Mixing
to show if f is measurable or not
$begingroup$
Given $X={1,2,3,4}$ and the sigma-Algebra $S={emptyset,{1},{2},{1,2},{3,4},{1,3,4},{2,3,4},{1,2,3,4}} $,
I've got to check out whether the function $f:(X,S)rightarrow (mathbb{R}, mathscr{B(mathbb{R}))}$is measurable or not for
i) $f(x)=(x-3)^2$ and ii) $f(x)=|x-frac{7}{2}|$
My solution would be, that both functions are not measurable, because you have that
${0}in mathscr{B(mathbb{R})}$ but $f^{-1}({0})=3$ for i) and $f^{-1}({0})=frac{7}{2}$. As $3 notin S$ and $frac{7}{2} notin S$, they would be not measurable. However this is only true for i) but does not count for ii). ii) is measurable and I don't know where and why I am wrong with my argument??
measure-theory measurable-functions
asked Jan 26 at 11:21
ThesinusThesinus
254210
$endgroup$
add a comment |
$begingroup$
Given $X={1,2,3,4}$ and the sigma-Algebra $S={emptyset,{1},{2},{1,2},{3,4},{1,3,4},{2,3,4},{1,2,3,4}} $,
I've got to check out whether the function $f:(X,S)rightarrow (mathbb{R}, mathscr{B(mathbb{R}))}$is measurable or not for
i) $f(x)=(x-3)^2$ and ii) $f(x)=|x-frac{7}{2}|$
My solution would be, that both functions are not measurable, because you have that
${0}in mathscr{B(mathbb{R})}$ but $f^{-1}({0})=3$ for i) and $f^{-1}({0})=frac{7}{2}$. As $3 notin S$ and $frac{7}{2} notin S$, they would be not measurable. However this is only true for i) but does not count for ii). ii) is measurable and I don't know where and why I am wrong with my argument??
measure-theory measurable-functions
asked Jan 26 at 11:21
ThesinusThesinus
254210
$endgroup$
1
$begingroup$
Because 7/2 is not in $X$ so your function never has the value 0
$endgroup$
– orange
Jan 26 at 11:31
$begingroup$
You can ofcourse use the same argument for the value of $f(3)$.
$endgroup$
– orange
Jan 26 at 11:33
add a comment |
$begingroup$
Given $X={1,2,3,4}$ and the sigma-Algebra $S={emptyset,{1},{2},{1,2},{3,4},{1,3,4},{2,3,4},{1,2,3,4}} $,
I've got to check out whether the function $f:(X,S)rightarrow (mathbb{R}, mathscr{B(mathbb{R}))}$is measurable or not for
i) $f(x)=(x-3)^2$ and ii) $f(x)=|x-frac{7}{2}|$
My solution would be, that both functions are not measurable, because you have that
${0}in mathscr{B(mathbb{R})}$ but $f^{-1}({0})=3$ for i) and $f^{-1}({0})=frac{7}{2}$. As $3 notin S$ and $frac{7}{2} notin S$, they would be not measurable. However this is only true for i) but does not count for ii). ii) is measurable and I don't know where and why I am wrong with my argument??
measure-theory measurable-functions
asked Jan 26 at 11:21
ThesinusThesinus
254210
$endgroup$
Given $X={1,2,3,4}$ and the sigma-Algebra $S={emptyset,{1},{2},{1,2},{3,4},{1,3,4},{2,3,4},{1,2,3,4}} $,
I've got to check out whether the function $f:(X,S)rightarrow (mathbb{R}, mathscr{B(mathbb{R}))}$is measurable or not for
i) $f(x)=(x-3)^2$ and ii) $f(x)=|x-frac{7}{2}|$
My solution would be, that both functions are not measurable, because you have that
${0}in mathscr{B(mathbb{R})}$ but $f^{-1}({0})=3$ for i) and $f^{-1}({0})=frac{7}{2}$. As $3 notin S$ and $frac{7}{2} notin S$, they would be not measurable. However this is only true for i) but does not count for ii). ii) is measurable and I don't know where and why I am wrong with my argument??
measure-theory measurable-functions
measure-theory measurable-functions
asked Jan 26 at 11:21
ThesinusThesinus
254210
asked Jan 26 at 11:21
ThesinusThesinus
254210
asked Jan 26 at 11:21
ThesinusThesinus
254210
asked Jan 26 at 11:21
ThesinusThesinus
254210
asked Jan 26 at 11:21
ThesinusThesinus
254210
254210
1
$begingroup$
Because 7/2 is not in $X$ so your function never has the value 0
$endgroup$
– orange
Jan 26 at 11:31
$begingroup$
You can ofcourse use the same argument for the value of $f(3)$.
