Mixing
Push-Forward Algebra of $mathcal{B}(mathbb{R})$ by $f(x) = tan(x)$
$begingroup$
Specifically $f(x)$ is defined to be $tan(x)$ when $cos(x) neq 0$, and $0$ elsewhere.
I am required to determine the $sigma$-algebra $mathcal{F} := {A subset mathbb{R}: f^{-1}(A) in mathcal{B}(mathbb{R}) }$
My intuition suggests that this push-forward algebra is $mathcal{B}$, as the preimage of any Borel subset of $mathbb{R}$ is countably many copies of what is essentially a "continuously deformed" version of that Borel subset.
It is easy to show that $mathcal{B}(mathbb{R})$ is a subset of the push-forward algebra, as the preimage of an open interval is a countable union of open intervals (with countably many isolated points $(..., -frac{pi}{2}, frac{pi}{2}, ...)$ thrown in if the open interval contains zero). Thus, as the Borel algebra on the reals is the smallest $sigma$-algebra containing the open intervals, and the push-forward algebra contains the open intervals, clearly $mathcal{B}(mathbb{R})$ is contained in the push-forward algebra.
For the reverse inclusion, however, I have no idea, and any help would be much appreciated!
integration measure-theory
edited Jan 31 at 13:21
Jordan Green
1,146410
asked Jan 31 at 12:24
QwertiopsQwertiops
185
$endgroup$
add a comment |
$begingroup$
Specifically $f(x)$ is defined to be $tan(x)$ when $cos(x) neq 0$, and $0$ elsewhere.
I am required to determine the $sigma$-algebra $mathcal{F} := {A subset mathbb{R}: f^{-1}(A) in mathcal{B}(mathbb{R}) }$
My intuition suggests that this push-forward algebra is $mathcal{B}$, as the preimage of any Borel subset of $mathbb{R}$ is countably many copies of what is essentially a "continuously deformed" version of that Borel subset.
It is easy to show that $mathcal{B}(mathbb{R})$ is a subset of the push-forward algebra, as the preimage of an open interval is a countable union of open intervals (with countably many isolated points $(..., -frac{pi}{2}, frac{pi}{2}, ...)$ thrown in if the open interval contains zero). Thus, as the Borel algebra on the reals is the smallest $sigma$-algebra containing the open intervals, and the push-forward algebra contains the open intervals, clearly $mathcal{B}(mathbb{R})$ is contained in the push-forward algebra.
For the reverse inclusion, however, I have no idea, and any help would be much appreciated!
integration measure-theory
edited Jan 31 at 13:21
Jordan Green
1,146410
asked Jan 31 at 12:24
QwertiopsQwertiops
185
$endgroup$
add a comment |
$begingroup$
Specifically $f(x)$ is defined to be $tan(x)$ when $cos(x) neq 0$, and $0$ elsewhere.
I am required to determine the $sigma$-algebra $mathcal{F} := {A subset mathbb{R}: f^{-1}(A) in mathcal{B}(mathbb{R}) }$
My intuition suggests that this push-forward algebra is $mathcal{B}$, as the preimage of any Borel subset of $mathbb{R}$ is countably many copies of what is essentially a "continuously deformed" version of that Borel subset.
It is easy to show that $mathcal{B}(mathbb{R})$ is a subset of the push-forward algebra, as the preimage of an open interval is a countable union of open intervals (with countably many isolated points $(..., -frac{pi}{2}, frac{pi}{2}, ...)$ thrown in if the open interval contains zero). Thus, as the Borel algebra on the reals is the smallest $sigma$-algebra containing the open intervals, and the push-forward algebra contains the open intervals, clearly $mathcal{B}(mathbb{R})$ is contained in the push-forward algebra.
For the reverse inclusion, however, I have no idea, and any help would be much appreciated!
integration measure-theory
edited Jan 31 at 13:21
Jordan Green
1,146410
asked Jan 31 at 12:24
QwertiopsQwertiops
185
$endgroup$
Specifically $f(x)$ is defined to be $tan(x)$ when $cos(x) neq 0$, and $0$ elsewhere.
I am required to determine the $sigma$-algebra $mathcal{F} := {A subset mathbb{R}: f^{-1}(A) in mathcal{B}(mathbb{R}) }$
My intuition suggests that this push-forward algebra is $mathcal{B}$, as the preimage of any Borel subset of $mathbb{R}$ is countably many copies of what is essentially a "continuously deformed" version of that Borel subset.
It is easy to show that $mathcal{B}(mathbb{R})$ is a subset of the push-forward algebra, as the preimage of an open interval is a countable union of open intervals (with countably many isolated points $(..., -frac{pi}{2}, frac{pi}{2}, ...)$ thrown in if the open interval contains zero). Thus, as the Borel algebra on the reals is the smallest $sigma$-algebra containing the open intervals, and the push-forward algebra contains the open intervals, clearly $mathcal{B}(mathbb{R})$ is contained in the push-forward algebra.
