A Borel set $E$ is a null set for the Borel measure that is induced by $sigma(x)=x^2$ if and only if $E$ is a...












0












$begingroup$



Ture or False? If $sigma(x)=x^2$ on an interval $[a,b]$ and if $sigma^*$ is the Borel measure on $[a,b]$ that is induced by $sigma$, then a Borel set $E$ in $[a,b]$ is a null set for $sigma^*$ if and only if $E$ is a Lebesgue null set.




Well, I don't quite get what $sigma^*$ really does. If $Asubseteq [a,b]$ then how does one define $sigma^*(A)$? Would it be $b^2-a^2$? I guess I need some explanation on the word "induced". Thanks in advance for any help.






real-analysis measure-theory lebesgue-measure







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$endgroup$












  • $begingroup$
    You need $ageq 0$ for this question to make sense right?
    $endgroup$
    – Shashi
    Jan 7 at 8:46










  • $begingroup$
    @Shashi Given KaviRamaMurthy's definition on $sigma^*$, I don't see a problem for $a$ being negative, though.
    $endgroup$
    – Alex
    Jan 7 at 9:26
















0












$begingroup$



Ture or False? If $sigma(x)=x^2$ on an interval $[a,b]$ and if $sigma^*$ is the Borel measure on $[a,b]$ that is induced by $sigma$, then a Borel set $E$ in $[a,b]$ is a null set for $sigma^*$ if and only if $E$ is a Lebesgue null set.




Well, I don't quite get what $sigma^*$ really does. If $Asubseteq [a,b]$ then how does one define $sigma^*(A)$? Would it be $b^2-a^2$? I guess I need some explanation on the word "induced". Thanks in advance for any help.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    You need $ageq 0$ for this question to make sense right?
    $endgroup$
    – Shashi
    Jan 7 at 8:46










  • $begingroup$
    @Shashi Given KaviRamaMurthy's definition on $sigma^*$, I don't see a problem for $a$ being negative, though.
    $endgroup$
    – Alex
    Jan 7 at 9:26














0












0








0





$begingroup$



Ture or False? If $sigma(x)=x^2$ on an interval $[a,b]$ and if $sigma^*$ is the Borel measure on $[a,b]$ that is induced by $sigma$, then a Borel set $E$ in $[a,b]$ is a null set for $sigma^*$ if and only if $E$ is a Lebesgue null set.




Well, I don't quite get what $sigma^*$ really does. If $Asubseteq [a,b]$ then how does one define $sigma^*(A)$? Would it be $b^2-a^2$? I guess I need some explanation on the word "induced". Thanks in advance for any help.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$





Ture or False? If $sigma(x)=x^2$ on an interval $[a,b]$ and if $sigma^*$ is the Borel measure on $[a,b]$ that is induced by $sigma$, then a Borel set $E$ in $[a,b]$ is a null set for $sigma^*$ if and only if $E$ is a Lebesgue null set.




Well, I don't quite get what $sigma^*$ really does. If $Asubseteq [a,b]$ then how does one define $sigma^*(A)$? Would it be $b^2-a^2$? I guess I need some explanation on the word "induced". Thanks in advance for any help.






real-analysis measure-theory lebesgue-measure




real-analysis measure-theory lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 8:27









AlexAlex

757




757












  • $begingroup$
    You need $ageq 0$ for this question to make sense right?
    $endgroup$
    – Shashi
    Jan 7 at 8:46










  • $begingroup$
    @Shashi Given KaviRamaMurthy's definition on $sigma^*$, I don't see a problem for $a$ being negative, though.
    $endgroup$
    – Alex
    Jan 7 at 9:26


















  • $begingroup$
    You need $ageq 0$ for this question to make sense right?
    $endgroup$
    – Shashi
    Jan 7 at 8:46










  • $begingroup$
    @Shashi Given KaviRamaMurthy's definition on $sigma^*$, I don't see a problem for $a$ being negative, though.
    $endgroup$
    – Alex
    Jan 7 at 9:26
















