$f circ | cdot |$ Lebesgue Integrable $iff$ g is Lebesgue Integrable
Define $f: [0, infty[ to bar{mathbb R}$ measurable
Show that:
$f circ | cdot |: mathbb R^d to bar{mathbb R}$ as a lebesgue integrable function $iff$ $g: mathbb R_{geq 0}to mathbb R, g(r):=r^{d-1}f(r)$ is lebesgue integrable
Ideas:
"$Rightarrow$" $g$ is simply $f$ scaled by a constant $r^{d-1}$ and therefore measurable and
$int_{[0,infty[}g(r)dr=int_{[0,infty[}r^{d-1}f(r)dr$
I would like to say that find a constant $c in mathbb R$ so that $|r^{d-1}|leq c$ but obviously $r in [0,infty[$ so this approach would not make sense. The only other solution that comes to mind would be substitution, so
set $|x| = r Rightarrow dx =dr$ (Is this even correct?) but here I do not get any further.
"$Leftarrow$"
$f circ |cdot|$ is by definition measurable
All I notice is that on $r in [1,infty[$: $f(r) leq g(r)$ and then using the monotonicity of integrals I could use $int_{[1,infty[}f(r)dr leq int_{[1,infty[}g(r)dr < infty$ but this still does not help me on $]0,1[$
Any help would be greatly appreciated.
real-analysis integration measure-theory lebesgue-integral
asked Jan 4 at 19:30
SABOYSABOY
508311
This question has an open bounty worth +50
reputation from SABOY ending in 3 days.
This question has not received enough attention.
add a comment |
Define $f: [0, infty[ to bar{mathbb R}$ measurable
Show that:
$f circ | cdot |: mathbb R^d to bar{mathbb R}$ as a lebesgue integrable function $iff$ $g: mathbb R_{geq 0}to mathbb R, g(r):=r^{d-1}f(r)$ is lebesgue integrable
Ideas:
"$Rightarrow$" $g$ is simply $f$ scaled by a constant $r^{d-1}$ and therefore measurable and
$int_{[0,infty[}g(r)dr=int_{[0,infty[}r^{d-1}f(r)dr$
I would like to say that find a constant $c in mathbb R$ so that $|r^{d-1}|leq c$ but obviously $r in [0,infty[$ so this approach would not make sense. The only other solution that comes to mind would be substitution, so
set $|x| = r Rightarrow dx =dr$ (Is this even correct?) but here I do not get any further.
"$Leftarrow$"
$f circ |cdot|$ is by definition measurable
All I notice is that on $r in [1,infty[$: $f(r) leq g(r)$ and then using the monotonicity of integrals I could use $int_{[1,infty[}f(r)dr leq int_{[1,infty[}g(r)dr < infty$ but this still does not help me on $]0,1[$
Any help would be greatly appreciated.
real-analysis integration measure-theory lebesgue-integral
asked Jan 4 at 19:30
SABOYSABOY
508311
This question has an open bounty worth +50
reputation from SABOY ending in 3 days.
This question has not received enough attention.
No, $f circ |cdot|: mathbb R^d to bar{mathbb R}$, while $f: [0, infty[ to bar{mathbb R}$
– SABOY
Jan 4 at 20:02
2
Do you know the "polar coordinates" formula? The case $d=2$ should be familiar.
– John Dawkins
Jan 4 at 23:05
See Rudin's RCA for polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Jan 5 at 0:03
add a comment |
Define $f: [0, infty[ to bar{mathbb R}$ measurable
Show that:
$f circ | cdot |: mathbb R^d to bar{mathbb R}$ as a lebesgue integrable function $iff$ $g: mathbb R_{geq 0}to mathbb R, g(r):=r^{d-1}f(r)$ is lebesgue integrable
Ideas:
"$Rightarrow$" $g$ is simply $f$ scaled by a constant $r^{d-1}$ and therefore measurable and
$int_{[0,infty[}g(r)dr=int_{[0,infty[}r^{d-1}f(r)dr$
I would like to say that find a constant $c in mathbb R$ so that $|r^{d-1}|leq c$ but obviously $r in [0,infty[$ so this approach would not make sense. The only other solution that comes to mind would be substitution, so
set $|x| = r Rightarrow dx =dr$ (Is this even correct?) but here I do not get any further.