For the reverse inclusion, however, I have no idea, and any help would be much appreciated!
integration measure-theory
integration measure-theory
edited Jan 31 at 13:21
Jordan Green
1,146410
asked Jan 31 at 12:24
QwertiopsQwertiops
185
edited Jan 31 at 13:21
Jordan Green
1,146410
asked Jan 31 at 12:24
QwertiopsQwertiops
185
edited Jan 31 at 13:21
Jordan Green
1,146410
edited Jan 31 at 13:21
Jordan Green
1,146410
edited Jan 31 at 13:21
Jordan Green
1,146410
1,146410
asked Jan 31 at 12:24
QwertiopsQwertiops
185
asked Jan 31 at 12:24
QwertiopsQwertiops
185
asked Jan 31 at 12:24
QwertiopsQwertiops
185
185
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To show that $mathcal{F} subset mathcal{B}$, we must show that if $f^{-1}(A)$ is Borel measurable, then $A$ is Borel measurable.
For this, suppose that $A subset mathbb{R}$ with $f^{-1}(A) in mathcal{B}(mathbb{R})$. We take two cases according to whether $A$ contains $0$. In either case, the key insight is that the continuous image of a Borel set under an injective function $g: mathbb{R} to mathbb{R}$ is again a Borel set. (For a proof, see Image of Borel set under continuous and injective map.) I guess it should be noted that the same still holds if the domain of $g$ is instead an interval.
Since $f$ is surjective, we have
$$
A = f(f^{-1}(A)), tag{1}.
$$
That is, $A$ is the image of $f^{-1}(A)$ under $f$. Of course, $f$ is not injective, but we will decompose the domain of $f$ into intervals so that the restriction of $f$ to each interval is injective.
Specifically, for each $j in mathbb{Z}$, let $f_j$ be the restriction of $f$ to the interval $(-frac{pi}{2}, frac{pi}{2}) + 2 pi j$. Each function $f_j$ is a continuous bijection of its domain onto $mathbb{R}$, and each of the sets $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$, because we can write each as the intersection of Borel sets; specifically, we have
$
f_j^{-1}(A) = f^{-1}(A) cap left(left(-frac{pi}{2}, frac{pi}{2} right) + 2 pi j right).
$
If $A$ does not contain $0$, then we can write
$
f^{-1}(A) = bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
in which case equation (1) becomes
$$
A = f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}} f left( f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Since each set $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$ and each function $f_j$ is injective on its domain, $f_j left( f_j^{-1}(A) right)$ is in $mathcal{B}(mathbb{R})$ for each $j$, which means that $A in mathcal{B}(mathbb{R})$.
In the case that $A$ does contain $0$, the proof works almost the same way. We let
$
S = left {dots, -frac{5 pi}{2}, -frac{3 pi}{2}, -frac{pi}{2}, -frac{pi}{2}, frac{3 pi}{2}, frac{5 pi}{2}, dots right }.
$
Then we have that
$
f^{-1}(A) = S cup bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
and, consequently,
$$
A = f(S) cup f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right) = { 0 } cup bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Thus, $A$ is a countable union of Borel measurable sets, which means that $A$ is Borel measurable.
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
$endgroup$
1
$begingroup$
Thank you for such a detailed and well-written response! All is clear now.
$endgroup$
– Qwertiops
Feb 1 at 21:01
add a comment |
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1 Answer
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$begingroup$
To show that $mathcal{F} subset mathcal{B}$, we must show that if $f^{-1}(A)$ is Borel measurable, then $A$ is Borel measurable.
For this, suppose that $A subset mathbb{R}$ with $f^{-1}(A) in mathcal{B}(mathbb{R})$. We take two cases according to whether $A$ contains $0$. In either case, the key insight is that the continuous image of a Borel set under an injective function $g: mathbb{R} to mathbb{R}$ is again a Borel set. (For a proof, see Image of Borel set under continuous and injective map.) I guess it should be noted that the same still holds if the domain of $g$ is instead an interval.
Since $f$ is surjective, we have
$$
A = f(f^{-1}(A)), tag{1}.
$$
That is, $A$ is the image of $f^{-1}(A)$ under $f$. Of course, $f$ is not injective, but we will decompose the domain of $f$ into intervals so that the restriction of $f$ to each interval is injective.
Specifically, for each $j in mathbb{Z}$, let $f_j$ be the restriction of $f$ to the interval $(-frac{pi}{2}, frac{pi}{2}) + 2 pi j$. Each function $f_j$ is a continuous bijection of its domain onto $mathbb{R}$, and each of the sets $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$, because we can write each as the intersection of Borel sets; specifically, we have
$
f_j^{-1}(A) = f^{-1}(A) cap left(left(-frac{pi}{2}, frac{pi}{2} right) + 2 pi j right).
