If $E$ open set. Why $left{(x,y):x-yin Eright}$ is Lebesgue measurable?
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Without using that $f(x,y)=x-y$ is continuous.
If $E$ is open set in $mathbb{R}$. Why $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is a Lebesgue measurable set? $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is open set?
real-analysis measure-theory
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
$endgroup$
add a comment |
$begingroup$
Without using that $f(x,y)=x-y$ is continuous.
If $E$ is open set in $mathbb{R}$. Why $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is a Lebesgue measurable set? $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is open set?
real-analysis measure-theory
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
$endgroup$
$begingroup$
Could you use that rotation of $Bbb R^2$ by $45°$ is continuous/measurable?
$endgroup$
– Hagen von Eitzen
Jan 6 at 23:16
add a comment |
$begingroup$
Without using that $f(x,y)=x-y$ is continuous.
If $E$ is open set in $mathbb{R}$. Why $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is a Lebesgue measurable set? $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is open set?
real-analysis measure-theory
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
$endgroup$
Without using that $f(x,y)=x-y$ is continuous.
If $E$ is open set in $mathbb{R}$. Why $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is a Lebesgue measurable set? $left{(x,y)inmathbb{R}^2:x-yin Eright}$ is open set?
real-analysis measure-theory
real-analysis measure-theory
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
asked Jan 6 at 23:13
eraldcoileraldcoil
403211
403211
$begingroup$
Could you use that rotation of $Bbb R^2$ by $45°$ is continuous/measurable?
$endgroup$
– Hagen von Eitzen
Jan 6 at 23:16
add a comment |
$begingroup$
Could you use that rotation of $Bbb R^2$ by $45°$ is continuous/measurable?
$endgroup$
– Hagen von Eitzen
Jan 6 at 23:16
$begingroup$
Could you use that rotation of $Bbb R^2$ by $45°$ is continuous/measurable?
$endgroup$
– Hagen von Eitzen
Jan 6 at 23:16
$begingroup$
Could you use that rotation of $Bbb R^2$ by $45°$ is continuous/measurable?
$endgroup$
– Hagen von Eitzen
Jan 6 at 23:16
add a comment |
1 Answer
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Let $S={(x,y)mid x-yin E}$.
Given a point $(a,b)in S$.
Then $a-bin E$, and hence, there is a neighbourhood of $a-b$, say $]a-b-r, a-b+r[subseteq E$ for some $r>0$.
Then the open disc with radius $r/2$ aroud the center $(a,b)$ is contained in $S$: if $(x,y)$ is in this disc, then $a-r/2<x<a+r/2$ and $b-r/2<y<b+r/2$, thus $a-b-r<x-y<a-b+r$, hence $x-yin E$.
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
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$begingroup$
Let $S={(x,y)mid x-yin E}$.
Given a point $(a,b)in S$.
Then $a-bin E$, and hence, there is a neighbourhood of $a-b$, say $]a-b-r, a-b+r[subseteq E$ for some $r>0$.
Then the open disc with radius $r/2$ aroud the center $(a,b)$ is contained in $S$: if $(x,y)$ is in this disc, then $a-r/2<x<a+r/2$ and $b-r/2<y<b+r/2$, thus $a-b-r<x-y<a-b+r$, hence $x-yin E$.
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
$endgroup$
add a comment |
$begingroup$
Let $S={(x,y)mid x-yin E}$.
Given a point $(a,b)in S$.
Then $a-bin E$, and hence, there is a neighbourhood of $a-b$, say $]a-b-r, a-b+r[subseteq E$ for some $r>0$.
Then the open disc with radius $r/2$ aroud the center $(a,b)$ is contained in $S$: if $(x,y)$ is in this disc, then $a-r/2<x<a+r/2$ and $b-r/2<y<b+r/2$, thus $a-b-r<x-y<a-b+r$, hence $x-yin E$.
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
$endgroup$
add a comment |
$begingroup$
Let $S={(x,y)mid x-yin E}$.
Given a point $(a,b)in S$.
Then $a-bin E$, and hence, there is a neighbourhood of $a-b$, say $]a-b-r, a-b+r[subseteq E$ for some $r>0$.
Then the open disc with radius $r/2$ aroud the center $(a,b)$ is contained in $S$: if $(x,y)$ is in this disc, then $a-r/2<x<a+r/2$ and $b-r/2<y<b+r/2$, thus $a-b-r<x-y<a-b+r$, hence $x-yin E$.
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
$endgroup$
Let $S={(x,y)mid x-yin E}$.
Given a point $(a,b)in S$.
Then $a-bin E$, and hence, there is a neighbourhood of $a-b$, say $]a-b-r, a-b+r[subseteq E$ for some $r>0$.
Then the open disc with radius $r/2$ aroud the center $(a,b)$ is contained in $S$: if $(x,y)$ is in this disc, then $a-r/2<x<a+r/2$ and $b-r/2<y<b+r/2$, thus $a-b-r<x-y<a-b+r$, hence $x-yin E$.
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
answered Jan 6 at 23:24
A. PongráczA. Pongrácz
6,0821929
6,0821929
add a comment |
add a comment |
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$begingroup$
Could you use that rotation of $Bbb R^2$ by $45°$ is continuous/measurable?
$endgroup$
– Hagen von Eitzen
Jan 6 at 23:16