"$Leftarrow$"
$f circ |cdot|$ is by definition measurable
All I notice is that on $r in [1,infty[$: $f(r) leq g(r)$ and then using the monotonicity of integrals I could use $int_{[1,infty[}f(r)dr leq int_{[1,infty[}g(r)dr < infty$ but this still does not help me on $]0,1[$
Any help would be greatly appreciated.
real-analysis integration measure-theory lebesgue-integral
asked Jan 4 at 19:30
SABOYSABOY
508311
Define $f: [0, infty[ to bar{mathbb R}$ measurable
Show that:
$f circ | cdot |: mathbb R^d to bar{mathbb R}$ as a lebesgue integrable function $iff$ $g: mathbb R_{geq 0}to mathbb R, g(r):=r^{d-1}f(r)$ is lebesgue integrable
Ideas:
"$Rightarrow$" $g$ is simply $f$ scaled by a constant $r^{d-1}$ and therefore measurable and
$int_{[0,infty[}g(r)dr=int_{[0,infty[}r^{d-1}f(r)dr$
I would like to say that find a constant $c in mathbb R$ so that $|r^{d-1}|leq c$ but obviously $r in [0,infty[$ so this approach would not make sense. The only other solution that comes to mind would be substitution, so
set $|x| = r Rightarrow dx =dr$ (Is this even correct?) but here I do not get any further.
"$Leftarrow$"
$f circ |cdot|$ is by definition measurable
All I notice is that on $r in [1,infty[$: $f(r) leq g(r)$ and then using the monotonicity of integrals I could use $int_{[1,infty[}f(r)dr leq int_{[1,infty[}g(r)dr < infty$ but this still does not help me on $]0,1[$
Any help would be greatly appreciated.
real-analysis integration measure-theory lebesgue-integral
real-analysis integration measure-theory lebesgue-integral
asked Jan 4 at 19:30
SABOYSABOY
508311
asked Jan 4 at 19:30
SABOYSABOY
508311
asked Jan 4 at 19:30
SABOYSABOY
508311
asked Jan 4 at 19:30
SABOYSABOY
508311
asked Jan 4 at 19:30
SABOYSABOY
508311
508311
This question has an open bounty worth +50
reputation from SABOY ending in 3 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from SABOY ending in 3 days.
This question has not received enough attention.
No, $f circ |cdot|: mathbb R^d to bar{mathbb R}$, while $f: [0, infty[ to bar{mathbb R}$
– SABOY
Jan 4 at 20:02
2
Do you know the "polar coordinates" formula? The case $d=2$ should be familiar.
– John Dawkins
Jan 4 at 23:05
See Rudin's RCA for polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Jan 5 at 0:03
add a comment |
No, $f circ |cdot|: mathbb R^d to bar{mathbb R}$, while $f: [0, infty[ to bar{mathbb R}$
– SABOY
Jan 4 at 20:02
2
Do you know the "polar coordinates" formula? The case $d=2$ should be familiar.
– John Dawkins
Jan 4 at 23:05
See Rudin's RCA for polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Jan 5 at 0:03
No, $f circ |cdot|: mathbb R^d to bar{mathbb R}$, while $f: [0, infty[ to bar{mathbb R}$
– SABOY
Jan 4 at 20:02
No, $f circ |cdot|: mathbb R^d to bar{mathbb R}$, while $f: [0, infty[ to bar{mathbb R}$
– SABOY
Jan 4 at 20:02
2
2
Do you know the "polar coordinates" formula? The case $d=2$ should be familiar.
– John Dawkins
Jan 4 at 23:05
Do you know the "polar coordinates" formula? The case $d=2$ should be familiar.
– John Dawkins
Jan 4 at 23:05
See Rudin's RCA for polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Jan 5 at 0:03
See Rudin's RCA for polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Jan 5 at 0:03
add a comment |
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No, $f circ |cdot|: mathbb R^d to bar{mathbb R}$, while $f: [0, infty[ to bar{mathbb R}$
– SABOY
Jan 4 at 20:02
2
Do you know the "polar coordinates" formula? The case $d=2$ should be familiar.
– John Dawkins
Jan 4 at 23:05
See Rudin's RCA for polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Jan 5 at 0:03