$begingroup$
You need $ageq 0$ for this question to make sense right?
$endgroup$
– Shashi
Jan 7 at 8:46




$begingroup$
You need $ageq 0$ for this question to make sense right?
$endgroup$
– Shashi
Jan 7 at 8:46












$begingroup$
@Shashi Given KaviRamaMurthy's definition on $sigma^*$, I don't see a problem for $a$ being negative, though.
$endgroup$
– Alex
Jan 7 at 9:26




$begingroup$
@Shashi Given KaviRamaMurthy's definition on $sigma^*$, I don't see a problem for $a$ being negative, though.
$endgroup$
– Alex
Jan 7 at 9:26










1 Answer
1






active

oldest

votes


















0












$begingroup$

Definition of $sigma^{*}$: $sigma^{*}(A)=m {x: x^{2} in A}$ where $m$ is Lebesgue measure. The fact that $ m(A)=0$ iff $sigma^{*} (A)=0$ is a consequence of the following theorem: if $f$ is absolutely continuous then $f$ maps sets of measure $0$ to sets of measure $0$. The function $f(x)=x^{2}$ is absolutely continuous because it is the indefinite integral of $x to 2x$. Simialrly, use the fact that the function $sqrt {|x|}$ is integrable for the converse part.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for your answer. However, I don't quite get your last line. Did you mean take $f(x)=2x$? So take $Asubseteq [a,b]$ then $m(A)=0implies int_A 2xdm=0implies int_A dsigma^*=0$. Is this what you were suggesting?
    $endgroup$
    – Alex
    Jan 7 at 22:22










  • $begingroup$
    @Alex I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 23:12










  • $begingroup$
    Thanks for updating. Now we have $m(A) = 0 implies m(f(A))=0 $ since $f(x)=x^2$ is abs. continuous. Then $m{x|x^2in A}=0implies sigma^*(A)=0$, which looks good. For the opposite direction, I still don't get how $sqrt{|x|}$ works. But $sigma^*(A)=0implies int_Adsigma^* = 0implies int_A 2xdx = 0implies m(A)=0$. Does this make sense?
    $endgroup$
    – Alex
    Jan 7 at 23:54












  • $begingroup$
    @Alex When $ageq 0$, we have $sqrt x =sqrt a +frac 1 2 int_a^{x} frac 1 {sqrt t}, dt$ so $sqrt x$ is absolutely continuous. I leave it to you to consider the case $a<0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 0:04













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Definition of $sigma^{*}$: $sigma^{*}(A)=m {x: x^{2} in A}$ where $m$ is Lebesgue measure. The fact that $ m(A)=0$ iff $sigma^{*} (A)=0$ is a consequence of the following theorem: if $f$ is absolutely continuous then $f$ maps sets of measure $0$ to sets of measure $0$. The function $f(x)=x^{2}$ is absolutely continuous because it is the indefinite integral of $x to 2x$. Simialrly, use the fact that the function $sqrt {|x|}$ is integrable for the converse part.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for your answer. However, I don't quite get your last line. Did you mean take $f(x)=2x$? So take $Asubseteq [a,b]$ then $m(A)=0implies int_A 2xdm=0implies int_A dsigma^*=0$. Is this what you were suggesting?
    $endgroup$
    – Alex
    Jan 7 at 22:22










  • $begingroup$
    @Alex I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 23:12










  • $begingroup$
    Thanks for updating. Now we have $m(A) = 0 implies m(f(A))=0 $ since $f(x)=x^2$ is abs. continuous. Then $m{x|x^2in A}=0implies sigma^*(A)=0$, which looks good. For the opposite direction, I still don't get how $sqrt{|x|}$ works. But $sigma^*(A)=0implies int_Adsigma^* = 0implies int_A 2xdx = 0implies m(A)=0$. Does this make sense?
    $endgroup$
    – Alex
    Jan 7 at 23:54












