$f circ | cdot |$ Lebesgue Integrable $iff$ g is Lebesgue Integrable












1














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







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This question has an open bounty worth +50
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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
















1














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







share|cite|improve this question















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














1












1








1







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










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