Mixing
![Do Random Variables have to be bijective? [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Measurability of a measurable function times a negative constant
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Given a measurable space $(X,scrA)$ where $scrA$ is a sigma algebra on $X$, I would like to know if $f:X to mathbbR$ is measurable then $cf$ where $c in mathbbR$ is also measurable.
A function $f:X to mathbbR$ is measurable if $[f > alpha] = x in X ::: f(x) > alpha in mathscrA$ for all $alpha in mathbbR$.
If $c = 0$, then $[cf > alpha] = X$ for all $alpha < 0$ and $[cf > alpha] = emptyset$ for all $alpha geq 0$, so $[cf > alpha] in scrA$.
If $c > 0$, then $[cf > alpha] = [f > fracalphac] in scrA$.
My question is when $c < 0$, my attempt is as follows.
Since $[cf > alpha] = [f < fracalphac] = x in X ::: f(x) < fracalphac $ and $[f > fracalphac] in scrA$, the complement of $[f > fracalphac] $ which is $[f leq fracalphac] in scrA$, but as $[f < fracalphac] neq [f leq fracalphac]$, I am stuck in showing $[f < fracalphac] in scrA$.
Any help will be greatly appreciated.
measure-theory
asked Jul 22 at 10:42
James
1688
add a comment |Â
up vote
0
down vote
favorite
Given a measurable space $(X,scrA)$ where $scrA$ is a sigma algebra on $X$, I would like to know if $f:X to mathbbR$ is measurable then $cf$ where $c in mathbbR$ is also measurable.
A function $f:X to mathbbR$ is measurable if $[f > alpha] = x in X ::: f(x) > alpha in mathscrA$ for all $alpha in mathbbR$.
If $c = 0$, then $[cf > alpha] = X$ for all $alpha < 0$ and $[cf > alpha] = emptyset$ for all $alpha geq 0$, so $[cf > alpha] in scrA$.
If $c > 0$, then $[cf > alpha] = [f > fracalphac] in scrA$.
My question is when $c < 0$, my attempt is as follows.
Since $[cf > alpha] = [f < fracalphac] = x in X ::: f(x) < fracalphac $ and $[f > fracalphac] in scrA$, the complement of $[f > fracalphac] $ which is $[f leq fracalphac] in scrA$, but as $[f < fracalphac] neq [f leq fracalphac]$, I am stuck in showing $[f < fracalphac] in scrA$.
Any help will be greatly appreciated.
measure-theory
asked Jul 22 at 10:42
James
1688
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a measurable space $(X,scrA)$ where $scrA$ is a sigma algebra on $X$, I would like to know if $f:X to mathbbR$ is measurable then $cf$ where $c in mathbbR$ is also measurable.
A function $f:X to mathbbR$ is measurable if $[f > alpha] = x in X ::: f(x) > alpha in mathscrA$ for all $alpha in mathbbR$.
If $c = 0$, then $[cf > alpha] = X$ for all $alpha < 0$ and $[cf > alpha] = emptyset$ for all $alpha geq 0$, so $[cf > alpha] in scrA$.
If $c > 0$, then $[cf > alpha] = [f > fracalphac] in scrA$.
My question is when $c < 0$, my attempt is as follows.
Since $[cf > alpha] = [f < fracalphac] = x in X ::: f(x) < fracalphac $ and $[f > fracalphac] in scrA$, the complement of $[f > fracalphac] $ which is $[f leq fracalphac] in scrA$, but as $[f < fracalphac] neq [f leq fracalphac]$, I am stuck in showing $[f < fracalphac] in scrA$.
Any help will be greatly appreciated.
measure-theory
asked Jul 22 at 10:42
James
1688
Given a measurable space $(X,scrA)$ where $scrA$ is a sigma algebra on $X$, I would like to know if $f:X to mathbbR$ is measurable then $cf$ where $c in mathbbR$ is also measurable.
A function $f:X to mathbbR$ is measurable if $[f > alpha] = x in X ::: f(x) > alpha in mathscrA$ for all $alpha in mathbbR$.
If $c = 0$, then $[cf > alpha] = X$ for all $alpha < 0$ and $[cf > alpha] = emptyset$ for all $alpha geq 0$, so $[cf > alpha] in scrA$.
