Mixing
Show $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$
$begingroup$
Let $alpha neq 0$
Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:
Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$
Ideas:
Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore
$int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:
$int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$
and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$
But this leads nowhere...
Any ideas?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
$endgroup$
add a comment |
$begingroup$
Let $alpha neq 0$
Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:
Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$
Ideas:
Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore
$int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:
$int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$
and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$
But this leads nowhere...
Any ideas?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
$endgroup$
add a comment |
$begingroup$
Let $alpha neq 0$
Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:
Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$
Ideas:
Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore
$int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:
$int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$
and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$
But this leads nowhere...
Any ideas?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
$endgroup$
Let $alpha neq 0$
Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:
Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$
Ideas:
Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore
$int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:
$int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$
and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$
But this leads nowhere...
Any ideas?
real-analysis measure-theory lebesgue-integral lebesgue-measure
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
asked Jan 13 at 19:43
MinaThumaMinaThuma
1508
1508
add a comment |
add a comment |
1 Answer
1
active
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votes
$begingroup$
You made a small mistake when integrating the indicator function.
$$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$
answered Jan 13 at 19:49
angryavianangryavian
41k23380
$endgroup$
$begingroup$
Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
$endgroup$
– MinaThuma
Jan 13 at 19:56
2
$begingroup$
@MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
$endgroup$
– angryavian
Jan 13 at 19:59
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You made a small mistake when integrating the indicator function.
$$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$
answered Jan 13 at 19:49
angryavianangryavian
41k23380
$endgroup$
$begingroup$
Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
$endgroup$
– MinaThuma
Jan 13 at 19:56
2
$begingroup$
@MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
$endgroup$
– angryavian
Jan 13 at 19:59
add a comment |
$begingroup$
You made a small mistake when integrating the indicator function.
$$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$
answered Jan 13 at 19:49
angryavianangryavian
41k23380
$endgroup$
$begingroup$
Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
$endgroup$
– MinaThuma
Jan 13 at 19:56
2
$begingroup$
@MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
$endgroup$
– angryavian
Jan 13 at 19:59
add a comment |
$begingroup$
You made a small mistake when integrating the indicator function.
$$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$
answered Jan 13 at 19:49
angryavianangryavian
41k23380
$endgroup$
You made a small mistake when integrating the indicator function.
$$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$
answered Jan 13 at 19:49
angryavianangryavian
41k23380
answered Jan 13 at 19:49
angryavianangryavian
41k23380
answered Jan 13 at 19:49
angryavianangryavian
41k23380
answered Jan 13 at 19:49
angryavianangryavian
41k23380
41k23380
$begingroup$
Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
$endgroup$
– MinaThuma
Jan 13 at 19:56
2
$begingroup$
@MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
$endgroup$
– angryavian
Jan 13 at 19:59
add a comment |
$begingroup$
Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
$endgroup$
– MinaThuma
Jan 13 at 19:56
2
$begingroup$
@MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
$endgroup$
– angryavian
Jan 13 at 19:59
$begingroup$
Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
$endgroup$
– MinaThuma
Jan 13 at 19:56
$begingroup$
Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
$endgroup$
– MinaThuma
Jan 13 at 19:56
2
2
$begingroup$
@MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
$endgroup$
– angryavian
Jan 13 at 19:59
$begingroup$
@MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
$endgroup$
– angryavian
Jan 13 at 19:59
add a comment |
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