Mixing
Why is the set ${x mid f(x) not= g(x)}$ measurable in a topological hausdorff space?
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assuming i have a measure space $(X, mathscr{M}_X, mu)$ and a measurable space $(Y,mathscr{M}_Y)$, two measurable mappings $f,g:X to Y$ and it holds that $Y$ is a topological hausdorff space with a countable base. (I think i am supposed to assume that $mathscr{M}_Y$ is the Borel-$sigma$-Algebra generated by the open/closed subsets of Y)
Now i am wonderung: Why exactly is the set ${x in X mid f(x) not= g(x)}$ measurable? in other words why is $${x in X mid f(x) not= g(x)} in mathscr{M}_X$$
I know that the diagonal $Delta Y = {(y,y)mid y in Y}subset Y times Y$ is closed in any hausdorff space.
I also know that since $Delta Y$ is closed in $Y times Y$, its complement $(Y times Y)setminus Delta Y$ is open.
However, ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is only a subset of $(Y times Y)setminus Delta Y$ for which i can't say whether it is open or closed.
The only idea that's left was to try this:
Pick an arbitrary point $p = (f(x),g(x))$ for which $f(x)not= g(x)$ in $Ytimes Y$. Since $Y$ is a Hausdorff space, we can find disjoint open neighbourhoods $U$ around $f(x)$ and $V$ around $g(x)$ such that $U times V$ is an open neighbourhood of $(f(x),g(x))$ with $$U times V cap Delta Y = emptyset$$
Since ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is a union of all such open neighbourhoods $Utimes V$ of points $(f(x),g(x))$ with $f(x) not= g(x)$ it is open in $Ytimes Y$.
(is this correct?)
If the last paragraph would be correct, i'd be finished.
Thank you very much for any of your help.
EDIT: New summary of what i gathered so far with the help of the recent comments and answers:
As pointed out in the comments, i did the mistake to purely focus on $$B:= {(f(x),g(x)mid f(x) not= g(x), x in X}$$ and tried to conclude that its preimage is measurable iff $B$ is open which was a wrong assumption in the first place.
Therefore, i do not need $B$ itself to be open. It's sufficient to show that the union of all open disjoint neighbourhoods $Utimes V$ of arbitrary points $f(x),g(x)$ for which $f(x) not= g(x)$ is an open set; because what matters is its preimage and not the set (image) itself. Let $$ B' := bigcup{U times V mid U, V in mathcal{B} ; U cap V = emptyset}$$
$B'$ is obviously an open set, therefore its preimage is mesaurable.
For the preimage of open disjoint neighbourhoods $U,V$ of $f(x)$ and $g(x)$ respectively, it holds that
$$(ftimes g)^{-1}(Utimes V) = {x mid f(x)in U wedge g(x) in V} = f^{-1}(U)cap g^{-1}(V)$$ (this is something i actually didnt know, it didn't know the definition of cartesian products of mappings, that's also the reason i didnt understand alex's question right away)
Since $Utimes V$ is open, $(ftimes g)^{-1}(U times V)$ is measurable. Since $mathcal{B}$ is a countable base the set
$${x in Xmid f(x) not= g(x) } = bigcup { (ftimes g)^{-1}(Utimes V) mid U, V in mathcal{B}, U cap V = emptyset} = bigcup{ f^{-1}(U)cap g^{-1}(V)mid U, V in mathcal{B}, U cap V = emptyset}$$ is a countable union of measurable sets and therefore measurable itself.
general-topology measure-theory measurable-functions measurable-sets
asked Jan 23 at 3:15
ZestZest
27213
$endgroup$
add a comment |
$begingroup$
assuming i have a measure space $(X, mathscr{M}_X, mu)$ and a measurable space $(Y,mathscr{M}_Y)$, two measurable mappings $f,g:X to Y$ and it holds that $Y$ is a topological hausdorff space with a countable base. (I think i am supposed to assume that $mathscr{M}_Y$ is the Borel-$sigma$-Algebra generated by the open/closed subsets of Y)
Now i am wonderung: Why exactly is the set ${x in X mid f(x) not= g(x)}$ measurable? in other words why is $${x in X mid f(x) not= g(x)} in mathscr{M}_X$$
I know that the diagonal $Delta Y = {(y,y)mid y in Y}subset Y times Y$ is closed in any hausdorff space.
I also know that since $Delta Y$ is closed in $Y times Y$, its complement $(Y times Y)setminus Delta Y$ is open.
However, ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is only a subset of $(Y times Y)setminus Delta Y$ for which i can't say whether it is open or closed.
The only idea that's left was to try this:
Pick an arbitrary point $p = (f(x),g(x))$ for which $f(x)not= g(x)$ in $Ytimes Y$. Since $Y$ is a Hausdorff space, we can find disjoint open neighbourhoods $U$ around $f(x)$ and $V$ around $g(x)$ such that $U times V$ is an open neighbourhood of $(f(x),g(x))$ with $$U times V cap Delta Y = emptyset$$
Since ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is a union of all such open neighbourhoods $Utimes V$ of points $(f(x),g(x))$ with $f(x) not= g(x)$ it is open in $Ytimes Y$.
(is this correct?)
If the last paragraph would be correct, i'd be finished.
Thank you very much for any of your help.
EDIT: New summary of what i gathered so far with the help of the recent comments and answers:
As pointed out in the comments, i did the mistake to purely focus on $$B:= {(f(x),g(x)mid f(x) not= g(x), x in X}$$ and tried to conclude that its preimage is measurable iff $B$ is open which was a wrong assumption in the first place.
Therefore, i do not need $B$ itself to be open. It's sufficient to show that the union of all open disjoint neighbourhoods $Utimes V$ of arbitrary points $f(x),g(x)$ for which $f(x) not= g(x)$ is an open set; because what matters is its preimage and not the set (image) itself. Let $$ B' := bigcup{U times V mid U, V in mathcal{B} ; U cap V = emptyset}$$
$B'$ is obviously an open set, therefore its preimage is mesaurable.
