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Extended Definition of Measurability
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Let $(X_1, mathcalA_1)$ and $(X_2, mathcalA_2)$ be measurable spaces. Given a map $fcolon X_1longrightarrow X_2$ we say that $f$ is measurable if $f^-1(V)$ is a measurable set in $X_1$ for every open set $V$ in $X_2$.
In Rudin's "Real and Complex Analysis, 3rd ed." the above definition is enlarged: We say that $f$ defined on a set $Ein mathcalA_1$ is measurable on $X_1$ if $mu(E^c)=0$ and if $f^-1(V)cap E$ is measurable for every open set $V$.
While I understand, that the secon definition is a real extension of the first, I don't understand what he writes next.
"If we define $f(x)=0$ for $xin E^c$, we obtain a measurable funciton on $X$, in the old sense" (Old sense = first definition). He then adds "If our measure happens to be complete, we can define $f$ on $E^c$ in a perfectly arbitrary manner, and we still get a measurable function"
I cannot see why defining $f(x)=0, xin E^c$ leads to the first defintion.
measure-theory
asked Aug 2 at 15:50
EpsilonDelta
4951513
add a comment |Â
up vote
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down vote
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Let $(X_1, mathcalA_1)$ and $(X_2, mathcalA_2)$ be measurable spaces. Given a map $fcolon X_1longrightarrow X_2$ we say that $f$ is measurable if $f^-1(V)$ is a measurable set in $X_1$ for every open set $V$ in $X_2$.
In Rudin's "Real and Complex Analysis, 3rd ed." the above definition is enlarged: We say that $f$ defined on a set $Ein mathcalA_1$ is measurable on $X_1$ if $mu(E^c)=0$ and if $f^-1(V)cap E$ is measurable for every open set $V$.
While I understand, that the secon definition is a real extension of the first, I don't understand what he writes next.
"If we define $f(x)=0$ for $xin E^c$, we obtain a measurable funciton on $X$, in the old sense" (Old sense = first definition). He then adds "If our measure happens to be complete, we can define $f$ on $E^c$ in a perfectly arbitrary manner, and we still get a measurable function"
I cannot see why defining $f(x)=0, xin E^c$ leads to the first defintion.
measure-theory
asked Aug 2 at 15:50
EpsilonDelta
4951513
There is no such thing as an open set in a measurable space, in general.
– Eric Wofsey
Aug 2 at 15:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X_1, mathcalA_1)$ and $(X_2, mathcalA_2)$ be measurable spaces. Given a map $fcolon X_1longrightarrow X_2$ we say that $f$ is measurable if $f^-1(V)$ is a measurable set in $X_1$ for every open set $V$ in $X_2$.
In Rudin's "Real and Complex Analysis, 3rd ed." the above definition is enlarged: We say that $f$ defined on a set $Ein mathcalA_1$ is measurable on $X_1$ if $mu(E^c)=0$ and if $f^-1(V)cap E$ is measurable for every open set $V$.
While I understand, that the secon definition is a real extension of the first, I don't understand what he writes next.
"If we define $f(x)=0$ for $xin E^c$, we obtain a measurable funciton on $X$, in the old sense" (Old sense = first definition). He then adds "If our measure happens to be complete, we can define $f$ on $E^c$ in a perfectly arbitrary manner, and we still get a measurable function"
I cannot see why defining $f(x)=0, xin E^c$ leads to the first defintion.
measure-theory
asked Aug 2 at 15:50
EpsilonDelta
4951513
Let $(X_1, mathcalA_1)$ and $(X_2, mathcalA_2)$ be measurable spaces. Given a map $fcolon X_1longrightarrow X_2$ we say that $f$ is measurable if $f^-1(V)$ is a measurable set in $X_1$ for every open set $V$ in $X_2$.
In Rudin's "Real and Complex Analysis, 3rd ed." the above definition is enlarged: We say that $f$ defined on a set $Ein mathcalA_1$ is measurable on $X_1$ if $mu(E^c)=0$ and if $f^-1(V)cap E$ is measurable for every open set $V$.
While I understand, that the secon definition is a real extension of the first, I don't understand what he writes next.