If $c > 0$, then $[cf > alpha] = [f > fracalphac] in scrA$.
My question is when $c < 0$, my attempt is as follows.
Since $[cf > alpha] = [f < fracalphac] = x in X ::: f(x) < fracalphac $ and $[f > fracalphac] in scrA$, the complement of $[f > fracalphac] $ which is $[f leq fracalphac] in scrA$, but as $[f < fracalphac] neq [f leq fracalphac]$, I am stuck in showing $[f < fracalphac] in scrA$.
Any help will be greatly appreciated.
measure-theory
asked Jul 22 at 10:42
James
1688
asked Jul 22 at 10:42
James
1688
asked Jul 22 at 10:42
James
1688
asked Jul 22 at 10:42
James
1688
1688
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Since $mathcal A$ is a $sigma$-algebra, asserting that$$leftxin X,middleinmathcal A$$is equivalent to asserting that its complement, which is$$leftxin X,middle,tag1$$belongs to $mathcal A$. And$$(1)=bigcap_ninmathbb Nleft,f(x)>fracalpha c-frac1nright$$Therefore, $(1)inmathcal A$.
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
Thank you very much for your enlightenment as always, greatly appreciated.
– James
Jul 22 at 12:07
I'm glad I could help.
– José Carlos Santos
Jul 22 at 12:08
add a comment |Â
up vote
1
down vote
Note that for $b=alpha /c$ we have $[f<b] = [fge b]^C$ and
$$[fge b] = cap_n=1^infty [f >b - 1/n]in mathscr A.$$
($A^C$ is the complement of the set $A$)
answered Jul 22 at 11:04
Epiousios
1,481522
Thank you so much it was really helpful.
– James
Jul 22 at 12:07
add a comment |Â
up vote
1
down vote
In general if $(X,mathscr A)$, $(Y,mathscr B)$ and $(Z,mathscr C)$ are measurable spaces and $f:Xto Y$ and $g:Yto Z$ are both measurable then also $gcirc f:Xto Z$ is measurable.
This because: $$(gcirc f)^-1(mathscr C)=f^-1(g^-1(mathscr C))subseteq f^-1(mathscr B)subseteqmathscr A$$
The first $subseteq$ because $g$ is measurable and the second $subseteq$ because $f$ is measurable.
This can be applied here for $(Y,mathscr B)=(mathbb R,mathcal B)=(Z,mathscr C)$ where $cf:Xtomathbb R$ can be recognized as $gcirc f$ if the function $g:mathbb Rtomathbb R$ is prescribed by $xmapsto cx$ where $cinmathbb R$. It is evident that this function $g$ is measurable.
answered Jul 22 at 11:39
drhab
86.4k541118
Thank you for your great wisdom.
– James
Jul 22 at 12:07
On my turn I must thank Someone else for that ;-).
– drhab
Jul 22 at 12:10
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $mathcal A$ is a $sigma$-algebra, asserting that$$leftxin X,middleinmathcal A$$is equivalent to asserting that its complement, which is$$leftxin X,middle,tag1$$belongs to $mathcal A$. And$$(1)=bigcap_ninmathbb Nleft,f(x)>fracalpha c-frac1nright$$Therefore, $(1)inmathcal A$.
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
Thank you very much for your enlightenment as always, greatly appreciated.
– James
Jul 22 at 12:07
I'm glad I could help.
– José Carlos Santos
Jul 22 at 12:08
add a comment |Â
up vote
2
down vote
accepted
Since $mathcal A$ is a $sigma$-algebra, asserting that$$leftxin X,middleinmathcal A$$is equivalent to asserting that its complement, which is$$leftxin X,middle,tag1$$belongs to $mathcal A$. And$$(1)=bigcap_ninmathbb Nleft,f(x)>fracalpha c-frac1nright$$Therefore, $(1)inmathcal A$.
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
Thank you very much for your enlightenment as always, greatly appreciated.
– James
Jul 22 at 12:07
I'm glad I could help.
– José Carlos Santos
Jul 22 at 12:08
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $mathcal A$ is a $sigma$-algebra, asserting that$$leftxin X,middleinmathcal A$$is equivalent to asserting that its complement, which is$$leftxin X,middle,tag1$$belongs to $mathcal A$. And$$(1)=bigcap_ninmathbb Nleft,f(x)>fracalpha c-frac1nright$$Therefore, $(1)inmathcal A$.