For the preimage of open disjoint neighbourhoods $U,V$ of $f(x)$ and $g(x)$ respectively, it holds that
$$(ftimes g)^{-1}(Utimes V) = {x mid f(x)in U wedge g(x) in V} = f^{-1}(U)cap g^{-1}(V)$$ (this is something i actually didnt know, it didn't know the definition of cartesian products of mappings, that's also the reason i didnt understand alex's question right away)
Since $Utimes V$ is open, $(ftimes g)^{-1}(U times V)$ is measurable. Since $mathcal{B}$ is a countable base the set
$${x in Xmid f(x) not= g(x) } = bigcup { (ftimes g)^{-1}(Utimes V) mid U, V in mathcal{B}, U cap V = emptyset} = bigcup{ f^{-1}(U)cap g^{-1}(V)mid U, V in mathcal{B}, U cap V = emptyset}$$ is a countable union of measurable sets and therefore measurable itself.
general-topology measure-theory measurable-functions measurable-sets
asked Jan 23 at 3:15
ZestZest
27213
$endgroup$
$begingroup$
There is obviously a typo in the expression ${(f(x),g(x)mid x in X}$ which occurs at the start of the 5th and 8th paragraphs. For one thing the opening parenthesis remains unclosed. If ${(f(x),g(x))mid x in X}$ is what was meant, then I don't see why it would necessarily be a subset of $(Y times Y)setminus Delta Y$, since there's nothing in the definition of the set to eliminate those $ x $ for which $ f(x) = g(x) $. Was it supposed to have been ${(f(x),g(x))mid x in X, f(x) ne g(x) } $ ?
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– lonza leggiera
Jan 26 at 11:00
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hello @lonzaleggiera, you're right, that's a typo. thanks for pointing out. i will edit my question immediately.
$endgroup$
– Zest
Jan 26 at 17:28
add a comment |
$begingroup$
assuming i have a measure space $(X, mathscr{M}_X, mu)$ and a measurable space $(Y,mathscr{M}_Y)$, two measurable mappings $f,g:X to Y$ and it holds that $Y$ is a topological hausdorff space with a countable base. (I think i am supposed to assume that $mathscr{M}_Y$ is the Borel-$sigma$-Algebra generated by the open/closed subsets of Y)
Now i am wonderung: Why exactly is the set ${x in X mid f(x) not= g(x)}$ measurable? in other words why is $${x in X mid f(x) not= g(x)} in mathscr{M}_X$$
I know that the diagonal $Delta Y = {(y,y)mid y in Y}subset Y times Y$ is closed in any hausdorff space.
I also know that since $Delta Y$ is closed in $Y times Y$, its complement $(Y times Y)setminus Delta Y$ is open.
However, ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is only a subset of $(Y times Y)setminus Delta Y$ for which i can't say whether it is open or closed.
The only idea that's left was to try this:
Pick an arbitrary point $p = (f(x),g(x))$ for which $f(x)not= g(x)$ in $Ytimes Y$. Since $Y$ is a Hausdorff space, we can find disjoint open neighbourhoods $U$ around $f(x)$ and $V$ around $g(x)$ such that $U times V$ is an open neighbourhood of $(f(x),g(x))$ with $$U times V cap Delta Y = emptyset$$
Since ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is a union of all such open neighbourhoods $Utimes V$ of points $(f(x),g(x))$ with $f(x) not= g(x)$ it is open in $Ytimes Y$.
(is this correct?)
If the last paragraph would be correct, i'd be finished.
Thank you very much for any of your help.
EDIT: New summary of what i gathered so far with the help of the recent comments and answers:
As pointed out in the comments, i did the mistake to purely focus on $$B:= {(f(x),g(x)mid f(x) not= g(x), x in X}$$ and tried to conclude that its preimage is measurable iff $B$ is open which was a wrong assumption in the first place.
Therefore, i do not need $B$ itself to be open. It's sufficient to show that the union of all open disjoint neighbourhoods $Utimes V$ of arbitrary points $f(x),g(x)$ for which $f(x) not= g(x)$ is an open set; because what matters is its preimage and not the set (image) itself. Let $$ B' := bigcup{U times V mid U, V in mathcal{B} ; U cap V = emptyset}$$
$B'$ is obviously an open set, therefore its preimage is mesaurable.
For the preimage of open disjoint neighbourhoods $U,V$ of $f(x)$ and $g(x)$ respectively, it holds that
$$(ftimes g)^{-1}(Utimes V) = {x mid f(x)in U wedge g(x) in V} = f^{-1}(U)cap g^{-1}(V)$$ (this is something i actually didnt know, it didn't know the definition of cartesian products of mappings, that's also the reason i didnt understand alex's question right away)
Since $Utimes V$ is open, $(ftimes g)^{-1}(U times V)$ is measurable. Since $mathcal{B}$ is a countable base the set
$${x in Xmid f(x) not= g(x) } = bigcup { (ftimes g)^{-1}(Utimes V) mid U, V in mathcal{B}, U cap V = emptyset} = bigcup{ f^{-1}(U)cap g^{-1}(V)mid U, V in mathcal{B}, U cap V = emptyset}$$ is a countable union of measurable sets and therefore measurable itself.
general-topology measure-theory measurable-functions measurable-sets
asked Jan 23 at 3:15
ZestZest
27213
$endgroup$
assuming i have a measure space $(X, mathscr{M}_X, mu)$ and a measurable space $(Y,mathscr{M}_Y)$, two measurable mappings $f,g:X to Y$ and it holds that $Y$ is a topological hausdorff space with a countable base. (I think i am supposed to assume that $mathscr{M}_Y$ is the Borel-$sigma$-Algebra generated by the open/closed subsets of Y)
Now i am wonderung: Why exactly is the set ${x in X mid f(x) not= g(x)}$ measurable? in other words why is $${x in X mid f(x) not= g(x)} in mathscr{M}_X$$
I know that the diagonal $Delta Y = {(y,y)mid y in Y}subset Y times Y$ is closed in any hausdorff space.
I also know that since $Delta Y$ is closed in $Y times Y$, its complement $(Y times Y)setminus Delta Y$ is open.
However, ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is only a subset of $(Y times Y)setminus Delta Y$ for which i can't say whether it is open or closed.
The only idea that's left was to try this:
Pick an arbitrary point $p = (f(x),g(x))$ for which $f(x)not= g(x)$ in $Ytimes Y$. Since $Y$ is a Hausdorff space, we can find disjoint open neighbourhoods $U$ around $f(x)$ and $V$ around $g(x)$ such that $U times V$ is an open neighbourhood of $(f(x),g(x))$ with $$U times V cap Delta Y = emptyset$$
Since ${(f(x),g(x)mid x in X, f(x) not= g(x) }$ is a union of all such open neighbourhoods $Utimes V$ of points $(f(x),g(x))$ with $f(x) not= g(x)$ it is open in $Ytimes Y$.
(is this correct?)
If the last paragraph would be correct, i'd be finished.
Thank you very much for any of your help.
EDIT: New summary of what i gathered so far with the help of the recent comments and answers:
As pointed out in the comments, i did the mistake to purely focus on $$B:= {(f(x),g(x)mid f(x) not= g(x), x in X}$$ and tried to conclude that its preimage is measurable iff $B$ is open which was a wrong assumption in the first place.
Therefore, i do not need $B$ itself to be open. It's sufficient to show that the union of all open disjoint neighbourhoods $Utimes V$ of arbitrary points $f(x),g(x)$ for which $f(x) not= g(x)$ is an open set; because what matters is its preimage and not the set (image) itself. Let $$ B' := bigcup{U times V mid U, V in mathcal{B} ; U cap V = emptyset}$$
$B'$ is obviously an open set, therefore its preimage is mesaurable.
For the preimage of open disjoint neighbourhoods $U,V$ of $f(x)$ and $g(x)$ respectively, it holds that
$$(ftimes g)^{-1}(Utimes V) = {x mid f(x)in U wedge g(x) in V} = f^{-1}(U)cap g^{-1}(V)$$ (this is something i actually didnt know, it didn't know the definition of cartesian products of mappings, that's also the reason i didnt understand alex's question right away)
Since $Utimes V$ is open, $(ftimes g)^{-1}(U times V)$ is measurable. Since $mathcal{B}$ is a countable base the set
$${x in Xmid f(x) not= g(x) } = bigcup { (ftimes g)^{-1}(Utimes V) mid U, V in mathcal{B}, U cap V = emptyset} = bigcup{ f^{-1}(U)cap g^{-1}(V)mid U, V in mathcal{B}, U cap V = emptyset}$$ is a countable union of measurable sets and therefore measurable itself.
general-topology measure-theory measurable-functions measurable-sets
general-topology measure-theory measurable-functions measurable-sets
asked Jan 23 at 3:15
ZestZest
27213
asked Jan 23 at 3:15
ZestZest
27213
edited Jan 28 at 2:09
Zest
asked Jan 23 at 3:15
ZestZest
27213
asked Jan 23 at 3:15
ZestZest
27213
asked Jan 23 at 3:15
ZestZest
27213
27213
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There is obviously a typo in the expression ${(f(x),g(x)mid x in X}$ which occurs at the start of the 5th and 8th paragraphs. For one thing the opening parenthesis remains unclosed. If ${(f(x),g(x))mid x in X}$ is what was meant, then I don't see why it would necessarily be a subset of $(Y times Y)setminus Delta Y$, since there's nothing in the definition of the set to eliminate those $ x $ for which $ f(x) = g(x) $. Was it supposed to have been ${(f(x),g(x))mid x in X, f(x) ne g(x) } $ ?
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– lonza leggiera
Jan 26 at 11:00
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hello @lonzaleggiera, you're right, that's a typo. thanks for pointing out. i will edit my question immediately.
$endgroup$
– Zest
Jan 26 at 17:28
add a comment |
$begingroup$
There is obviously a typo in the expression ${(f(x),g(x)mid x in X}$ which occurs at the start of the 5th and 8th paragraphs. For one thing the opening parenthesis remains unclosed. If ${(f(x),g(x))mid x in X}$ is what was meant, then I don't see why it would necessarily be a subset of $(Y times Y)setminus Delta Y$, since there's nothing in the definition of the set to eliminate those $ x $ for which $ f(x) = g(x) $. Was it supposed to have been ${(f(x),g(x))mid x in X, f(x) ne g(x) } $ ?
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– lonza leggiera
Jan 26 at 11:00
$begingroup$
hello @lonzaleggiera, you're right, that's a typo. thanks for pointing out. i will edit my question immediately.
$endgroup$
– Zest
Jan 26 at 17:28
$begingroup$
There is obviously a typo in the expression ${(f(x),g(x)mid x in X}$ which occurs at the start of the 5th and 8th paragraphs. For one thing the opening parenthesis remains unclosed. If ${(f(x),g(x))mid x in X}$ is what was meant, then I don't see why it would necessarily be a subset of $(Y times Y)setminus Delta Y$, since there's nothing in the definition of the set to eliminate those $ x $ for which $ f(x) = g(x) $. Was it supposed to have been ${(f(x),g(x))mid x in X, f(x) ne g(x) } $ ?
$endgroup$
– lonza leggiera
Jan 26 at 11:00
$begingroup$
There is obviously a typo in the expression ${(f(x),g(x)mid x in X}$ which occurs at the start of the 5th and 8th paragraphs. For one thing the opening parenthesis remains unclosed. If ${(f(x),g(x))mid x in X}$ is what was meant, then I don't see why it would necessarily be a subset of $(Y times Y)setminus Delta Y$, since there's nothing in the definition of the set to eliminate those $ x $ for which $ f(x) = g(x) $. Was it supposed to have been ${(f(x),g(x))mid x in X, f(x) ne g(x) } $ ?
$endgroup$
– lonza leggiera
Jan 26 at 11:00
$begingroup$
hello @lonzaleggiera, you're right, that's a typo. thanks for pointing out. i will edit my question immediately.
$endgroup$
– Zest
Jan 26 at 17:28
$begingroup$
hello @lonzaleggiera, you're right, that's a typo. thanks for pointing out. i will edit my question immediately.