"If we define $f(x)=0$ for $xin E^c$, we obtain a measurable funciton on $X$, in the old sense" (Old sense = first definition). He then adds "If our measure happens to be complete, we can define $f$ on $E^c$ in a perfectly arbitrary manner, and we still get a measurable function"
I cannot see why defining $f(x)=0, xin E^c$ leads to the first defintion.
measure-theory
asked Aug 2 at 15:50
EpsilonDelta
4951513
asked Aug 2 at 15:50
EpsilonDelta
4951513
asked Aug 2 at 15:50
EpsilonDelta
4951513
asked Aug 2 at 15:50
EpsilonDelta
4951513
4951513
There is no such thing as an open set in a measurable space, in general.
– Eric Wofsey
Aug 2 at 15:55
add a comment |Â
There is no such thing as an open set in a measurable space, in general.
– Eric Wofsey
Aug 2 at 15:55
There is no such thing as an open set in a measurable space, in general.
– Eric Wofsey
Aug 2 at 15:55
There is no such thing as an open set in a measurable space, in general.
– Eric Wofsey
Aug 2 at 15:55
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
First of all, presumably you mean for $X_2$ to be $mathbbR$ (or $mathbbC$) and for $mathcalA_2$ to be the Borel $sigma$-algebra.
Otherwise it does not make sense to talk about open sets or to define $f(x)=0$.
Now, if you extend $f$ by defining $f(x)=0$ for $xin E^c$, it becomes defined on all of $X_1$. Moreover, for any open set $V$, $f^-1(V)=f^-1(V)cap E$ if $0notin V$ and $f^-1(V)=(f^-1(V)cap E)cup E^c$ if $0in V$. Either way, $f^-1(V)$ is measurable. Thus this extended version of $f$ satisfies the first definition of measurability.
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First of all, presumably you mean for $X_2$ to be $mathbbR$ (or $mathbbC$) and for $mathcalA_2$ to be the Borel $sigma$-algebra.
Otherwise it does not make sense to talk about open sets or to define $f(x)=0$.
Now, if you extend $f$ by defining $f(x)=0$ for $xin E^c$, it becomes defined on all of $X_1$. Moreover, for any open set $V$, $f^-1(V)=f^-1(V)cap E$ if $0notin V$ and $f^-1(V)=(f^-1(V)cap E)cup E^c$ if $0in V$. Either way, $f^-1(V)$ is measurable. Thus this extended version of $f$ satisfies the first definition of measurability.
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
add a comment |Â
up vote
1
down vote
accepted
First of all, presumably you mean for $X_2$ to be $mathbbR$ (or $mathbbC$) and for $mathcalA_2$ to be the Borel $sigma$-algebra.
Otherwise it does not make sense to talk about open sets or to define $f(x)=0$.
Now, if you extend $f$ by defining $f(x)=0$ for $xin E^c$, it becomes defined on all of $X_1$. Moreover, for any open set $V$, $f^-1(V)=f^-1(V)cap E$ if $0notin V$ and $f^-1(V)=(f^-1(V)cap E)cup E^c$ if $0in V$. Either way, $f^-1(V)$ is measurable. Thus this extended version of $f$ satisfies the first definition of measurability.
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First of all, presumably you mean for $X_2$ to be $mathbbR$ (or $mathbbC$) and for $mathcalA_2$ to be the Borel $sigma$-algebra.
Otherwise it does not make sense to talk about open sets or to define $f(x)=0$.
Now, if you extend $f$ by defining $f(x)=0$ for $xin E^c$, it becomes defined on all of $X_1$. Moreover, for any open set $V$, $f^-1(V)=f^-1(V)cap E$ if $0notin V$ and $f^-1(V)=(f^-1(V)cap E)cup E^c$ if $0in V$. Either way, $f^-1(V)$ is measurable. Thus this extended version of $f$ satisfies the first definition of measurability.
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
First of all, presumably you mean for $X_2$ to be $mathbbR$ (or $mathbbC$) and for $mathcalA_2$ to be the Borel $sigma$-algebra.
Otherwise it does not make sense to talk about open sets or to define $f(x)=0$.
Now, if you extend $f$ by defining $f(x)=0$ for $xin E^c$, it becomes defined on all of $X_1$. Moreover, for any open set $V$, $f^-1(V)=f^-1(V)cap E$ if $0notin V$ and $f^-1(V)=(f^-1(V)cap E)cup E^c$ if $0in V$. Either way, $f^-1(V)$ is measurable. Thus this extended version of $f$ satisfies the first definition of measurability.
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
answered Aug 2 at 16:12
Eric Wofsey
161k12188297
161k12188297
add a comment |Â
add a comment |Â
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There is no such thing as an open set in a measurable space, in general.
– Eric Wofsey
Aug 2 at 15:55