Mixing
Question regarding measurable set, Hausdorff-Space and almost everywhere properties of measurable functions...
$begingroup$
I've been given the following task
Let $(X,mathscr{M}_X,mu)$ a measure space. Two measurable mappings $f,g:X to
Y$ into a measurable space $(Y,mathscr{M}_Y)$ are called equal
almost everywhere if $$ N:= {x in X mid f(x) not= g(x)}$$ is a
null set.
Show that the almost everywhere equality is an equivalence relation "$sim$".
So i've been working a bit on that proof and i am supposed to assume $Y$ is a Hausdorff space and $mathscr{M}_Y = mathscr{B}(mathscr{O}_Y) = mathscr{B}(mathscr{A}_Y)$ in other words, the $sigma-$Algebra $mathscr{M}_Y$ on $Y$ is is the Borel-$sigma-$Algebra generated by the open (or equivalently the closed) subsets of $Y$.
So my attempt would have been like this:
- $f sim f$
The set $N_1 := {x in X mid f(x) not= f(x)} = emptyset$ is obviously measurable with $emptyset in mathscr{M}_X$
- $f sim g Rightarrow g sim f$
And here is my main issue. My first attempt was working on this step by simply assuming $$ N:= {x in X mid f(x) not= g(x)}$$ is a null set.
However, as far as i'm concerned, i am supposed to prove that $N$ is measurable, i.e. $N in mathscr{M}_X$
That's where i'm a bit lost. I know that $Y$ is Hausdorff iff $$ Delta Y := {(y,y) mid y in Y} $$ is closed in $Ytimes Y$.
Therefore $(Y times Y) setminus Delta Y$ is an open set. However, i have yet to figure out how to prove that $N in mathscr{M}_X$
Unfortunately i am stuck. Can anyone help me ?
Thank you very much for every help!
general-topology measure-theory almost-everywhere measurable-functions hausdorff-measure
asked Jan 22 at 3:42
ZestZest
26513
$endgroup$
add a comment |
$begingroup$
I've been given the following task
Let $(X,mathscr{M}_X,mu)$ a measure space. Two measurable mappings $f,g:X to
Y$ into a measurable space $(Y,mathscr{M}_Y)$ are called equal
almost everywhere if $$ N:= {x in X mid f(x) not= g(x)}$$ is a
null set.
Show that the almost everywhere equality is an equivalence relation "$sim$".
So i've been working a bit on that proof and i am supposed to assume $Y$ is a Hausdorff space and $mathscr{M}_Y = mathscr{B}(mathscr{O}_Y) = mathscr{B}(mathscr{A}_Y)$ in other words, the $sigma-$Algebra $mathscr{M}_Y$ on $Y$ is is the Borel-$sigma-$Algebra generated by the open (or equivalently the closed) subsets of $Y$.
So my attempt would have been like this:
- $f sim f$
The set $N_1 := {x in X mid f(x) not= f(x)} = emptyset$ is obviously measurable with $emptyset in mathscr{M}_X$
- $f sim g Rightarrow g sim f$
And here is my main issue. My first attempt was working on this step by simply assuming $$ N:= {x in X mid f(x) not= g(x)}$$ is a null set.
However, as far as i'm concerned, i am supposed to prove that $N$ is measurable, i.e. $N in mathscr{M}_X$
That's where i'm a bit lost. I know that $Y$ is Hausdorff iff $$ Delta Y := {(y,y) mid y in Y} $$ is closed in $Ytimes Y$.
Therefore $(Y times Y) setminus Delta Y$ is an open set. However, i have yet to figure out how to prove that $N in mathscr{M}_X$
Unfortunately i am stuck. Can anyone help me ?
Thank you very much for every help!
general-topology measure-theory almost-everywhere measurable-functions hausdorff-measure
asked Jan 22 at 3:42
ZestZest
26513
$endgroup$
1
$begingroup$
Why do you think you need to prove $N$ is measurable?
$endgroup$
– Eric Wofsey
Jan 22 at 5:51
add a comment |
$begingroup$
I've been given the following task
Let $(X,mathscr{M}_X,mu)$ a measure space. Two measurable mappings $f,g:X to
Y$ into a measurable space $(Y,mathscr{M}_Y)$ are called equal
almost everywhere if $$ N:= {x in X mid f(x) not= g(x)}$$ is a
null set.