$endgroup$
– Zest
Jan 26 at 17:28
add a comment |
3 Answers
3
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oldest
votes
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In answer to the question asked, about whether the argument in the last paragraph is correct, I don't believe that it is at it stands.
The problem is with the assertion $" {(
,f(x),g(x) )mid x in X, f(x) not= g(x) } $ is a union of all such open neighbourhoods $ dots $ ." If you take the union of all the open neighbourhoods you describe you will certainly get an open set containing $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $, but I don't see why it should be equal to it, and I believe I can give a counterexample to the general statement that $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ must be open under the hypotheses given
Nevertheless, I think Alex Ravsky's proof shows that the basic idea underlying the argument was sound.
Addendum: It may be helpful for me to add one of the counterexamples I was referring to in my above comments.
Take $ X = Y = mathbb R $—the set of real numbers—, and $ mathscr{M}_X ,mathscr{M}_Y $ both to be the Borel sets of $ mathbb R $. Let $ fleft(xright)= 0 mbox{ and } gleft(xright)= x mbox{ for all } xinmathbb R $. Then $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ is the punctured line segment $ left{ left(,0,x,right) mid xne0 right} $ of $ mathbb R^2 $, which is not an open set.
Comment on proposed new proof:
Apart from some probable typos, itemised below, the new revised proof by Zest seems to me to be correct.
The following appear likely to me to be typos:
$$B' = bigcup left{ Utimes Vmid U,Vin B, Ucap V=emptysetright} .$$
This should be
$$B' = bigcup left{ Utimes Vmid U,Vin{mathcal B}, Ucap V=emptysetright}$$
- I'm not sure what the observation "Since $ B $ is a countable base $ mathcal B dots $" is trying to say. As far as I can see, all that's necessary here is the observation that $ mathcal B $ is a countable base.
- A union sign is missing from the last term on the left of the final series of equations. That is, $ left{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $ should be $ bigcupleft{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $
The nice observation Alex Ravsky and Zest have made here, and which I had missed, is that $ left{,xin X mid fleft(xright)neq gleft(xright),right} $ is actually equal to $ left(ftimes gright)^{-1}left(B'right) $ rather than being a strict subset of it. Since this isn't immediately obvious—at least it wasn't to me, although the proof is fairly straightforward once the fact is realised—I'd like to see a brief proof of that fact included in any formal writeup.
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
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thanks for your help. well, the union of open sets is open, thus if in fact the set ${(f(x),g(x)) mid x in X, f(x) not= g(x)}$ is a union of disjoint open neighbourhood, it is open.
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– Zest
Jan 26 at 23:53
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But how did you get that the set ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) }$ is a union of disjoint open neighbourhoods? If $ U $ and $ V $ are disjoint open neighbourhoods around $ f(x) $ and $ g(x) $ respectively, $ U $ may well contain an element $ u $ that isn't in the image of $ f $, and $ V $ an element $ v $ that isn't in the image of $ g $. The point $ (u, v) $ will be in any union that includes $ Utimes V $ but *not* in ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) } $, because there is no $ x in X $ with $ (f(x),g(x)) = (u,v)$ .
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– lonza leggiera
Jan 27 at 0:18
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you're right. i didnt consider that. i will think about it again! thanks for the clarification.
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– Zest
Jan 27 at 0:25
1
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i have another question. is it possible that we don't even need $ B := { f(x),g(x) mid x in X, f(x)not= g(x)}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $Bsubset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $fnot=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
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– Zest
Jan 27 at 9:36
1
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Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done
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– lonza leggiera
Jan 27 at 13:33
|
show 1 more comment
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$h(x)=(f(x),g(x))$ is measurable and ${x:f(x)neq g(x)}=h^{-1}(Ytimes Y-Delta)$ is measurable since $Delta$ is closed, $(Ytimes Y)-Delta$ is open.
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
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$h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
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– Zest
Jan 23 at 3:39
add a comment |
$begingroup$
I think all presented proofs are incomplete (for instance, they don’t use the fact that $Y$ has a countable base $mathcal B$). I also assume that $mathcal M_X$ is closed with respect to intersections and countable unions.
Let $A={xin X|f(x)ne g(x)}$. Let $xin A$ be any point. Since $f(x)ne g(x)$ and $Y$ is Hausdorff, there exists disjoint neighborhoods $U,Vinmathcal B$ of $f(x)$ and $g(x)$, respectively. Thus
$$A=bigcup {f^{-1}(U)cap g^{-1}(V): U,Vinmathcal B, Ucap V=varnothing}$$
is measurable.
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
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hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
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– Zest
Jan 26 at 23:59
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@Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $mathcal M_X$ is closed with respect to (finite) intersections and countable unions.
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– Alex Ravsky
Jan 27 at 0:11
1
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i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
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– Zest
Jan 27 at 0:23
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hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
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– Zest
Jan 27 at 10:42
add a comment |
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$begingroup$
In answer to the question asked, about whether the argument in the last paragraph is correct, I don't believe that it is at it stands.
The problem is with the assertion $" {(
,f(x),g(x) )mid x in X, f(x) not= g(x) } $ is a union of all such open neighbourhoods $ dots $ ." If you take the union of all the open neighbourhoods you describe you will certainly get an open set containing $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $, but I don't see why it should be equal to it, and I believe I can give a counterexample to the general statement that $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ must be open under the hypotheses given
Nevertheless, I think Alex Ravsky's proof shows that the basic idea underlying the argument was sound.
Addendum: It may be helpful for me to add one of the counterexamples I was referring to in my above comments.
Take $ X = Y = mathbb R $—the set of real numbers—, and $ mathscr{M}_X ,mathscr{M}_Y $ both to be the Borel sets of $ mathbb R $. Let $ fleft(xright)= 0 mbox{ and } gleft(xright)= x mbox{ for all } xinmathbb R $. Then $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ is the punctured line segment $ left{ left(,0,x,right) mid xne0 right} $ of $ mathbb R^2 $, which is not an open set.
Comment on proposed new proof:
Apart from some probable typos, itemised below, the new revised proof by Zest seems to me to be correct.