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
Since $mathcal A$ is a $sigma$-algebra, asserting that$$leftxin X,middleinmathcal A$$is equivalent to asserting that its complement, which is$$leftxin X,middle,tag1$$belongs to $mathcal A$. And$$(1)=bigcap_ninmathbb Nleft,f(x)>fracalpha c-frac1nright$$Therefore, $(1)inmathcal A$.
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
answered Jul 22 at 11:05
José Carlos Santos
113k1698177
113k1698177
Thank you very much for your enlightenment as always, greatly appreciated.
– James
Jul 22 at 12:07
I'm glad I could help.
– José Carlos Santos
Jul 22 at 12:08
add a comment |Â
Thank you very much for your enlightenment as always, greatly appreciated.
– James
Jul 22 at 12:07
I'm glad I could help.
– José Carlos Santos
Jul 22 at 12:08
Thank you very much for your enlightenment as always, greatly appreciated.
– James
Jul 22 at 12:07
Thank you very much for your enlightenment as always, greatly appreciated.
– James
Jul 22 at 12:07
I'm glad I could help.
– José Carlos Santos
Jul 22 at 12:08
I'm glad I could help.
– José Carlos Santos
Jul 22 at 12:08
add a comment |Â
up vote
1
down vote
Note that for $b=alpha /c$ we have $[f<b] = [fge b]^C$ and
$$[fge b] = cap_n=1^infty [f >b - 1/n]in mathscr A.$$
($A^C$ is the complement of the set $A$)
answered Jul 22 at 11:04
Epiousios
1,481522
Thank you so much it was really helpful.
– James
Jul 22 at 12:07
add a comment |Â
up vote
1
down vote
Note that for $b=alpha /c$ we have $[f<b] = [fge b]^C$ and
$$[fge b] = cap_n=1^infty [f >b - 1/n]in mathscr A.$$
($A^C$ is the complement of the set $A$)
answered Jul 22 at 11:04
Epiousios
1,481522
Thank you so much it was really helpful.
– James
Jul 22 at 12:07
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that for $b=alpha /c$ we have $[f<b] = [fge b]^C$ and
$$[fge b] = cap_n=1^infty [f >b - 1/n]in mathscr A.$$
($A^C$ is the complement of the set $A$)
answered Jul 22 at 11:04
Epiousios
1,481522
Note that for $b=alpha /c$ we have $[f<b] = [fge b]^C$ and
$$[fge b] = cap_n=1^infty [f >b - 1/n]in mathscr A.$$
($A^C$ is the complement of the set $A$)
answered Jul 22 at 11:04
Epiousios
1,481522
answered Jul 22 at 11:04
Epiousios
1,481522
answered Jul 22 at 11:04
Epiousios
1,481522
answered Jul 22 at 11:04
Epiousios
1,481522
1,481522
Thank you so much it was really helpful.
– James
Jul 22 at 12:07
add a comment |Â
Thank you so much it was really helpful.
– James
Jul 22 at 12:07
Thank you so much it was really helpful.
– James
Jul 22 at 12:07
Thank you so much it was really helpful.
– James
Jul 22 at 12:07
add a comment |Â
up vote
1
down vote
In general if $(X,mathscr A)$, $(Y,mathscr B)$ and $(Z,mathscr C)$ are measurable spaces and $f:Xto Y$ and $g:Yto Z$ are both measurable then also $gcirc f:Xto Z$ is measurable.
This because: $$(gcirc f)^-1(mathscr C)=f^-1(g^-1(mathscr C))subseteq f^-1(mathscr B)subseteqmathscr A$$
The first $subseteq$ because $g$ is measurable and the second $subseteq$ because $f$ is measurable.
This can be applied here for $(Y,mathscr B)=(mathbb R,mathcal B)=(Z,mathscr C)$ where $cf:Xtomathbb R$ can be recognized as $gcirc f$ if the function $g:mathbb Rtomathbb R$ is prescribed by $xmapsto cx$ where $cinmathbb R$. It is evident that this function $g$ is measurable.
answered Jul 22 at 11:39
drhab
86.4k541118
Thank you for your great wisdom.
