Dilatation of a measurable set is a measurable set.












3












$begingroup$


Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.



Show that $lambda E$ is measurable.



I have this.



Let $f:Eto lambda E$



$f(a)=lambda a$ continuos and bijective.



Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:28












  • $begingroup$
    If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
    $endgroup$
    – eraldcoil
    Dec 3 '18 at 23:41






  • 1




    $begingroup$
    Yes, because Lipschitz functions map null sets to null sets.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:43
















3












$begingroup$


Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.



Show that $lambda E$ is measurable.



I have this.



Let $f:Eto lambda E$



$f(a)=lambda a$ continuos and bijective.



Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:28












  • $begingroup$
    If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
    $endgroup$
    – eraldcoil
    Dec 3 '18 at 23:41






  • 1




    $begingroup$
    Yes, because Lipschitz functions map null sets to null sets.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:43














3












3








3





$begingroup$


Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.



Show that $lambda E$ is measurable.



I have this.



Let $f:Eto lambda E$



$f(a)=lambda a$ continuos and bijective.



Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$




Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.



Show that $lambda E$ is measurable.



I have this.



Let $f:Eto lambda E$



$f(a)=lambda a$ continuos and bijective.



Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?






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








edited Dec 3 '18 at 23:34







eraldcoil

















asked Dec 3 '18 at 23:18









eraldcoileraldcoil

386211




386211








  • 1




    $begingroup$
    You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:28












  • $begingroup$
    If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
    $endgroup$
    – eraldcoil
    Dec 3 '18 at 23:41






  • 1




    $begingroup$
    Yes, because Lipschitz functions map null sets to null sets.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:43














  • 1




    $begingroup$
    You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:28












  • $begingroup$
    If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
    $endgroup$
    – eraldcoil
    Dec 3 '18 at 23:41






  • 1




    $begingroup$
    Yes, because Lipschitz functions map null sets to null sets.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 23:43








1




1




$begingroup$
You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:28






$begingroup$
You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:28














$begingroup$
If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
$endgroup$
– eraldcoil
Dec 3 '18 at 23:41




$begingroup$
If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
$endgroup$
– eraldcoil
Dec 3 '18 at 23:41




1




1




$begingroup$
Yes, because Lipschitz functions map null sets to null sets.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:43




$begingroup$
Yes, because Lipschitz functions map null sets to null sets.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:43










2 Answers
2






active

oldest

votes


















2












$begingroup$

First we show that if $A$ has measure zero, then $lambda A$ has measure zero.



Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:



Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see



$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see



$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$



Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.



Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Why $lambda E=g^{-1}(G)$?
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 2:24






  • 1




    $begingroup$
    Must be $lambda G=g^{-1}(G)$
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 4:25










  • $begingroup$
    Yes, you're absolutely right!
    $endgroup$
    – Sean Haight
    Dec 7 '18 at 5:58



















0












$begingroup$

If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$



Therefore $lambda E$ is measurable.



It is correct?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is $lambda G - lambda E$ measurable?
    $endgroup$
    – Sean Haight
    Dec 6 '18 at 22:35










  • $begingroup$
    Oh. Sorry. Should be $m^{ast}$ outer measure
    $endgroup$
    – eraldcoil
    Dec 6 '18 at 22:37











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

First we show that if $A$ has measure zero, then $lambda A$ has measure zero.



Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:



Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see



$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see



$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$



Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.



Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Why $lambda E=g^{-1}(G)$?
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 2:24






  • 1




    $begingroup$
    Must be $lambda G=g^{-1}(G)$
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 4:25










  • $begingroup$
    Yes, you're absolutely right!
    $endgroup$
    – Sean Haight
    Dec 7 '18 at 5:58
















2












$begingroup$

First we show that if $A$ has measure zero, then $lambda A$ has measure zero.



Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:



Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see



$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see



$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$



Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.



Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Why $lambda E=g^{-1}(G)$?
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 2:24






  • 1




    $begingroup$
    Must be $lambda G=g^{-1}(G)$
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 4:25










  • $begingroup$
    Yes, you're absolutely right!
    $endgroup$
    – Sean Haight
    Dec 7 '18 at 5:58














2












2








2





$begingroup$

First we show that if $A$ has measure zero, then $lambda A$ has measure zero.



Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:



Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see



$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see



$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$



Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.



Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.






share|cite|improve this answer











$endgroup$



First we show that if $A$ has measure zero, then $lambda A$ has measure zero.



Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:



Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see



$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see



$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$



Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$



Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.



Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 5:58

























answered Dec 4 '18 at 6:11









Sean HaightSean Haight

675519




675519








  • 1




    $begingroup$
    Why $lambda E=g^{-1}(G)$?
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 2:24






  • 1




    $begingroup$
    Must be $lambda G=g^{-1}(G)$
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 4:25










  • $begingroup$
    Yes, you're absolutely right!
    $endgroup$
    – Sean Haight
    Dec 7 '18 at 5:58














  • 1




    $begingroup$
    Why $lambda E=g^{-1}(G)$?
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 2:24






  • 1




    $begingroup$
    Must be $lambda G=g^{-1}(G)$
    $endgroup$
    – eraldcoil
    Dec 7 '18 at 4:25










  • $begingroup$
    Yes, you're absolutely right!
    $endgroup$
    – Sean Haight
    Dec 7 '18 at 5:58








1




1




$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24




$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24




1




1




$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25




$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25












$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58




$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58











0












$begingroup$

If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$



Therefore $lambda E$ is measurable.



It is correct?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is $lambda G - lambda E$ measurable?
    $endgroup$
    – Sean Haight
    Dec 6 '18 at 22:35










  • $begingroup$
    Oh. Sorry. Should be $m^{ast}$ outer measure
    $endgroup$
    – eraldcoil
    Dec 6 '18 at 22:37
















0












$begingroup$

If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$



Therefore $lambda E$ is measurable.



It is correct?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is $lambda G - lambda E$ measurable?
    $endgroup$
    – Sean Haight
    Dec 6 '18 at 22:35










  • $begingroup$
    Oh. Sorry. Should be $m^{ast}$ outer measure
    $endgroup$
    – eraldcoil
    Dec 6 '18 at 22:37














0












0








0





$begingroup$

If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$



Therefore $lambda E$ is measurable.



It is correct?






share|cite|improve this answer









$endgroup$



If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$



Therefore $lambda E$ is measurable.



It is correct?







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answered Dec 6 '18 at 22:28









eraldcoileraldcoil

386211




386211












  • $begingroup$
    Why is $lambda G - lambda E$ measurable?
    $endgroup$
    – Sean Haight
    Dec 6 '18 at 22:35










  • $begingroup$
    Oh. Sorry. Should be $m^{ast}$ outer measure
    $endgroup$
    – eraldcoil
    Dec 6 '18 at 22:37


















  • $begingroup$
    Why is $lambda G - lambda E$ measurable?
    $endgroup$
    – Sean Haight
    Dec 6 '18 at 22:35










  • $begingroup$
    Oh. Sorry. Should be $m^{ast}$ outer measure
    $endgroup$
    – eraldcoil
    Dec 6 '18 at 22:37
















$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35




$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35












$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37




$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37


















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