$
If $A$ does not contain $0$, then we can write
$
f^{-1}(A) = bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
in which case equation (1) becomes
$$
A = f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}} f left( f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Since each set $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$ and each function $f_j$ is injective on its domain, $f_j left( f_j^{-1}(A) right)$ is in $mathcal{B}(mathbb{R})$ for each $j$, which means that $A in mathcal{B}(mathbb{R})$.
In the case that $A$ does contain $0$, the proof works almost the same way. We let
$
S = left {dots, -frac{5 pi}{2}, -frac{3 pi}{2}, -frac{pi}{2}, -frac{pi}{2}, frac{3 pi}{2}, frac{5 pi}{2}, dots right }.
$
Then we have that
$
f^{-1}(A) = S cup bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
and, consequently,
$$
A = f(S) cup f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right) = { 0 } cup bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Thus, $A$ is a countable union of Borel measurable sets, which means that $A$ is Borel measurable.
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
$endgroup$
1
$begingroup$
Thank you for such a detailed and well-written response! All is clear now.
$endgroup$
– Qwertiops
Feb 1 at 21:01
add a comment |
$begingroup$
To show that $mathcal{F} subset mathcal{B}$, we must show that if $f^{-1}(A)$ is Borel measurable, then $A$ is Borel measurable.
For this, suppose that $A subset mathbb{R}$ with $f^{-1}(A) in mathcal{B}(mathbb{R})$. We take two cases according to whether $A$ contains $0$. In either case, the key insight is that the continuous image of a Borel set under an injective function $g: mathbb{R} to mathbb{R}$ is again a Borel set. (For a proof, see Image of Borel set under continuous and injective map.) I guess it should be noted that the same still holds if the domain of $g$ is instead an interval.
Since $f$ is surjective, we have
$$
A = f(f^{-1}(A)), tag{1}.
$$
That is, $A$ is the image of $f^{-1}(A)$ under $f$. Of course, $f$ is not injective, but we will decompose the domain of $f$ into intervals so that the restriction of $f$ to each interval is injective.
Specifically, for each $j in mathbb{Z}$, let $f_j$ be the restriction of $f$ to the interval $(-frac{pi}{2}, frac{pi}{2}) + 2 pi j$. Each function $f_j$ is a continuous bijection of its domain onto $mathbb{R}$, and each of the sets $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$, because we can write each as the intersection of Borel sets; specifically, we have
$
f_j^{-1}(A) = f^{-1}(A) cap left(left(-frac{pi}{2}, frac{pi}{2} right) + 2 pi j right).
$
If $A$ does not contain $0$, then we can write
$
f^{-1}(A) = bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
in which case equation (1) becomes
$$
A = f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}} f left( f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Since each set $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$ and each function $f_j$ is injective on its domain, $f_j left( f_j^{-1}(A) right)$ is in $mathcal{B}(mathbb{R})$ for each $j$, which means that $A in mathcal{B}(mathbb{R})$.
In the case that $A$ does contain $0$, the proof works almost the same way. We let
$
S = left {dots, -frac{5 pi}{2}, -frac{3 pi}{2}, -frac{pi}{2}, -frac{pi}{2}, frac{3 pi}{2}, frac{5 pi}{2}, dots right }.
$
Then we have that
$
f^{-1}(A) = S cup bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
and, consequently,
$$
A = f(S) cup f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right) = { 0 } cup bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Thus, $A$ is a countable union of Borel measurable sets, which means that $A$ is Borel measurable.
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
$endgroup$
1
$begingroup$
Thank you for such a detailed and well-written response! All is clear now.
$endgroup$
– Qwertiops
Feb 1 at 21:01
add a comment |
$begingroup$
To show that $mathcal{F} subset mathcal{B}$, we must show that if $f^{-1}(A)$ is Borel measurable, then $A$ is Borel measurable.
For this, suppose that $A subset mathbb{R}$ with $f^{-1}(A) in mathcal{B}(mathbb{R})$. We take two cases according to whether $A$ contains $0$. In either case, the key insight is that the continuous image of a Borel set under an injective function $g: mathbb{R} to mathbb{R}$ is again a Borel set. (For a proof, see Image of Borel set under continuous and injective map.) I guess it should be noted that the same still holds if the domain of $g$ is instead an interval.
Since $f$ is surjective, we have
$$
A = f(f^{-1}(A)), tag{1}.
$$
That is, $A$ is the image of $f^{-1}(A)$ under $f$. Of course, $f$ is not injective, but we will decompose the domain of $f$ into intervals so that the restriction of $f$ to each interval is injective.
