x) = vert x vert.$ Let $b in mathbb{R}$:
- If $b < 0$:
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < 0} = varnothing$$
Since the empty set is an open (closed) it is a borel set so $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b < 0$.
- if $ b geq 0:$
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < b} = (-b,b)$$
Since $(-b,b)$ is an open set it is a borel set and $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b geq 0.$
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
$endgroup$
$begingroup$
Where did you take into consideration $|x|=x^++x^-$?
$endgroup$
– Atstovas
Dec 11 '18 at 13:29
$begingroup$
@Atstovas I didn't. I thought I'd only do one. Are you having trouble for $vert xvert$ ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:01
$begingroup$
Yesss... accually I new how to show that $x^2$ is Borel function. But I didn’t know what to do with given conditions
$endgroup$
– Atstovas
Dec 11 '18 at 14:10
$begingroup$
Which given conditions ? For $ f = vert x vert$ you should do exactly the same show that $f^{-1}(-infty,b)$ is a borel set for all $b$. You're making me doubt if you even understood the solution for $f = x^2$. Are you sure you understood that one ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:26
$begingroup$
@Atstovas I've edited my answer. Any questions ?
$endgroup$
– Digitalis
Dec 11 '18 at 15:15
|
show 1 more comment
$begingroup$
If you really want to run the details with $x to x^2$, I suggest that you take a look at this
Show that continuous functions on $mathbb R$ are Borel-measurable
Probably though, it would be smarter to really understand the answer given in this link, and thus to be able to show that this holds for any continuous function on $mathbb{R}$, which includes the case you are looking for.
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$f: mathbb{R} rightarrow mathbb{R}$ is borel $iff ; forall B in mathcal{B}(mathbb{R}) : f^{-1}(B) in mathcal{B}(mathbb{R}) iff ; forall b in mathbb{R} : f^{-1}(-infty,b) in mathcal{B}(mathbb{R}).$
$$ f^{-1}((-infty,b)) = {x : x^2 in (-infty,b) } = { x : x^2 < b }$$
If $b < 0 :$ then ${ x : x^2 < b } = varnothing.$
If $b geq 0$ then
$$ x^2 < b iff - sqrt b < x < sqrt b.$$
So $$ f^{-1}((-infty,b))= begin{cases} varnothing &in mathcal{B}(mathbb{R}) & b < 0 \ (-sqrt b, sqrt b)& in mathcal{B}(mathbb{R}) & b geq 0end{cases}$$
and $f$ is borel.
Consider $f(x) = vert x vert.$ Let $b in mathbb{R}$:
- If $b < 0$:
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < 0} = varnothing$$
Since the empty set is an open (closed) it is a borel set so $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b < 0$.
- if $ b geq 0:$
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < b} = (-b,b)$$
Since $(-b,b)$ is an open set it is a borel set and $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b geq 0.$
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
$endgroup$
$begingroup$
Where did you take into consideration $|x|=x^++x^-$?
$endgroup$
– Atstovas
Dec 11 '18 at 13:29
$begingroup$
@Atstovas I didn't. I thought I'd only do one. Are you having trouble for $vert xvert$ ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:01
$begingroup$
Yesss... accually I new how to show that $x^2$ is Borel function. But I didn’t know what to do with given conditions
$endgroup$
– Atstovas
Dec 11 '18 at 14:10
$begingroup$
Which given conditions ? For $ f = vert x vert$ you should do exactly the same show that $f^{-1}(-infty,b)$ is a borel set for all $b$. You're making me doubt if you even understood the solution for $f = x^2$. Are you sure you understood that one ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:26
$begingroup$
@Atstovas I've edited my answer. Any questions ?
$endgroup$
– Digitalis
Dec 11 '18 at 15:15
|
show 1 more comment
$begingroup$
$f: mathbb{R} rightarrow mathbb{R}$ is borel $iff ; forall B in mathcal{B}(mathbb{R}) : f^{-1}(B) in mathcal{B}(mathbb{R}) iff ; forall b in mathbb{R} : f^{-1}(-infty,b) in mathcal{B}(mathbb{R}).$
$$ f^{-1}((-infty,b)) = {x : x^2 in (-infty,b) } = { x : x^2 < b }$$
If $b < 0 :$ then ${ x : x^2 < b } = varnothing.$
If $b geq 0$ then
$$ x^2 < b iff - sqrt b < x < sqrt b.$$
So $$ f^{-1}((-infty,b))= begin{cases} varnothing &in mathcal{B}(mathbb{R}) & b < 0 \ (-sqrt b, sqrt b)& in mathcal{B}(mathbb{R}) & b geq 0end{cases}$$
and $f$ is borel.
Consider $f(x) = vert x vert.$ Let $b in mathbb{R}$:
- If $b < 0$:
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < 0} = varnothing$$
Since the empty set is an open (closed) it is a borel set so $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b < 0$.