The following appear likely to me to be typos:
$$B' = bigcup left{ Utimes Vmid U,Vin B, Ucap V=emptysetright} .$$
This should be
$$B' = bigcup left{ Utimes Vmid U,Vin{mathcal B}, Ucap V=emptysetright}$$
- I'm not sure what the observation "Since $ B $ is a countable base $ mathcal B dots $" is trying to say. As far as I can see, all that's necessary here is the observation that $ mathcal B $ is a countable base.
- A union sign is missing from the last term on the left of the final series of equations. That is, $ left{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $ should be $ bigcupleft{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $
The nice observation Alex Ravsky and Zest have made here, and which I had missed, is that $ left{,xin X mid fleft(xright)neq gleft(xright),right} $ is actually equal to $ left(ftimes gright)^{-1}left(B'right) $ rather than being a strict subset of it. Since this isn't immediately obvious—at least it wasn't to me, although the proof is fairly straightforward once the fact is realised—I'd like to see a brief proof of that fact included in any formal writeup.
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
$endgroup$
$begingroup$
thanks for your help. well, the union of open sets is open, thus if in fact the set ${(f(x),g(x)) mid x in X, f(x) not= g(x)}$ is a union of disjoint open neighbourhood, it is open.
$endgroup$
– Zest
Jan 26 at 23:53
$begingroup$
But how did you get that the set ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) }$ is a union of disjoint open neighbourhoods? If $ U $ and $ V $ are disjoint open neighbourhoods around $ f(x) $ and $ g(x) $ respectively, $ U $ may well contain an element $ u $ that isn't in the image of $ f $, and $ V $ an element $ v $ that isn't in the image of $ g $. The point $ (u, v) $ will be in any union that includes $ Utimes V $ but *not* in ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) } $, because there is no $ x in X $ with $ (f(x),g(x)) = (u,v)$ .
$endgroup$
– lonza leggiera
Jan 27 at 0:18
$begingroup$
you're right. i didnt consider that. i will think about it again! thanks for the clarification.
$endgroup$
– Zest
Jan 27 at 0:25
1
$begingroup$
i have another question. is it possible that we don't even need $ B := { f(x),g(x) mid x in X, f(x)not= g(x)}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $Bsubset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $fnot=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
$endgroup$
– Zest
Jan 27 at 9:36
1
$begingroup$
Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done
$endgroup$
– lonza leggiera
Jan 27 at 13:33
|
show 1 more comment
$begingroup$
In answer to the question asked, about whether the argument in the last paragraph is correct, I don't believe that it is at it stands.
The problem is with the assertion $" {(
,f(x),g(x) )mid x in X, f(x) not= g(x) } $ is a union of all such open neighbourhoods $ dots $ ." If you take the union of all the open neighbourhoods you describe you will certainly get an open set containing $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $, but I don't see why it should be equal to it, and I believe I can give a counterexample to the general statement that $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ must be open under the hypotheses given
Nevertheless, I think Alex Ravsky's proof shows that the basic idea underlying the argument was sound.
Addendum: It may be helpful for me to add one of the counterexamples I was referring to in my above comments.
Take $ X = Y = mathbb R $—the set of real numbers—, and $ mathscr{M}_X ,mathscr{M}_Y $ both to be the Borel sets of $ mathbb R $. Let $ fleft(xright)= 0 mbox{ and } gleft(xright)= x mbox{ for all } xinmathbb R $. Then $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ is the punctured line segment $ left{ left(,0,x,right) mid xne0 right} $ of $ mathbb R^2 $, which is not an open set.
Comment on proposed new proof:
Apart from some probable typos, itemised below, the new revised proof by Zest seems to me to be correct.
The following appear likely to me to be typos:
$$B' = bigcup left{ Utimes Vmid U,Vin B, Ucap V=emptysetright} .$$
This should be
$$B' = bigcup left{ Utimes Vmid U,Vin{mathcal B}, Ucap V=emptysetright}$$
- I'm not sure what the observation "Since $ B $ is a countable base $ mathcal B dots $" is trying to say. As far as I can see, all that's necessary here is the observation that $ mathcal B $ is a countable base.
- A union sign is missing from the last term on the left of the final series of equations. That is, $ left{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $ should be $ bigcupleft{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $
The nice observation Alex Ravsky and Zest have made here, and which I had missed, is that $ left{,xin X mid fleft(xright)neq gleft(xright),right} $ is actually equal to $ left(ftimes gright)^{-1}left(B'right) $ rather than being a strict subset of it. Since this isn't immediately obvious—at least it wasn't to me, although the proof is fairly straightforward once the fact is realised—I'd like to see a brief proof of that fact included in any formal writeup.
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
$endgroup$
$begingroup$
thanks for your help. well, the union of open sets is open, thus if in fact the set ${(f(x),g(x)) mid x in X, f(x) not= g(x)}$ is a union of disjoint open neighbourhood, it is open.
$endgroup$
– Zest
Jan 26 at 23:53
$begingroup$
But how did you get that the set ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) }$ is a union of disjoint open neighbourhoods? If $ U $ and $ V $ are disjoint open neighbourhoods around $ f(x) $ and $ g(x) $ respectively, $ U $ may well contain an element $ u $ that isn't in the image of $ f $, and $ V $ an element $ v $ that isn't in the image of $ g $. The point $ (u, v) $ will be in any union that includes $ Utimes V $ but *not* in ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) } $, because there is no $ x in X $ with $ (f(x),g(x)) = (u,v)$ .
$endgroup$
– lonza leggiera
Jan 27 at 0:18
$begingroup$
you're right. i didnt consider that. i will think about it again! thanks for the clarification.
$endgroup$
– Zest
Jan 27 at 0:25
1
$begingroup$
i have another question. is it possible that we don't even need $ B := { f(x),g(x) mid x in X, f(x)not= g(x)}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $Bsubset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $fnot=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
$endgroup$
– Zest
Jan 27 at 9:36
1
$begingroup$
Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done
$endgroup$
– lonza leggiera
Jan 27 at 13:33
|
show 1 more comment
$begingroup$
In answer to the question asked, about whether the argument in the last paragraph is correct, I don't believe that it is at it stands.
