Fubini Thm to prove $int _{[0,infty)^n setminus [0,1]^n} frac{1}{sum_1^n x_i^{a_i}} dlambda <infty...
$begingroup$
This question have an answer which doesn't use Fibini-Tonelli here:
A problem in n-dimesnsional Lebesgue measure
The proposition states that $a_1,a_2,...,a_n >0$:
$int _{[0,infty)^n setminus [0,1]^n} frac{1}{sum_1^n x_i^{a_i}} dlambda <infty Leftrightarrow sum_1^nfrac{1}{a_i} <1$
I tried to work it out using Fubini Tonelli theorem, and by induction, by splitting the integral to 2 integral of $l,m$ dimensions such that $l+m=n$. I tried to work out how to split that $sum_1 ^{n} frac{1}{a_i}$ in order to apply the induction assumption. But I didn't succeed to find a way to do so.
I wish for a solution using Fubini theorem, I also hope it is not considered a duplicate, please let me know if so.
real-analysis lebesgue-integral
asked Jan 20 at 9:06
dandan
571613
$endgroup$
add a comment |
$begingroup$
This question have an answer which doesn't use Fibini-Tonelli here:
A problem in n-dimesnsional Lebesgue measure
The proposition states that $a_1,a_2,...,a_n >0$:
$int _{[0,infty)^n setminus [0,1]^n} frac{1}{sum_1^n x_i^{a_i}} dlambda <infty Leftrightarrow sum_1^nfrac{1}{a_i} <1$
I tried to work it out using Fubini Tonelli theorem, and by induction, by splitting the integral to 2 integral of $l,m$ dimensions such that $l+m=n$. I tried to work out how to split that $sum_1 ^{n} frac{1}{a_i}$ in order to apply the induction assumption. But I didn't succeed to find a way to do so.
I wish for a solution using Fubini theorem, I also hope it is not considered a duplicate, please let me know if so.
real-analysis lebesgue-integral
asked Jan 20 at 9:06
dandan
571613
$endgroup$
add a comment |
$begingroup$
This question have an answer which doesn't use Fibini-Tonelli here:
A problem in n-dimesnsional Lebesgue measure
The proposition states that $a_1,a_2,...,a_n >0$:
$int _{[0,infty)^n setminus [0,1]^n} frac{1}{sum_1^n x_i^{a_i}} dlambda <infty Leftrightarrow sum_1^nfrac{1}{a_i} <1$
I tried to work it out using Fubini Tonelli theorem, and by induction, by splitting the integral to 2 integral of $l,m$ dimensions such that $l+m=n$. I tried to work out how to split that $sum_1 ^{n} frac{1}{a_i}$ in order to apply the induction assumption. But I didn't succeed to find a way to do so.
I wish for a solution using Fubini theorem, I also hope it is not considered a duplicate, please let me know if so.
real-analysis lebesgue-integral
asked Jan 20 at 9:06
dandan
571613
$endgroup$
This question have an answer which doesn't use Fibini-Tonelli here:
A problem in n-dimesnsional Lebesgue measure
The proposition states that $a_1,a_2,...,a_n >0$:
$int _{[0,infty)^n setminus [0,1]^n} frac{1}{sum_1^n x_i^{a_i}} dlambda <infty Leftrightarrow sum_1^nfrac{1}{a_i} <1$
I tried to work it out using Fubini Tonelli theorem, and by induction, by splitting the integral to 2 integral of $l,m$ dimensions such that $l+m=n$. I tried to work out how to split that $sum_1 ^{n} frac{1}{a_i}$ in order to apply the induction assumption. But I didn't succeed to find a way to do so.
I wish for a solution using Fubini theorem, I also hope it is not considered a duplicate, please let me know if so.
real-analysis lebesgue-integral
real-analysis lebesgue-integral
asked Jan 20 at 9:06
dandan
571613
asked Jan 20 at 9:06
dandan
571613
edited Jan 20 at 16:37
dan
asked Jan 20 at 9:06
dandan
571613
asked Jan 20 at 9:06
dandan
571613
asked Jan 20 at 9:06
dandan
571613
571613
add a comment |
add a comment |
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