Show that the almost everywhere equality is an equivalence relation "$sim$".
So i've been working a bit on that proof and i am supposed to assume $Y$ is a Hausdorff space and $mathscr{M}_Y = mathscr{B}(mathscr{O}_Y) = mathscr{B}(mathscr{A}_Y)$ in other words, the $sigma-$Algebra $mathscr{M}_Y$ on $Y$ is is the Borel-$sigma-$Algebra generated by the open (or equivalently the closed) subsets of $Y$.
So my attempt would have been like this:
- $f sim f$
The set $N_1 := {x in X mid f(x) not= f(x)} = emptyset$ is obviously measurable with $emptyset in mathscr{M}_X$
- $f sim g Rightarrow g sim f$
And here is my main issue. My first attempt was working on this step by simply assuming $$ N:= {x in X mid f(x) not= g(x)}$$ is a null set.
However, as far as i'm concerned, i am supposed to prove that $N$ is measurable, i.e. $N in mathscr{M}_X$
That's where i'm a bit lost. I know that $Y$ is Hausdorff iff $$ Delta Y := {(y,y) mid y in Y} $$ is closed in $Ytimes Y$.
Therefore $(Y times Y) setminus Delta Y$ is an open set. However, i have yet to figure out how to prove that $N in mathscr{M}_X$
Unfortunately i am stuck. Can anyone help me ?
Thank you very much for every help!
general-topology measure-theory almost-everywhere measurable-functions hausdorff-measure
asked Jan 22 at 3:42
ZestZest
26513
$endgroup$
I've been given the following task
Let $(X,mathscr{M}_X,mu)$ a measure space. Two measurable mappings $f,g:X to
Y$ into a measurable space $(Y,mathscr{M}_Y)$ are called equal
almost everywhere if $$ N:= {x in X mid f(x) not= g(x)}$$ is a
null set.
Show that the almost everywhere equality is an equivalence relation "$sim$".
So i've been working a bit on that proof and i am supposed to assume $Y$ is a Hausdorff space and $mathscr{M}_Y = mathscr{B}(mathscr{O}_Y) = mathscr{B}(mathscr{A}_Y)$ in other words, the $sigma-$Algebra $mathscr{M}_Y$ on $Y$ is is the Borel-$sigma-$Algebra generated by the open (or equivalently the closed) subsets of $Y$.
So my attempt would have been like this:
- $f sim f$
The set $N_1 := {x in X mid f(x) not= f(x)} = emptyset$ is obviously measurable with $emptyset in mathscr{M}_X$
- $f sim g Rightarrow g sim f$
And here is my main issue. My first attempt was working on this step by simply assuming $$ N:= {x in X mid f(x) not= g(x)}$$ is a null set.
However, as far as i'm concerned, i am supposed to prove that $N$ is measurable, i.e. $N in mathscr{M}_X$
That's where i'm a bit lost. I know that $Y$ is Hausdorff iff $$ Delta Y := {(y,y) mid y in Y} $$ is closed in $Ytimes Y$.
Therefore $(Y times Y) setminus Delta Y$ is an open set. However, i have yet to figure out how to prove that $N in mathscr{M}_X$
Unfortunately i am stuck. Can anyone help me ?
Thank you very much for every help!
general-topology measure-theory almost-everywhere measurable-functions hausdorff-measure
general-topology measure-theory almost-everywhere measurable-functions hausdorff-measure
asked Jan 22 at 3:42
ZestZest
26513
asked Jan 22 at 3:42
ZestZest
26513
asked Jan 22 at 3:42
ZestZest
26513
asked Jan 22 at 3:42
ZestZest
26513
asked Jan 22 at 3:42
ZestZest
26513
26513
1
$begingroup$
Why do you think you need to prove $N$ is measurable?
$endgroup$
– Eric Wofsey
Jan 22 at 5:51
add a comment |
1
$begingroup$
Why do you think you need to prove $N$ is measurable?
$endgroup$
– Eric Wofsey
Jan 22 at 5:51
1
1
$begingroup$
Why do you think you need to prove $N$ is measurable?
$endgroup$
– Eric Wofsey
Jan 22 at 5:51
$begingroup$
Why do you think you need to prove $N$ is measurable?