The problem is with the assertion $" {(
,f(x),g(x) )mid x in X, f(x) not= g(x) } $ is a union of all such open neighbourhoods $ dots $ ." If you take the union of all the open neighbourhoods you describe you will certainly get an open set containing $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $, but I don't see why it should be equal to it, and I believe I can give a counterexample to the general statement that $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ must be open under the hypotheses given
Nevertheless, I think Alex Ravsky's proof shows that the basic idea underlying the argument was sound.
Addendum: It may be helpful for me to add one of the counterexamples I was referring to in my above comments.
Take $ X = Y = mathbb R $—the set of real numbers—, and $ mathscr{M}_X ,mathscr{M}_Y $ both to be the Borel sets of $ mathbb R $. Let $ fleft(xright)= 0 mbox{ and } gleft(xright)= x mbox{ for all } xinmathbb R $. Then $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ is the punctured line segment $ left{ left(,0,x,right) mid xne0 right} $ of $ mathbb R^2 $, which is not an open set.
Comment on proposed new proof:
Apart from some probable typos, itemised below, the new revised proof by Zest seems to me to be correct.
The following appear likely to me to be typos:
$$B' = bigcup left{ Utimes Vmid U,Vin B, Ucap V=emptysetright} .$$
This should be
$$B' = bigcup left{ Utimes Vmid U,Vin{mathcal B}, Ucap V=emptysetright}$$
- I'm not sure what the observation "Since $ B $ is a countable base $ mathcal B dots $" is trying to say. As far as I can see, all that's necessary here is the observation that $ mathcal B $ is a countable base.
- A union sign is missing from the last term on the left of the final series of equations. That is, $ left{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $ should be $ bigcupleft{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $
The nice observation Alex Ravsky and Zest have made here, and which I had missed, is that $ left{,xin X mid fleft(xright)neq gleft(xright),right} $ is actually equal to $ left(ftimes gright)^{-1}left(B'right) $ rather than being a strict subset of it. Since this isn't immediately obvious—at least it wasn't to me, although the proof is fairly straightforward once the fact is realised—I'd like to see a brief proof of that fact included in any formal writeup.
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
$endgroup$
In answer to the question asked, about whether the argument in the last paragraph is correct, I don't believe that it is at it stands.
The problem is with the assertion $" {(
,f(x),g(x) )mid x in X, f(x) not= g(x) } $ is a union of all such open neighbourhoods $ dots $ ." If you take the union of all the open neighbourhoods you describe you will certainly get an open set containing $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $, but I don't see why it should be equal to it, and I believe I can give a counterexample to the general statement that $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ must be open under the hypotheses given
Nevertheless, I think Alex Ravsky's proof shows that the basic idea underlying the argument was sound.
Addendum: It may be helpful for me to add one of the counterexamples I was referring to in my above comments.
Take $ X = Y = mathbb R $—the set of real numbers—, and $ mathscr{M}_X ,mathscr{M}_Y $ both to be the Borel sets of $ mathbb R $. Let $ fleft(xright)= 0 mbox{ and } gleft(xright)= x mbox{ for all } xinmathbb R $. Then $ {(
,f(x),g(x) )mid x in X,, f(x) not= g(x) } $ is the punctured line segment $ left{ left(,0,x,right) mid xne0 right} $ of $ mathbb R^2 $, which is not an open set.
Comment on proposed new proof:
Apart from some probable typos, itemised below, the new revised proof by Zest seems to me to be correct.
The following appear likely to me to be typos:
$$B' = bigcup left{ Utimes Vmid U,Vin B, Ucap V=emptysetright} .$$
This should be
$$B' = bigcup left{ Utimes Vmid U,Vin{mathcal B}, Ucap V=emptysetright}$$
- I'm not sure what the observation "Since $ B $ is a countable base $ mathcal B dots $" is trying to say. As far as I can see, all that's necessary here is the observation that $ mathcal B $ is a countable base.
- A union sign is missing from the last term on the left of the final series of equations. That is, $ left{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $ should be $ bigcupleft{,f^{-1}left(Uright)cap g^{-1}left(Vright)mid U,Vin{mathcal B}, Ucap V=emptysetright} $
The nice observation Alex Ravsky and Zest have made here, and which I had missed, is that $ left{,xin X mid fleft(xright)neq gleft(xright),right} $ is actually equal to $ left(ftimes gright)^{-1}left(B'right) $ rather than being a strict subset of it. Since this isn't immediately obvious—at least it wasn't to me, although the proof is fairly straightforward once the fact is realised—I'd like to see a brief proof of that fact included in any formal writeup.
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
edited Jan 28 at 0:11
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
answered Jan 26 at 23:11
lonza leggieralonza leggiera
1,03228
1,03228
$begingroup$
thanks for your help. well, the union of open sets is open, thus if in fact the set ${(f(x),g(x)) mid x in X, f(x) not= g(x)}$ is a union of disjoint open neighbourhood, it is open.
$endgroup$
– Zest
Jan 26 at 23:53
$begingroup$
But how did you get that the set ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) }$ is a union of disjoint open neighbourhoods? If $ U $ and $ V $ are disjoint open neighbourhoods around $ f(x) $ and $ g(x) $ respectively, $ U $ may well contain an element $ u $ that isn't in the image of $ f $, and $ V $ an element $ v $ that isn't in the image of $ g $. The point $ (u, v) $ will be in any union that includes $ Utimes V $ but *not* in ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) } $, because there is no $ x in X $ with $ (f(x),g(x)) = (u,v)$ .
$endgroup$
– lonza leggiera
Jan 27 at 0:18
$begingroup$
you're right. i didnt consider that. i will think about it again! thanks for the clarification.