Specifically, for each $j in mathbb{Z}$, let $f_j$ be the restriction of $f$ to the interval $(-frac{pi}{2}, frac{pi}{2}) + 2 pi j$. Each function $f_j$ is a continuous bijection of its domain onto $mathbb{R}$, and each of the sets $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$, because we can write each as the intersection of Borel sets; specifically, we have
$
f_j^{-1}(A) = f^{-1}(A) cap left(left(-frac{pi}{2}, frac{pi}{2} right) + 2 pi j right).
$
If $A$ does not contain $0$, then we can write
$
f^{-1}(A) = bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
in which case equation (1) becomes
$$
A = f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}} f left( f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Since each set $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$ and each function $f_j$ is injective on its domain, $f_j left( f_j^{-1}(A) right)$ is in $mathcal{B}(mathbb{R})$ for each $j$, which means that $A in mathcal{B}(mathbb{R})$.
In the case that $A$ does contain $0$, the proof works almost the same way. We let
$
S = left {dots, -frac{5 pi}{2}, -frac{3 pi}{2}, -frac{pi}{2}, -frac{pi}{2}, frac{3 pi}{2}, frac{5 pi}{2}, dots right }.
$
Then we have that
$
f^{-1}(A) = S cup bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
and, consequently,
$$
A = f(S) cup f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right) = { 0 } cup bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Thus, $A$ is a countable union of Borel measurable sets, which means that $A$ is Borel measurable.
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
$endgroup$
To show that $mathcal{F} subset mathcal{B}$, we must show that if $f^{-1}(A)$ is Borel measurable, then $A$ is Borel measurable.
For this, suppose that $A subset mathbb{R}$ with $f^{-1}(A) in mathcal{B}(mathbb{R})$. We take two cases according to whether $A$ contains $0$. In either case, the key insight is that the continuous image of a Borel set under an injective function $g: mathbb{R} to mathbb{R}$ is again a Borel set. (For a proof, see Image of Borel set under continuous and injective map.) I guess it should be noted that the same still holds if the domain of $g$ is instead an interval.
Since $f$ is surjective, we have
$$
A = f(f^{-1}(A)), tag{1}.
$$
That is, $A$ is the image of $f^{-1}(A)$ under $f$. Of course, $f$ is not injective, but we will decompose the domain of $f$ into intervals so that the restriction of $f$ to each interval is injective.
Specifically, for each $j in mathbb{Z}$, let $f_j$ be the restriction of $f$ to the interval $(-frac{pi}{2}, frac{pi}{2}) + 2 pi j$. Each function $f_j$ is a continuous bijection of its domain onto $mathbb{R}$, and each of the sets $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$, because we can write each as the intersection of Borel sets; specifically, we have
$
f_j^{-1}(A) = f^{-1}(A) cap left(left(-frac{pi}{2}, frac{pi}{2} right) + 2 pi j right).
$
If $A$ does not contain $0$, then we can write
$
f^{-1}(A) = bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
in which case equation (1) becomes
$$
A = f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}} f left( f_j^{-1}(A) right)
= bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Since each set $f_j^{-1}(A)$ is in $mathcal{B}(mathbb{R})$ and each function $f_j$ is injective on its domain, $f_j left( f_j^{-1}(A) right)$ is in $mathcal{B}(mathbb{R})$ for each $j$, which means that $A in mathcal{B}(mathbb{R})$.
In the case that $A$ does contain $0$, the proof works almost the same way. We let
$
S = left {dots, -frac{5 pi}{2}, -frac{3 pi}{2}, -frac{pi}{2}, -frac{pi}{2}, frac{3 pi}{2}, frac{5 pi}{2}, dots right }.
$
Then we have that
$
f^{-1}(A) = S cup bigcup_{j in mathbb{Z}} f_j^{-1}(A),
$
and, consequently,
$$
A = f(S) cup f left( bigcup_{j in mathbb{Z}} f_j^{-1}(A) right) = { 0 } cup bigcup_{j in mathbb{Z}}f_j left( f_j^{-1}(A) right).
$$
Thus, $A$ is a countable union of Borel measurable sets, which means that $A$ is Borel measurable.
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
edited Jan 31 at 19:12
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
answered Jan 31 at 13:37
Jordan GreenJordan Green
1,146410
1,146410
1
$begingroup$
Thank you for such a detailed and well-written response! All is clear now.
$endgroup$
– Qwertiops
Feb 1 at 21:01
add a comment |
1
$begingroup$
Thank you for such a detailed and well-written response! All is clear now.
$endgroup$
– Qwertiops
Feb 1 at 21:01
1
1
$begingroup$
Thank you for such a detailed and well-written response! All is clear now.
$endgroup$
– Qwertiops
Feb 1 at 21:01
$begingroup$
Thank you for such a detailed and well-written response! All is clear now.
$endgroup$
– Qwertiops
Feb 1 at 21:01
add a comment |
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