  • $begingroup$
    @Alex When $ageq 0$, we have $sqrt x =sqrt a +frac 1 2 int_a^{x} frac 1 {sqrt t}, dt$ so $sqrt x$ is absolutely continuous. I leave it to you to consider the case $a<0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 0:04


















0












$begingroup$

Definition of $sigma^{*}$: $sigma^{*}(A)=m {x: x^{2} in A}$ where $m$ is Lebesgue measure. The fact that $ m(A)=0$ iff $sigma^{*} (A)=0$ is a consequence of the following theorem: if $f$ is absolutely continuous then $f$ maps sets of measure $0$ to sets of measure $0$. The function $f(x)=x^{2}$ is absolutely continuous because it is the indefinite integral of $x to 2x$. Simialrly, use the fact that the function $sqrt {|x|}$ is integrable for the converse part.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for your answer. However, I don't quite get your last line. Did you mean take $f(x)=2x$? So take $Asubseteq [a,b]$ then $m(A)=0implies int_A 2xdm=0implies int_A dsigma^*=0$. Is this what you were suggesting?
    $endgroup$
    – Alex
    Jan 7 at 22:22










  • $begingroup$
    @Alex I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 23:12










  • $begingroup$
    Thanks for updating. Now we have $m(A) = 0 implies m(f(A))=0 $ since $f(x)=x^2$ is abs. continuous. Then $m{x|x^2in A}=0implies sigma^*(A)=0$, which looks good. For the opposite direction, I still don't get how $sqrt{|x|}$ works. But $sigma^*(A)=0implies int_Adsigma^* = 0implies int_A 2xdx = 0implies m(A)=0$. Does this make sense?
    $endgroup$
    – Alex
    Jan 7 at 23:54












  • $begingroup$
    @Alex When $ageq 0$, we have $sqrt x =sqrt a +frac 1 2 int_a^{x} frac 1 {sqrt t}, dt$ so $sqrt x$ is absolutely continuous. I leave it to you to consider the case $a<0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 0:04
















0












0








0





$begingroup$

Definition of $sigma^{*}$: $sigma^{*}(A)=m {x: x^{2} in A}$ where $m$ is Lebesgue measure. The fact that $ m(A)=0$ iff $sigma^{*} (A)=0$ is a consequence of the following theorem: if $f$ is absolutely continuous then $f$ maps sets of measure $0$ to sets of measure $0$. The function $f(x)=x^{2}$ is absolutely continuous because it is the indefinite integral of $x to 2x$. Simialrly, use the fact that the function $sqrt {|x|}$ is integrable for the converse part.






share|cite|improve this answer











$endgroup$



Definition of $sigma^{*}$: $sigma^{*}(A)=m {x: x^{2} in A}$ where $m$ is Lebesgue measure. The fact that $ m(A)=0$ iff $sigma^{*} (A)=0$ is a consequence of the following theorem: if $f$ is absolutely continuous then $f$ maps sets of measure $0$ to sets of measure $0$. The function $f(x)=x^{2}$ is absolutely continuous because it is the indefinite integral of $x to 2x$. Simialrly, use the fact that the function $sqrt {|x|}$ is integrable for the converse part.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 7 at 23:12

























answered Jan 7 at 8:37









Kavi Rama MurthyKavi Rama Murthy

70k53170




70k53170












  • $begingroup$
    Thanks for your answer. However, I don't quite get your last line. Did you mean take $f(x)=2x$? So take $Asubseteq [a,b]$ then $m(A)=0implies int_A 2xdm=0implies int_A dsigma^*=0$. Is this what you were suggesting?
    $endgroup$
    – Alex
    Jan 7 at 22:22










  • $begingroup$
    @Alex I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 23:12










  • $begingroup$
    Thanks for updating. Now we have $m(A) = 0 implies m(f(A))=0 $ since $f(x)=x^2$ is abs. continuous. Then $m{x|x^2in A}=0implies sigma^*(A)=0$, which looks good. For the opposite direction, I still don't get how $sqrt{|x|}$ works. But $sigma^*(A)=0implies int_Adsigma^* = 0implies int_A 2xdx = 0implies m(A)=0$. Does this make sense?
    $endgroup$
    – Alex
    Jan 7 at 23:54












