Hints on $lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$
$begingroup$
Let $B in mathcal{B}^{d}$ and $alphaneq 0$
Any ideas on how to show:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$
My idea using "simple sets":
Let $B:=[a_{1},b_{1}[times...times[a_{d},b_{d}[$ this means:
Note without loss of generality $alpha > 0$ otherwise switch intervals $[a,b[$ to $[alpha b,alpha a[$
$alpha B=[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[$
Therefore $lambda^{d}(alpha B)=prod_{i=1}^{d}(alpha b_{i}-alpha a_{i})=alpha^{d}prod_{i=1}^{d}( b_{i}- a_{i})=alphalambda^{d}(B)$
Since ${[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[}$ is a $cap-$stable generator of $mathcal{B}^{d}$ it follows:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$ for any $B in mathcal{B}^{d}$
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
$endgroup$
add a comment |
$begingroup$
Let $B in mathcal{B}^{d}$ and $alphaneq 0$
Any ideas on how to show:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$
My idea using "simple sets":
Let $B:=[a_{1},b_{1}[times...times[a_{d},b_{d}[$ this means:
Note without loss of generality $alpha > 0$ otherwise switch intervals $[a,b[$ to $[alpha b,alpha a[$
$alpha B=[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[$
Therefore $lambda^{d}(alpha B)=prod_{i=1}^{d}(alpha b_{i}-alpha a_{i})=alpha^{d}prod_{i=1}^{d}( b_{i}- a_{i})=alphalambda^{d}(B)$
Since ${[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[}$ is a $cap-$stable generator of $mathcal{B}^{d}$ it follows:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$ for any $B in mathcal{B}^{d}$
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
$endgroup$
$begingroup$
Hint: Show that the result holds for simple sets, then consider the class of sets such that it holds.
$endgroup$
– Did
Jan 13 at 17:57
$begingroup$
@Did So $lambda^{d}(alpha B)=int_{mathbb R^{d}}chi_{alpha B}dlambda^{d}$ does not help? I am confused as to how to use your hint
$endgroup$
– MinaThuma
Jan 13 at 18:43
$begingroup$
No it does not. Re my hint, it has two parts. Let me suggest to first concentrate on the first part, and then to turn to the second part.
$endgroup$
– Did
Jan 13 at 18:45
$begingroup$
@Did I have edited my question, is it correct?
$endgroup$
– MinaThuma
Jan 13 at 19:01
$begingroup$
Yes, although you might be more explicit about how you deduce the result for the class of Borel sets from the result for the smaller class.
$endgroup$
– Did
Jan 13 at 19:03
add a comment |
$begingroup$
Let $B in mathcal{B}^{d}$ and $alphaneq 0$
Any ideas on how to show:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$
My idea using "simple sets":
Let $B:=[a_{1},b_{1}[times...times[a_{d},b_{d}[$ this means:
Note without loss of generality $alpha > 0$ otherwise switch intervals $[a,b[$ to $[alpha b,alpha a[$
$alpha B=[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[$
Therefore $lambda^{d}(alpha B)=prod_{i=1}^{d}(alpha b_{i}-alpha a_{i})=alpha^{d}prod_{i=1}^{d}( b_{i}- a_{i})=alphalambda^{d}(B)$
Since ${[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[}$ is a $cap-$stable generator of $mathcal{B}^{d}$ it follows:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$ for any $B in mathcal{B}^{d}$
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
$endgroup$
Let $B in mathcal{B}^{d}$ and $alphaneq 0$
Any ideas on how to show:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$
My idea using "simple sets":
Let $B:=[a_{1},b_{1}[times...times[a_{d},b_{d}[$ this means:
Note without loss of generality $alpha > 0$ otherwise switch intervals $[a,b[$ to $[alpha b,alpha a[$
$alpha B=[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[$
Therefore $lambda^{d}(alpha B)=prod_{i=1}^{d}(alpha b_{i}-alpha a_{i})=alpha^{d}prod_{i=1}^{d}( b_{i}- a_{i})=alphalambda^{d}(B)$
Since ${[alpha a_{1},alpha b_{1}[times...times[alpha a_{d},alpha b_{d}[}$ is a $cap-$stable generator of $mathcal{B}^{d}$ it follows:
$lambda^{d}(alpha B)=|alpha|^dlambda^{d}(B)$ for any $B in mathcal{B}^{d}$
real-analysis measure-theory lebesgue-integral lebesgue-measure
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
edited Jan 13 at 19:00
MinaThuma
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
asked Jan 13 at 17:51
MinaThumaMinaThuma
27010
27010
$begingroup$
Hint: Show that the result holds for simple sets, then consider the class of sets such that it holds.