$endgroup$
– Zest
Jan 27 at 0:25
1
$begingroup$
i have another question. is it possible that we don't even need $ B := { f(x),g(x) mid x in X, f(x)not= g(x)}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $Bsubset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $fnot=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
$endgroup$
– Zest
Jan 27 at 9:36
1
$begingroup$
Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done
$endgroup$
– lonza leggiera
Jan 27 at 13:33
|
show 1 more comment
$begingroup$
thanks for your help. well, the union of open sets is open, thus if in fact the set ${(f(x),g(x)) mid x in X, f(x) not= g(x)}$ is a union of disjoint open neighbourhood, it is open.
$endgroup$
– Zest
Jan 26 at 23:53
$begingroup$
But how did you get that the set ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) }$ is a union of disjoint open neighbourhoods? If $ U $ and $ V $ are disjoint open neighbourhoods around $ f(x) $ and $ g(x) $ respectively, $ U $ may well contain an element $ u $ that isn't in the image of $ f $, and $ V $ an element $ v $ that isn't in the image of $ g $. The point $ (u, v) $ will be in any union that includes $ Utimes V $ but *not* in ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) } $, because there is no $ x in X $ with $ (f(x),g(x)) = (u,v)$ .
$endgroup$
– lonza leggiera
Jan 27 at 0:18
$begingroup$
you're right. i didnt consider that. i will think about it again! thanks for the clarification.
$endgroup$
– Zest
Jan 27 at 0:25
1
$begingroup$
i have another question. is it possible that we don't even need $ B := { f(x),g(x) mid x in X, f(x)not= g(x)}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $Bsubset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $fnot=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
$endgroup$
– Zest
Jan 27 at 9:36
1
$begingroup$
Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done
$endgroup$
– lonza leggiera
Jan 27 at 13:33
$begingroup$
thanks for your help. well, the union of open sets is open, thus if in fact the set ${(f(x),g(x)) mid x in X, f(x) not= g(x)}$ is a union of disjoint open neighbourhood, it is open.
$endgroup$
– Zest
Jan 26 at 23:53
$begingroup$
thanks for your help. well, the union of open sets is open, thus if in fact the set ${(f(x),g(x)) mid x in X, f(x) not= g(x)}$ is a union of disjoint open neighbourhood, it is open.
$endgroup$
– Zest
Jan 26 at 23:53
$begingroup$
But how did you get that the set ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) }$ is a union of disjoint open neighbourhoods? If $ U $ and $ V $ are disjoint open neighbourhoods around $ f(x) $ and $ g(x) $ respectively, $ U $ may well contain an element $ u $ that isn't in the image of $ f $, and $ V $ an element $ v $ that isn't in the image of $ g $. The point $ (u, v) $ will be in any union that includes $ Utimes V $ but *not* in ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) } $, because there is no $ x in X $ with $ (f(x),g(x)) = (u,v)$ .
$endgroup$
– lonza leggiera
Jan 27 at 0:18
$begingroup$
But how did you get that the set ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) }$ is a union of disjoint open neighbourhoods? If $ U $ and $ V $ are disjoint open neighbourhoods around $ f(x) $ and $ g(x) $ respectively, $ U $ may well contain an element $ u $ that isn't in the image of $ f $, and $ V $ an element $ v $ that isn't in the image of $ g $. The point $ (u, v) $ will be in any union that includes $ Utimes V $ but *not* in ${( ,f(x),g(x) )mid x in X, f(x) not= g(x) } $, because there is no $ x in X $ with $ (f(x),g(x)) = (u,v)$ .
$endgroup$
– lonza leggiera
Jan 27 at 0:18
$begingroup$
you're right. i didnt consider that. i will think about it again! thanks for the clarification.
$endgroup$
– Zest
Jan 27 at 0:25
$begingroup$
you're right. i didnt consider that. i will think about it again! thanks for the clarification.
$endgroup$
– Zest
Jan 27 at 0:25
1
1
$begingroup$
i have another question. is it possible that we don't even need $ B := { f(x),g(x) mid x in X, f(x)not= g(x)}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $Bsubset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $fnot=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
$endgroup$
– Zest
Jan 27 at 9:36
$begingroup$
i have another question. is it possible that we don't even need $ B := { f(x),g(x) mid x in X, f(x)not= g(x)}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $Bsubset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $fnot=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
$endgroup$
– Zest
Jan 27 at 9:36
1
1
$begingroup$
Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done
$endgroup$
– lonza leggiera
Jan 27 at 13:33
$begingroup$
Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done
$endgroup$
– lonza leggiera
Jan 27 at 13:33
|
show 1 more comment
$begingroup$
$h(x)=(f(x),g(x))$ is measurable and ${x:f(x)neq g(x)}=h^{-1}(Ytimes Y-Delta)$ is measurable since $Delta$ is closed, $(Ytimes Y)-Delta$ is open.
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
$endgroup$
$begingroup$
$h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
$endgroup$
– Zest
Jan 23 at 3:39
add a comment |
$begingroup$
$h(x)=(f(x),g(x))$ is measurable and ${x:f(x)neq g(x)}=h^{-1}(Ytimes Y-Delta)$ is measurable since $Delta$ is closed, $(Ytimes Y)-Delta$ is open.
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
$endgroup$
$begingroup$
$h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
$endgroup$
– Zest
Jan 23 at 3:39
add a comment |
$begingroup$
$h(x)=(f(x),g(x))$ is measurable and ${x:f(x)neq g(x)}=h^{-1}(Ytimes Y-Delta)$ is measurable since $Delta$ is closed, $(Ytimes Y)-Delta$ is open.
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
$endgroup$
$h(x)=(f(x),g(x))$ is measurable and ${x:f(x)neq g(x)}=h^{-1}(Ytimes Y-Delta)$ is measurable since $Delta$ is closed, $(Ytimes Y)-Delta$ is open.
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
answered Jan 23 at 3:25
Tsemo AristideTsemo Aristide
59.3k11445
59.3k11445
$begingroup$
$h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
$endgroup$
– Zest
Jan 23 at 3:39
add a comment |
$begingroup$
$h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
$endgroup$
– Zest
Jan 23 at 3:39
$begingroup$
$h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
$endgroup$
– Zest
Jan 23 at 3:39
$begingroup$
$h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
$endgroup$
– Zest
Jan 23 at 3:39
add a comment |
$begingroup$
I think all presented proofs are incomplete (for instance, they don’t use the fact that $Y$ has a countable base $mathcal B$). I also assume that $mathcal M_X$ is closed with respect to intersections and countable unions.