- if $ b geq 0:$
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < b} = (-b,b)$$
Since $(-b,b)$ is an open set it is a borel set and $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b geq 0.$
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
$endgroup$
$begingroup$
Where did you take into consideration $|x|=x^++x^-$?
$endgroup$
– Atstovas
Dec 11 '18 at 13:29
$begingroup$
@Atstovas I didn't. I thought I'd only do one. Are you having trouble for $vert xvert$ ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:01
$begingroup$
Yesss... accually I new how to show that $x^2$ is Borel function. But I didn’t know what to do with given conditions
$endgroup$
– Atstovas
Dec 11 '18 at 14:10
$begingroup$
Which given conditions ? For $ f = vert x vert$ you should do exactly the same show that $f^{-1}(-infty,b)$ is a borel set for all $b$. You're making me doubt if you even understood the solution for $f = x^2$. Are you sure you understood that one ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:26
$begingroup$
@Atstovas I've edited my answer. Any questions ?
$endgroup$
– Digitalis
Dec 11 '18 at 15:15
|
show 1 more comment
$begingroup$
$f: mathbb{R} rightarrow mathbb{R}$ is borel $iff ; forall B in mathcal{B}(mathbb{R}) : f^{-1}(B) in mathcal{B}(mathbb{R}) iff ; forall b in mathbb{R} : f^{-1}(-infty,b) in mathcal{B}(mathbb{R}).$
$$ f^{-1}((-infty,b)) = {x : x^2 in (-infty,b) } = { x : x^2 < b }$$
If $b < 0 :$ then ${ x : x^2 < b } = varnothing.$
If $b geq 0$ then
$$ x^2 < b iff - sqrt b < x < sqrt b.$$
So $$ f^{-1}((-infty,b))= begin{cases} varnothing &in mathcal{B}(mathbb{R}) & b < 0 \ (-sqrt b, sqrt b)& in mathcal{B}(mathbb{R}) & b geq 0end{cases}$$
and $f$ is borel.
Consider $f(x) = vert x vert.$ Let $b in mathbb{R}$:
- If $b < 0$:
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < 0} = varnothing$$
Since the empty set is an open (closed) it is a borel set so $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b < 0$.
- if $ b geq 0:$
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < b} = (-b,b)$$
Since $(-b,b)$ is an open set it is a borel set and $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b geq 0.$
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
$endgroup$
$f: mathbb{R} rightarrow mathbb{R}$ is borel $iff ; forall B in mathcal{B}(mathbb{R}) : f^{-1}(B) in mathcal{B}(mathbb{R}) iff ; forall b in mathbb{R} : f^{-1}(-infty,b) in mathcal{B}(mathbb{R}).$
$$ f^{-1}((-infty,b)) = {x : x^2 in (-infty,b) } = { x : x^2 < b }$$
If $b < 0 :$ then ${ x : x^2 < b } = varnothing.$
If $b geq 0$ then
$$ x^2 < b iff - sqrt b < x < sqrt b.$$
So $$ f^{-1}((-infty,b))= begin{cases} varnothing &in mathcal{B}(mathbb{R}) & b < 0 \ (-sqrt b, sqrt b)& in mathcal{B}(mathbb{R}) & b geq 0end{cases}$$
and $f$ is borel.
Consider $f(x) = vert x vert.$ Let $b in mathbb{R}$:
- If $b < 0$:
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < 0} = varnothing$$
Since the empty set is an open (closed) it is a borel set so $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b < 0$.
- if $ b geq 0:$
$$ f^{-1}(-infty ,b) = {x in mathbb{R} : f(x) = vert x vert < b} = (-b,b)$$
Since $(-b,b)$ is an open set it is a borel set and $ f^{-1}(-infty ,b) in mathcal{B}(mathbb{R}) in forall b geq 0.$
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
edited Dec 11 '18 at 15:14
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
answered Dec 10 '18 at 14:55
DigitalisDigitalis
528216
528216
$begingroup$
Where did you take into consideration $|x|=x^++x^-$?
$endgroup$
– Atstovas
Dec 11 '18 at 13:29
$begingroup$
@Atstovas I didn't. I thought I'd only do one. Are you having trouble for $vert xvert$ ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:01
$begingroup$
Yesss... accually I new how to show that $x^2$ is Borel function. But I didn’t know what to do with given conditions
$endgroup$
– Atstovas
Dec 11 '18 at 14:10
$begingroup$
Which given conditions ? For $ f = vert x vert$ you should do exactly the same show that $f^{-1}(-infty,b)$ is a borel set for all $b$. You're making me doubt if you even understood the solution for $f = x^2$. Are you sure you understood that one ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:26
$begingroup$
@Atstovas I've edited my answer. Any questions ?