$endgroup$
– Did
Jan 13 at 17:57
$begingroup$
@Did So $lambda^{d}(alpha B)=int_{mathbb R^{d}}chi_{alpha B}dlambda^{d}$ does not help? I am confused as to how to use your hint
$endgroup$
– MinaThuma
Jan 13 at 18:43
$begingroup$
No it does not. Re my hint, it has two parts. Let me suggest to first concentrate on the first part, and then to turn to the second part.
$endgroup$
– Did
Jan 13 at 18:45
$begingroup$
@Did I have edited my question, is it correct?
$endgroup$
– MinaThuma
Jan 13 at 19:01
$begingroup$
Yes, although you might be more explicit about how you deduce the result for the class of Borel sets from the result for the smaller class.
$endgroup$
– Did
Jan 13 at 19:03
add a comment |
$begingroup$
Hint: Show that the result holds for simple sets, then consider the class of sets such that it holds.
$endgroup$
– Did
Jan 13 at 17:57
$begingroup$
@Did So $lambda^{d}(alpha B)=int_{mathbb R^{d}}chi_{alpha B}dlambda^{d}$ does not help? I am confused as to how to use your hint
$endgroup$
– MinaThuma
Jan 13 at 18:43
$begingroup$
No it does not. Re my hint, it has two parts. Let me suggest to first concentrate on the first part, and then to turn to the second part.
$endgroup$
– Did
Jan 13 at 18:45
$begingroup$
@Did I have edited my question, is it correct?
$endgroup$
– MinaThuma
Jan 13 at 19:01
$begingroup$
Yes, although you might be more explicit about how you deduce the result for the class of Borel sets from the result for the smaller class.
$endgroup$
– Did
Jan 13 at 19:03
$begingroup$
Hint: Show that the result holds for simple sets, then consider the class of sets such that it holds.
$endgroup$
– Did
Jan 13 at 17:57
$begingroup$
Hint: Show that the result holds for simple sets, then consider the class of sets such that it holds.
$endgroup$
– Did
Jan 13 at 17:57
$begingroup$
@Did So $lambda^{d}(alpha B)=int_{mathbb R^{d}}chi_{alpha B}dlambda^{d}$ does not help? I am confused as to how to use your hint
$endgroup$
– MinaThuma
Jan 13 at 18:43
$begingroup$
@Did So $lambda^{d}(alpha B)=int_{mathbb R^{d}}chi_{alpha B}dlambda^{d}$ does not help? I am confused as to how to use your hint
$endgroup$
– MinaThuma
Jan 13 at 18:43
$begingroup$
No it does not. Re my hint, it has two parts. Let me suggest to first concentrate on the first part, and then to turn to the second part.
$endgroup$
– Did
Jan 13 at 18:45
$begingroup$
No it does not. Re my hint, it has two parts. Let me suggest to first concentrate on the first part, and then to turn to the second part.
$endgroup$
– Did
Jan 13 at 18:45
$begingroup$
@Did I have edited my question, is it correct?
$endgroup$
– MinaThuma
Jan 13 at 19:01
$begingroup$
@Did I have edited my question, is it correct?
$endgroup$
– MinaThuma
Jan 13 at 19:01
$begingroup$
Yes, although you might be more explicit about how you deduce the result for the class of Borel sets from the result for the smaller class.
$endgroup$
– Did
Jan 13 at 19:03
$begingroup$
Yes, although you might be more explicit about how you deduce the result for the class of Borel sets from the result for the smaller class.
$endgroup$
– Did
Jan 13 at 19:03
add a comment |
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$begingroup$
Hint: Show that the result holds for simple sets, then consider the class of sets such that it holds.
$endgroup$
– Did
Jan 13 at 17:57
$begingroup$
@Did So $lambda^{d}(alpha B)=int_{mathbb R^{d}}chi_{alpha B}dlambda^{d}$ does not help? I am confused as to how to use your hint
$endgroup$
– MinaThuma
Jan 13 at 18:43
$begingroup$
No it does not. Re my hint, it has two parts. Let me suggest to first concentrate on the first part, and then to turn to the second part.
$endgroup$
– Did
Jan 13 at 18:45
$begingroup$
@Did I have edited my question, is it correct?
$endgroup$
– MinaThuma
Jan 13 at 19:01
$begingroup$
Yes, although you might be more explicit about how you deduce the result for the class of Borel sets from the result for the smaller class.
$endgroup$
– Did
Jan 13 at 19:03