Mixing
Mass of $ A := { x in R^2 | x<y , y1/x }$
$begingroup$
How does one calculate the Lebesgue Measure of the above set?
I tried the following:
$ lambda^2(A) = int_{R^2} 1_A d(lambda^2(x,y)) = int_Rint_x^2 1_{ {1/x < y} } dlambda(y)dlambda(x)$
but then I have no idea how to reformulate the inequality $1/x < y $.
integration measure-theory lebesgue-measure
asked Jan 30 at 13:23
SlyderSlyder
316
$endgroup$
add a comment |
$begingroup$
How does one calculate the Lebesgue Measure of the above set?
I tried the following:
$ lambda^2(A) = int_{R^2} 1_A d(lambda^2(x,y)) = int_Rint_x^2 1_{ {1/x < y} } dlambda(y)dlambda(x)$
but then I have no idea how to reformulate the inequality $1/x < y $.
integration measure-theory lebesgue-measure
asked Jan 30 at 13:23
SlyderSlyder
316
$endgroup$
$begingroup$
Your set contains ${(x,y) in mathbb{R}^2 : x < 0, 0 < y < 2}$, which has infinite measure. Therefore there is either a mistake in the question or the answer is obviously $infty$.
$endgroup$
– Klaus
Jan 30 at 13:31
$begingroup$
Perhaps the OP meant to include an additional constraint - maybe something like $x > 0$?
$endgroup$
– Jordan Green
Jan 30 at 17:01
add a comment |
$begingroup$
How does one calculate the Lebesgue Measure of the above set?
I tried the following:
$ lambda^2(A) = int_{R^2} 1_A d(lambda^2(x,y)) = int_Rint_x^2 1_{ {1/x < y} } dlambda(y)dlambda(x)$
but then I have no idea how to reformulate the inequality $1/x < y $.
integration measure-theory lebesgue-measure
asked Jan 30 at 13:23
SlyderSlyder
316
$endgroup$
How does one calculate the Lebesgue Measure of the above set?
I tried the following:
$ lambda^2(A) = int_{R^2} 1_A d(lambda^2(x,y)) = int_Rint_x^2 1_{ {1/x < y} } dlambda(y)dlambda(x)$
but then I have no idea how to reformulate the inequality $1/x < y $.
integration measure-theory lebesgue-measure
integration measure-theory lebesgue-measure
asked Jan 30 at 13:23
SlyderSlyder
316
asked Jan 30 at 13:23
SlyderSlyder
316
edited Jan 30 at 13:29
Slyder
asked Jan 30 at 13:23
SlyderSlyder
316
asked Jan 30 at 13:23
SlyderSlyder
316
asked Jan 30 at 13:23
SlyderSlyder
316
316
$begingroup$
Your set contains ${(x,y) in mathbb{R}^2 : x < 0, 0 < y < 2}$, which has infinite measure. Therefore there is either a mistake in the question or the answer is obviously $infty$.
$endgroup$
– Klaus
Jan 30 at 13:31
$begingroup$
Perhaps the OP meant to include an additional constraint - maybe something like $x > 0$?
$endgroup$
– Jordan Green
Jan 30 at 17:01
add a comment |
$begingroup$
Your set contains ${(x,y) in mathbb{R}^2 : x < 0, 0 < y < 2}$, which has infinite measure. Therefore there is either a mistake in the question or the answer is obviously $infty$.
$endgroup$
– Klaus
Jan 30 at 13:31
$begingroup$
Perhaps the OP meant to include an additional constraint - maybe something like $x > 0$?
$endgroup$
– Jordan Green
Jan 30 at 17:01
$begingroup$
Your set contains ${(x,y) in mathbb{R}^2 : x < 0, 0 < y < 2}$, which has infinite measure. Therefore there is either a mistake in the question or the answer is obviously $infty$.
$endgroup$
– Klaus
Jan 30 at 13:31
$begingroup$
Your set contains ${(x,y) in mathbb{R}^2 : x < 0, 0 < y < 2}$, which has infinite measure. Therefore there is either a mistake in the question or the answer is obviously $infty$.
$endgroup$
– Klaus
Jan 30 at 13:31
$begingroup$
Perhaps the OP meant to include an additional constraint - maybe something like $x > 0$?
$endgroup$
– Jordan Green
Jan 30 at 17:01
$begingroup$
Perhaps the OP meant to include an additional constraint - maybe something like $x > 0$?
$endgroup$
– Jordan Green
Jan 30 at 17:01
add a comment |
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$begingroup$
Your set contains ${(x,y) in mathbb{R}^2 : x < 0, 0 < y < 2}$, which has infinite measure. Therefore there is either a mistake in the question or the answer is obviously $infty$.
$endgroup$
– Klaus
Jan 30 at 13:31
$begingroup$
Perhaps the OP meant to include an additional constraint - maybe something like $x > 0$?
$endgroup$
– Jordan Green
Jan 30 at 17:01