$endgroup$
– Digitalis
Dec 11 '18 at 15:15
|
show 1 more comment
$begingroup$
Where did you take into consideration $|x|=x^++x^-$?
$endgroup$
– Atstovas
Dec 11 '18 at 13:29
$begingroup$
@Atstovas I didn't. I thought I'd only do one. Are you having trouble for $vert xvert$ ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:01
$begingroup$
Yesss... accually I new how to show that $x^2$ is Borel function. But I didn’t know what to do with given conditions
$endgroup$
– Atstovas
Dec 11 '18 at 14:10
$begingroup$
Which given conditions ? For $ f = vert x vert$ you should do exactly the same show that $f^{-1}(-infty,b)$ is a borel set for all $b$. You're making me doubt if you even understood the solution for $f = x^2$. Are you sure you understood that one ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:26
$begingroup$
@Atstovas I've edited my answer. Any questions ?
$endgroup$
– Digitalis
Dec 11 '18 at 15:15
$begingroup$
Where did you take into consideration $|x|=x^++x^-$?
$endgroup$
– Atstovas
Dec 11 '18 at 13:29
$begingroup$
Where did you take into consideration $|x|=x^++x^-$?
$endgroup$
– Atstovas
Dec 11 '18 at 13:29
$begingroup$
@Atstovas I didn't. I thought I'd only do one. Are you having trouble for $vert xvert$ ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:01
$begingroup$
@Atstovas I didn't. I thought I'd only do one. Are you having trouble for $vert xvert$ ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:01
$begingroup$
Yesss... accually I new how to show that $x^2$ is Borel function. But I didn’t know what to do with given conditions
$endgroup$
– Atstovas
Dec 11 '18 at 14:10
$begingroup$
Yesss... accually I new how to show that $x^2$ is Borel function. But I didn’t know what to do with given conditions
$endgroup$
– Atstovas
Dec 11 '18 at 14:10
$begingroup$
Which given conditions ? For $ f = vert x vert$ you should do exactly the same show that $f^{-1}(-infty,b)$ is a borel set for all $b$. You're making me doubt if you even understood the solution for $f = x^2$. Are you sure you understood that one ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:26
$begingroup$
Which given conditions ? For $ f = vert x vert$ you should do exactly the same show that $f^{-1}(-infty,b)$ is a borel set for all $b$. You're making me doubt if you even understood the solution for $f = x^2$. Are you sure you understood that one ?
$endgroup$
– Digitalis
Dec 11 '18 at 14:26
$begingroup$
@Atstovas I've edited my answer. Any questions ?
$endgroup$
– Digitalis
Dec 11 '18 at 15:15
$begingroup$
@Atstovas I've edited my answer. Any questions ?
$endgroup$
– Digitalis
Dec 11 '18 at 15:15
|
show 1 more comment
$begingroup$
If you really want to run the details with $x to x^2$, I suggest that you take a look at this
Show that continuous functions on $mathbb R$ are Borel-measurable
Probably though, it would be smarter to really understand the answer given in this link, and thus to be able to show that this holds for any continuous function on $mathbb{R}$, which includes the case you are looking for.
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
$endgroup$
add a comment |
$begingroup$
If you really want to run the details with $x to x^2$, I suggest that you take a look at this
Show that continuous functions on $mathbb R$ are Borel-measurable
Probably though, it would be smarter to really understand the answer given in this link, and thus to be able to show that this holds for any continuous function on $mathbb{R}$, which includes the case you are looking for.
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
$endgroup$
add a comment |
$begingroup$
If you really want to run the details with $x to x^2$, I suggest that you take a look at this
Show that continuous functions on $mathbb R$ are Borel-measurable
Probably though, it would be smarter to really understand the answer given in this link, and thus to be able to show that this holds for any continuous function on $mathbb{R}$, which includes the case you are looking for.
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
$endgroup$
If you really want to run the details with $x to x^2$, I suggest that you take a look at this
Show that continuous functions on $mathbb R$ are Borel-measurable
Probably though, it would be smarter to really understand the answer given in this link, and thus to be able to show that this holds for any continuous function on $mathbb{R}$, which includes the case you are looking for.
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
answered Dec 10 '18 at 14:44
Gâteau-GalloisGâteau-Gallois
362112
362112
add a comment |
add a comment |
$begingroup$
Not answered but this question has useful tips in the comments.
$endgroup$
– Boshu
Dec 10 '18 at 13:53
$begingroup$
An easier thing to do might be to prove that all continuous functions are Borel, thus the results for all the functions you mention will follow
$endgroup$
– Gâteau-Gallois
Dec 10 '18 at 14:42
$begingroup$
@Gâteau-Gallois Isn't this kind of cyclic though ? To show continuity he must show that the pre image of any open set is an open set.
$endgroup$
– Digitalis
Dec 10 '18 at 14:52