$endgroup$
– Eric Wofsey
Jan 22 at 5:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As $a neq b$ iff $b neq a$ for any numbers, $N= {x: f(x) neq g(x)} = {x: g(x) neq f(x)}$. It's just the same set, so if the former is a null set, so is the latter trivially. Nothing to prove: if $f sim g$ by definition this is a null set.
For transitivity, assume $f sim g$ and $g sim h$.
Note that ${x: f(x) = g(x)} cap {x: g(x)=h(x)} subseteq {x: f(x)=h(x)}$ so that (taking complements):
$${x: f(x)neq h(x)} subseteq {x: f(x)neq g(x)} cup {x: g(x) neq h(x)}$$
so that ${x: f(x)neq h(x)}$ is a subset of the union of two null sets, hence a null set.
Hausdorffness or the precise $sigma$-algebra's are irrelevant.
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
$begingroup$
thank you very much. assuming i'd want to use the hausdorffness in order to prove that ${x in X mid f(x) not= g(x)}$ is measurable. how exactly would that work? i am pretty sure there is a connection but i don't get it.
$endgroup$
– Zest
Jan 22 at 12:03
1
$begingroup$
@Zest to show the set is always measurable when $f$ and $g$ are and $Y$ is Hausdorff, you can use that the diagonal is closed. But it’s irrelevant for this proof because $f sim g$ means that the set is measurable and has measure 0, so we can just use that in the proof.
$endgroup$
– Henno Brandsma
Jan 23 at 5:59
add a comment |
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1 Answer
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$begingroup$
As $a neq b$ iff $b neq a$ for any numbers, $N= {x: f(x) neq g(x)} = {x: g(x) neq f(x)}$. It's just the same set, so if the former is a null set, so is the latter trivially. Nothing to prove: if $f sim g$ by definition this is a null set.
For transitivity, assume $f sim g$ and $g sim h$.
Note that ${x: f(x) = g(x)} cap {x: g(x)=h(x)} subseteq {x: f(x)=h(x)}$ so that (taking complements):
$${x: f(x)neq h(x)} subseteq {x: f(x)neq g(x)} cup {x: g(x) neq h(x)}$$
so that ${x: f(x)neq h(x)}$ is a subset of the union of two null sets, hence a null set.
Hausdorffness or the precise $sigma$-algebra's are irrelevant.
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
$begingroup$
thank you very much. assuming i'd want to use the hausdorffness in order to prove that ${x in X mid f(x) not= g(x)}$ is measurable. how exactly would that work? i am pretty sure there is a connection but i don't get it.
$endgroup$
– Zest
Jan 22 at 12:03
1
$begingroup$
@Zest to show the set is always measurable when $f$ and $g$ are and $Y$ is Hausdorff, you can use that the diagonal is closed. But it’s irrelevant for this proof because $f sim g$ means that the set is measurable and has measure 0, so we can just use that in the proof.
$endgroup$
– Henno Brandsma
Jan 23 at 5:59
add a comment |
$begingroup$
As $a neq b$ iff $b neq a$ for any numbers, $N= {x: f(x) neq g(x)} = {x: g(x) neq f(x)}$. It's just the same set, so if the former is a null set, so is the latter trivially. Nothing to prove: if $f sim g$ by definition this is a null set.
For transitivity, assume $f sim g$ and $g sim h$.
Note that ${x: f(x) = g(x)} cap {x: g(x)=h(x)} subseteq {x: f(x)=h(x)}$ so that (taking complements):
$${x: f(x)neq h(x)} subseteq {x: f(x)neq g(x)} cup {x: g(x) neq h(x)}$$
so that ${x: f(x)neq h(x)}$ is a subset of the union of two null sets, hence a null set.
Hausdorffness or the precise $sigma$-algebra's are irrelevant.
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
$begingroup$
thank you very much. assuming i'd want to use the hausdorffness in order to prove that ${x in X mid f(x) not= g(x)}$ is measurable. how exactly would that work? i am pretty sure there is a connection but i don't get it.
$endgroup$
– Zest
Jan 22 at 12:03
1
$begingroup$
@Zest to show the set is always measurable when $f$ and $g$ are and $Y$ is Hausdorff, you can use that the diagonal is closed. But it’s irrelevant for this proof because $f sim g$ means that the set is measurable and has measure 0, so we can just use that in the proof.