  • $begingroup$
    @Alex When $ageq 0$, we have $sqrt x =sqrt a +frac 1 2 int_a^{x} frac 1 {sqrt t}, dt$ so $sqrt x$ is absolutely continuous. I leave it to you to consider the case $a<0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 0:04




















  • $begingroup$
    Thanks for your answer. However, I don't quite get your last line. Did you mean take $f(x)=2x$? So take $Asubseteq [a,b]$ then $m(A)=0implies int_A 2xdm=0implies int_A dsigma^*=0$. Is this what you were suggesting?
    $endgroup$
    – Alex
    Jan 7 at 22:22










  • $begingroup$
    @Alex I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 23:12










  • $begingroup$
    Thanks for updating. Now we have $m(A) = 0 implies m(f(A))=0 $ since $f(x)=x^2$ is abs. continuous. Then $m{x|x^2in A}=0implies sigma^*(A)=0$, which looks good. For the opposite direction, I still don't get how $sqrt{|x|}$ works. But $sigma^*(A)=0implies int_Adsigma^* = 0implies int_A 2xdx = 0implies m(A)=0$. Does this make sense?
    $endgroup$
    – Alex
    Jan 7 at 23:54












  • $begingroup$
    @Alex When $ageq 0$, we have $sqrt x =sqrt a +frac 1 2 int_a^{x} frac 1 {sqrt t}, dt$ so $sqrt x$ is absolutely continuous. I leave it to you to consider the case $a<0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 0:04


















$begingroup$
Thanks for your answer. However, I don't quite get your last line. Did you mean take $f(x)=2x$? So take $Asubseteq [a,b]$ then $m(A)=0implies int_A 2xdm=0implies int_A dsigma^*=0$. Is this what you were suggesting?
$endgroup$
– Alex
Jan 7 at 22:22




$begingroup$
Thanks for your answer. However, I don't quite get your last line. Did you mean take $f(x)=2x$? So take $Asubseteq [a,b]$ then $m(A)=0implies int_A 2xdm=0implies int_A dsigma^*=0$. Is this what you were suggesting?
$endgroup$
– Alex
Jan 7 at 22:22












$begingroup$
@Alex I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:12




$begingroup$
@Alex I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:12












$begingroup$
Thanks for updating. Now we have $m(A) = 0 implies m(f(A))=0 $ since $f(x)=x^2$ is abs. continuous. Then $m{x|x^2in A}=0implies sigma^*(A)=0$, which looks good. For the opposite direction, I still don't get how $sqrt{|x|}$ works. But $sigma^*(A)=0implies int_Adsigma^* = 0implies int_A 2xdx = 0implies m(A)=0$. Does this make sense?
$endgroup$
– Alex
Jan 7 at 23:54






$begingroup$
Thanks for updating. Now we have $m(A) = 0 implies m(f(A))=0 $ since $f(x)=x^2$ is abs. continuous. Then $m{x|x^2in A}=0implies sigma^*(A)=0$, which looks good. For the opposite direction, I still don't get how $sqrt{|x|}$ works. But $sigma^*(A)=0implies int_Adsigma^* = 0implies int_A 2xdx = 0implies m(A)=0$. Does this make sense?
$endgroup$
– Alex
Jan 7 at 23:54














$begingroup$
@Alex When $ageq 0$, we have $sqrt x =sqrt a +frac 1 2 int_a^{x} frac 1 {sqrt t}, dt$ so $sqrt x$ is absolutely continuous. I leave it to you to consider the case $a<0$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:04






$begingroup$
@Alex When $ageq 0$, we have $sqrt x =sqrt a +frac 1 2 int_a^{x} frac 1 {sqrt t}, dt$ so $sqrt x$ is absolutely continuous. I leave it to you to consider the case $a<0$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:04




















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