– James
Jul 22 at 12:07
On my turn I must thank Someone else for that ;-).
– drhab
Jul 22 at 12:10
add a comment |Â
up vote
1
down vote
In general if $(X,mathscr A)$, $(Y,mathscr B)$ and $(Z,mathscr C)$ are measurable spaces and $f:Xto Y$ and $g:Yto Z$ are both measurable then also $gcirc f:Xto Z$ is measurable.
This because: $$(gcirc f)^-1(mathscr C)=f^-1(g^-1(mathscr C))subseteq f^-1(mathscr B)subseteqmathscr A$$
The first $subseteq$ because $g$ is measurable and the second $subseteq$ because $f$ is measurable.
This can be applied here for $(Y,mathscr B)=(mathbb R,mathcal B)=(Z,mathscr C)$ where $cf:Xtomathbb R$ can be recognized as $gcirc f$ if the function $g:mathbb Rtomathbb R$ is prescribed by $xmapsto cx$ where $cinmathbb R$. It is evident that this function $g$ is measurable.
answered Jul 22 at 11:39
drhab
86.4k541118
Thank you for your great wisdom.
– James
Jul 22 at 12:07
On my turn I must thank Someone else for that ;-).
– drhab
Jul 22 at 12:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In general if $(X,mathscr A)$, $(Y,mathscr B)$ and $(Z,mathscr C)$ are measurable spaces and $f:Xto Y$ and $g:Yto Z$ are both measurable then also $gcirc f:Xto Z$ is measurable.
This because: $$(gcirc f)^-1(mathscr C)=f^-1(g^-1(mathscr C))subseteq f^-1(mathscr B)subseteqmathscr A$$
The first $subseteq$ because $g$ is measurable and the second $subseteq$ because $f$ is measurable.
This can be applied here for $(Y,mathscr B)=(mathbb R,mathcal B)=(Z,mathscr C)$ where $cf:Xtomathbb R$ can be recognized as $gcirc f$ if the function $g:mathbb Rtomathbb R$ is prescribed by $xmapsto cx$ where $cinmathbb R$. It is evident that this function $g$ is measurable.
answered Jul 22 at 11:39
drhab
86.4k541118
In general if $(X,mathscr A)$, $(Y,mathscr B)$ and $(Z,mathscr C)$ are measurable spaces and $f:Xto Y$ and $g:Yto Z$ are both measurable then also $gcirc f:Xto Z$ is measurable.
This because: $$(gcirc f)^-1(mathscr C)=f^-1(g^-1(mathscr C))subseteq f^-1(mathscr B)subseteqmathscr A$$
The first $subseteq$ because $g$ is measurable and the second $subseteq$ because $f$ is measurable.
This can be applied here for $(Y,mathscr B)=(mathbb R,mathcal B)=(Z,mathscr C)$ where $cf:Xtomathbb R$ can be recognized as $gcirc f$ if the function $g:mathbb Rtomathbb R$ is prescribed by $xmapsto cx$ where $cinmathbb R$. It is evident that this function $g$ is measurable.
answered Jul 22 at 11:39
drhab
86.4k541118
edited Jul 22 at 12:11
answered Jul 22 at 11:39
drhab
86.4k541118
answered Jul 22 at 11:39
drhab
86.4k541118
answered Jul 22 at 11:39
drhab
86.4k541118
86.4k541118
Thank you for your great wisdom.
– James
Jul 22 at 12:07
On my turn I must thank Someone else for that ;-).
– drhab
Jul 22 at 12:10
add a comment |Â
Thank you for your great wisdom.
– James
Jul 22 at 12:07
On my turn I must thank Someone else for that ;-).
– drhab
Jul 22 at 12:10
Thank you for your great wisdom.
– James
Jul 22 at 12:07
Thank you for your great wisdom.
– James
Jul 22 at 12:07
On my turn I must thank Someone else for that ;-).
– drhab
Jul 22 at 12:10
On my turn I must thank Someone else for that ;-).
– drhab
Jul 22 at 12:10
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859276%2fmeasurability-of-a-measurable-function-times-a-negative-constant%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password