$endgroup$
– Henno Brandsma
Jan 23 at 5:59
add a comment |
$begingroup$
As $a neq b$ iff $b neq a$ for any numbers, $N= {x: f(x) neq g(x)} = {x: g(x) neq f(x)}$. It's just the same set, so if the former is a null set, so is the latter trivially. Nothing to prove: if $f sim g$ by definition this is a null set.
For transitivity, assume $f sim g$ and $g sim h$.
Note that ${x: f(x) = g(x)} cap {x: g(x)=h(x)} subseteq {x: f(x)=h(x)}$ so that (taking complements):
$${x: f(x)neq h(x)} subseteq {x: f(x)neq g(x)} cup {x: g(x) neq h(x)}$$
so that ${x: f(x)neq h(x)}$ is a subset of the union of two null sets, hence a null set.
Hausdorffness or the precise $sigma$-algebra's are irrelevant.
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
As $a neq b$ iff $b neq a$ for any numbers, $N= {x: f(x) neq g(x)} = {x: g(x) neq f(x)}$. It's just the same set, so if the former is a null set, so is the latter trivially. Nothing to prove: if $f sim g$ by definition this is a null set.
For transitivity, assume $f sim g$ and $g sim h$.
Note that ${x: f(x) = g(x)} cap {x: g(x)=h(x)} subseteq {x: f(x)=h(x)}$ so that (taking complements):
$${x: f(x)neq h(x)} subseteq {x: f(x)neq g(x)} cup {x: g(x) neq h(x)}$$
so that ${x: f(x)neq h(x)}$ is a subset of the union of two null sets, hence a null set.
Hausdorffness or the precise $sigma$-algebra's are irrelevant.
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
answered Jan 22 at 6:34
Henno BrandsmaHenno Brandsma
112k348120
112k348120
$begingroup$
thank you very much. assuming i'd want to use the hausdorffness in order to prove that ${x in X mid f(x) not= g(x)}$ is measurable. how exactly would that work? i am pretty sure there is a connection but i don't get it.
$endgroup$
– Zest
Jan 22 at 12:03
1
$begingroup$
@Zest to show the set is always measurable when $f$ and $g$ are and $Y$ is Hausdorff, you can use that the diagonal is closed. But it’s irrelevant for this proof because $f sim g$ means that the set is measurable and has measure 0, so we can just use that in the proof.
$endgroup$
– Henno Brandsma
Jan 23 at 5:59
add a comment |
$begingroup$
thank you very much. assuming i'd want to use the hausdorffness in order to prove that ${x in X mid f(x) not= g(x)}$ is measurable. how exactly would that work? i am pretty sure there is a connection but i don't get it.
$endgroup$
– Zest
Jan 22 at 12:03
1
$begingroup$
@Zest to show the set is always measurable when $f$ and $g$ are and $Y$ is Hausdorff, you can use that the diagonal is closed. But it’s irrelevant for this proof because $f sim g$ means that the set is measurable and has measure 0, so we can just use that in the proof.
$endgroup$
– Henno Brandsma
Jan 23 at 5:59
$begingroup$
thank you very much. assuming i'd want to use the hausdorffness in order to prove that ${x in X mid f(x) not= g(x)}$ is measurable. how exactly would that work? i am pretty sure there is a connection but i don't get it.
$endgroup$
– Zest
Jan 22 at 12:03
$begingroup$
thank you very much. assuming i'd want to use the hausdorffness in order to prove that ${x in X mid f(x) not= g(x)}$ is measurable. how exactly would that work? i am pretty sure there is a connection but i don't get it.
$endgroup$
– Zest
Jan 22 at 12:03
1
1
$begingroup$
@Zest to show the set is always measurable when $f$ and $g$ are and $Y$ is Hausdorff, you can use that the diagonal is closed. But it’s irrelevant for this proof because $f sim g$ means that the set is measurable and has measure 0, so we can just use that in the proof.
$endgroup$
– Henno Brandsma
Jan 23 at 5:59
$begingroup$
@Zest to show the set is always measurable when $f$ and $g$ are and $Y$ is Hausdorff, you can use that the diagonal is closed. But it’s irrelevant for this proof because $f sim g$ means that the set is measurable and has measure 0, so we can just use that in the proof.
$endgroup$
– Henno Brandsma
Jan 23 at 5:59
add a comment |
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$begingroup$
Why do you think you need to prove $N$ is measurable?
$endgroup$
– Eric Wofsey
Jan 22 at 5:51