Let $A={xin X|f(x)ne g(x)}$. Let $xin A$ be any point. Since $f(x)ne g(x)$ and $Y$ is Hausdorff, there exists disjoint neighborhoods $U,Vinmathcal B$ of $f(x)$ and $g(x)$, respectively. Thus
$$A=bigcup {f^{-1}(U)cap g^{-1}(V): U,Vinmathcal B, Ucap V=varnothing}$$
is measurable.
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
$begingroup$
hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
$endgroup$
– Zest
Jan 26 at 23:59
$begingroup$
@Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $mathcal M_X$ is closed with respect to (finite) intersections and countable unions.
$endgroup$
– Alex Ravsky
Jan 27 at 0:11
1
$begingroup$
i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
$endgroup$
– Zest
Jan 27 at 0:23
$begingroup$
hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
$endgroup$
– Zest
Jan 27 at 10:42
add a comment |
$begingroup$
I think all presented proofs are incomplete (for instance, they don’t use the fact that $Y$ has a countable base $mathcal B$). I also assume that $mathcal M_X$ is closed with respect to intersections and countable unions.
Let $A={xin X|f(x)ne g(x)}$. Let $xin A$ be any point. Since $f(x)ne g(x)$ and $Y$ is Hausdorff, there exists disjoint neighborhoods $U,Vinmathcal B$ of $f(x)$ and $g(x)$, respectively. Thus
$$A=bigcup {f^{-1}(U)cap g^{-1}(V): U,Vinmathcal B, Ucap V=varnothing}$$
is measurable.
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
$begingroup$
hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
$endgroup$
– Zest
Jan 26 at 23:59
$begingroup$
@Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $mathcal M_X$ is closed with respect to (finite) intersections and countable unions.
$endgroup$
– Alex Ravsky
Jan 27 at 0:11
1
$begingroup$
i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
$endgroup$
– Zest
Jan 27 at 0:23
$begingroup$
hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
$endgroup$
– Zest
Jan 27 at 10:42
add a comment |
$begingroup$
I think all presented proofs are incomplete (for instance, they don’t use the fact that $Y$ has a countable base $mathcal B$). I also assume that $mathcal M_X$ is closed with respect to intersections and countable unions.
Let $A={xin X|f(x)ne g(x)}$. Let $xin A$ be any point. Since $f(x)ne g(x)$ and $Y$ is Hausdorff, there exists disjoint neighborhoods $U,Vinmathcal B$ of $f(x)$ and $g(x)$, respectively. Thus
$$A=bigcup {f^{-1}(U)cap g^{-1}(V): U,Vinmathcal B, Ucap V=varnothing}$$
is measurable.
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
I think all presented proofs are incomplete (for instance, they don’t use the fact that $Y$ has a countable base $mathcal B$). I also assume that $mathcal M_X$ is closed with respect to intersections and countable unions.
Let $A={xin X|f(x)ne g(x)}$. Let $xin A$ be any point. Since $f(x)ne g(x)$ and $Y$ is Hausdorff, there exists disjoint neighborhoods $U,Vinmathcal B$ of $f(x)$ and $g(x)$, respectively. Thus
$$A=bigcup {f^{-1}(U)cap g^{-1}(V): U,Vinmathcal B, Ucap V=varnothing}$$
is measurable.
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 26 at 5:13
Alex RavskyAlex Ravsky
42.7k32383
42.7k32383
$begingroup$
hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
$endgroup$
– Zest
Jan 26 at 23:59
$begingroup$
@Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $mathcal M_X$ is closed with respect to (finite) intersections and countable unions.
$endgroup$
– Alex Ravsky
Jan 27 at 0:11
1
$begingroup$
i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
$endgroup$
– Zest
Jan 27 at 0:23
$begingroup$
hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
$endgroup$
– Zest
Jan 27 at 10:42
add a comment |
$begingroup$
hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
$endgroup$
– Zest
Jan 26 at 23:59
$begingroup$
@Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $mathcal M_X$ is closed with respect to (finite) intersections and countable unions.
$endgroup$
– Alex Ravsky
Jan 27 at 0:11
1
$begingroup$
i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
$endgroup$
– Zest
Jan 27 at 0:23
$begingroup$
hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
$endgroup$
– Zest
Jan 27 at 10:42
$begingroup$
hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
$endgroup$
– Zest
Jan 26 at 23:59
$begingroup$
hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
$endgroup$
– Zest
Jan 26 at 23:59
$begingroup$
@Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $mathcal M_X$ is closed with respect to (finite) intersections and countable unions.
$endgroup$
– Alex Ravsky
Jan 27 at 0:11
$begingroup$
@Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $mathcal M_X$ is closed with respect to (finite) intersections and countable unions.
$endgroup$
– Alex Ravsky
Jan 27 at 0:11
1
1
$begingroup$
i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
$endgroup$
– Zest
Jan 27 at 0:23
$begingroup$
i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
$endgroup$
– Zest
Jan 27 at 0:23
$begingroup$
hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
$endgroup$
– Zest
Jan 27 at 10:42
$begingroup$
hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
$endgroup$
– Zest
Jan 27 at 10:42
add a comment |
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$begingroup$
There is obviously a typo in the expression ${(f(x),g(x)mid x in X}$ which occurs at the start of the 5th and 8th paragraphs. For one thing the opening parenthesis remains unclosed. If ${(f(x),g(x))mid x in X}$ is what was meant, then I don't see why it would necessarily be a subset of $(Y times Y)setminus Delta Y$, since there's nothing in the definition of the set to eliminate those $ x $ for which $ f(x) = g(x) $. Was it supposed to have been ${(f(x),g(x))mid x in X, f(x) ne g(x) } $ ?
$endgroup$
– lonza leggiera
Jan 26 at 11:00
$begingroup$
hello @lonzaleggiera, you're right, that's a typo. thanks for pointing out. i will edit my question immediately.
$endgroup$
– Zest
Jan 26 at 17:28