$endgroup$
– orange
Jan 26 at 11:33
add a comment |
1
$begingroup$
Because 7/2 is not in $X$ so your function never has the value 0
$endgroup$
– orange
Jan 26 at 11:31
$begingroup$
You can ofcourse use the same argument for the value of $f(3)$.
$endgroup$
– orange
Jan 26 at 11:33
1
1
$begingroup$
Because 7/2 is not in $X$ so your function never has the value 0
$endgroup$
– orange
Jan 26 at 11:31
$begingroup$
Because 7/2 is not in $X$ so your function never has the value 0
$endgroup$
– orange
Jan 26 at 11:31
$begingroup$
You can ofcourse use the same argument for the value of $f(3)$.
$endgroup$
– orange
Jan 26 at 11:33
$begingroup$
You can ofcourse use the same argument for the value of $f(3)$.
$endgroup$
– orange
Jan 26 at 11:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You stated that $f^{-1}({0})=frac72$ in case ii, which is not true.
We have $f^{-1}({0})=varnothingin S$.
The image of $f$ in case ii is $I:={frac52,frac32,frac12}$ and this with
- $f^{-1}left({frac52}right)={1}in S$
- $f^{-1}left({frac32}right)={2}in S$
- $f^{-1}left({frac12}right)={3,4}in S$
Also $varnothingin S$ so from this we are allowed to conclude that $f^{-1}(A)=f^{-1}(Acap I)in S$ for every subset of $mathbb R$, hence is measurable.
answered Jan 26 at 11:47
drhabdrhab
103k545136
$endgroup$
add a comment |
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$begingroup$
You stated that $f^{-1}({0})=frac72$ in case ii, which is not true.
We have $f^{-1}({0})=varnothingin S$.
The image of $f$ in case ii is $I:={frac52,frac32,frac12}$ and this with
- $f^{-1}left({frac52}right)={1}in S$
- $f^{-1}left({frac32}right)={2}in S$
- $f^{-1}left({frac12}right)={3,4}in S$
Also $varnothingin S$ so from this we are allowed to conclude that $f^{-1}(A)=f^{-1}(Acap I)in S$ for every subset of $mathbb R$, hence is measurable.
answered Jan 26 at 11:47
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
You stated that $f^{-1}({0})=frac72$ in case ii, which is not true.
We have $f^{-1}({0})=varnothingin S$.
The image of $f$ in case ii is $I:={frac52,frac32,frac12}$ and this with
- $f^{-1}left({frac52}right)={1}in S$
- $f^{-1}left({frac32}right)={2}in S$
- $f^{-1}left({frac12}right)={3,4}in S$
Also $varnothingin S$ so from this we are allowed to conclude that $f^{-1}(A)=f^{-1}(Acap I)in S$ for every subset of $mathbb R$, hence is measurable.
answered Jan 26 at 11:47
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
You stated that $f^{-1}({0})=frac72$ in case ii, which is not true.
We have $f^{-1}({0})=varnothingin S$.
The image of $f$ in case ii is $I:={frac52,frac32,frac12}$ and this with
- $f^{-1}left({frac52}right)={1}in S$
- $f^{-1}left({frac32}right)={2}in S$
- $f^{-1}left({frac12}right)={3,4}in S$
Also $varnothingin S$ so from this we are allowed to conclude that $f^{-1}(A)=f^{-1}(Acap I)in S$ for every subset of $mathbb R$, hence is measurable.
answered Jan 26 at 11:47
drhabdrhab
103k545136
$endgroup$
You stated that $f^{-1}({0})=frac72$ in case ii, which is not true.
We have $f^{-1}({0})=varnothingin S$.
The image of $f$ in case ii is $I:={frac52,frac32,frac12}$ and this with
- $f^{-1}left({frac52}right)={1}in S$
- $f^{-1}left({frac32}right)={2}in S$
- $f^{-1}left({frac12}right)={3,4}in S$
Also $varnothingin S$ so from this we are allowed to conclude that $f^{-1}(A)=f^{-1}(Acap I)in S$ for every subset of $mathbb R$, hence is measurable.
answered Jan 26 at 11:47
drhabdrhab
103k545136
answered Jan 26 at 11:47
drhabdrhab
103k545136
answered Jan 26 at 11:47
drhabdrhab
103k545136
answered Jan 26 at 11:47
drhabdrhab
103k545136
103k545136
add a comment |
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1
$begingroup$
Because 7/2 is not in $X$ so your function never has the value 0
$endgroup$
– orange
Jan 26 at 11:31
$begingroup$
You can ofcourse use the same argument for the value of $f(3)$.
$endgroup$
– orange
Jan 26 at 11:33