Mixing
Rewriting an integral of an absolutely continuous function
$begingroup$
Let $(X,M, mu)$ be a $sigma$-finite measure space and $f: X rightarrow [0, infty)$ be measurable. For each $alpha ge 0$ Define $E_alpha = { x in X : f(x) > alpha }$ and $lambda(alpha) = mu(E_alpha)$.
Suppose that $phi : [0, infty) rightarrow [0,infty)$ is an increasing function which is absolutely continuous on [0,T] for every T $in (0, infty)$. Prove that for each $beta ge 0$,
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x) = int_beta^infty phi '(alpha)lambda(alpha) dalpha$.
What I have so far:
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x)=int_{E_beta} phi(beta)-int_beta^{f(x)} phi ' (t) dt-phi(beta)dmu(x)$ because $phi$ is absolutely continuous.
$=-int_{E_beta}int_beta^{f(x)}phi ' (t)dt dmu(x) = int_beta^inftyint_{E_t} phi ' (t) dmu dt$ (I'm not really sure why this is true?? Fubini-Tonelli allows us to switch the integrals, but I don't understand why the bounds change)
$=int_beta^inftyphi ' (t) mu(E_t)dt=int_beta^infty phi '(t)lambda(t)dt$
Could someone please explain how the bounds change? Thanks!
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
asked Jan 6 at 20:24
Math LadyMath Lady
1196
$endgroup$
add a comment |
$begingroup$
Let $(X,M, mu)$ be a $sigma$-finite measure space and $f: X rightarrow [0, infty)$ be measurable. For each $alpha ge 0$ Define $E_alpha = { x in X : f(x) > alpha }$ and $lambda(alpha) = mu(E_alpha)$.
Suppose that $phi : [0, infty) rightarrow [0,infty)$ is an increasing function which is absolutely continuous on [0,T] for every T $in (0, infty)$. Prove that for each $beta ge 0$,
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x) = int_beta^infty phi '(alpha)lambda(alpha) dalpha$.
What I have so far:
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x)=int_{E_beta} phi(beta)-int_beta^{f(x)} phi ' (t) dt-phi(beta)dmu(x)$ because $phi$ is absolutely continuous.
$=-int_{E_beta}int_beta^{f(x)}phi ' (t)dt dmu(x) = int_beta^inftyint_{E_t} phi ' (t) dmu dt$ (I'm not really sure why this is true?? Fubini-Tonelli allows us to switch the integrals, but I don't understand why the bounds change)
$=int_beta^inftyphi ' (t) mu(E_t)dt=int_beta^infty phi '(t)lambda(t)dt$
Could someone please explain how the bounds change? Thanks!
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
asked Jan 6 at 20:24
Math LadyMath Lady
1196
$endgroup$
add a comment |
$begingroup$
Let $(X,M, mu)$ be a $sigma$-finite measure space and $f: X rightarrow [0, infty)$ be measurable. For each $alpha ge 0$ Define $E_alpha = { x in X : f(x) > alpha }$ and $lambda(alpha) = mu(E_alpha)$.
Suppose that $phi : [0, infty) rightarrow [0,infty)$ is an increasing function which is absolutely continuous on [0,T] for every T $in (0, infty)$. Prove that for each $beta ge 0$,
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x) = int_beta^infty phi '(alpha)lambda(alpha) dalpha$.
What I have so far:
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x)=int_{E_beta} phi(beta)-int_beta^{f(x)} phi ' (t) dt-phi(beta)dmu(x)$ because $phi$ is absolutely continuous.
$=-int_{E_beta}int_beta^{f(x)}phi ' (t)dt dmu(x) = int_beta^inftyint_{E_t} phi ' (t) dmu dt$ (I'm not really sure why this is true?? Fubini-Tonelli allows us to switch the integrals, but I don't understand why the bounds change)
$=int_beta^inftyphi ' (t) mu(E_t)dt=int_beta^infty phi '(t)lambda(t)dt$
Could someone please explain how the bounds change? Thanks!
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
asked Jan 6 at 20:24
Math LadyMath Lady
1196
$endgroup$
Let $(X,M, mu)$ be a $sigma$-finite measure space and $f: X rightarrow [0, infty)$ be measurable. For each $alpha ge 0$ Define $E_alpha = { x in X : f(x) > alpha }$ and $lambda(alpha) = mu(E_alpha)$.
Suppose that $phi : [0, infty) rightarrow [0,infty)$ is an increasing function which is absolutely continuous on [0,T] for every T $in (0, infty)$. Prove that for each $beta ge 0$,
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x) = int_beta^infty phi '(alpha)lambda(alpha) dalpha$.
What I have so far:
$int_{E_beta}(phi(f(x))-phi(beta))dmu(x)=int_{E_beta} phi(beta)-int_beta^{f(x)} phi ' (t) dt-phi(beta)dmu(x)$ because $phi$ is absolutely continuous.
$=-int_{E_beta}int_beta^{f(x)}phi ' (t)dt dmu(x) = int_beta^inftyint_{E_t} phi ' (t) dmu dt$ (I'm not really sure why this is true?? Fubini-Tonelli allows us to switch the integrals, but I don't understand why the bounds change)
$=int_beta^inftyphi ' (t) mu(E_t)dt=int_beta^infty phi '(t)lambda(t)dt$
Could someone please explain how the bounds change? Thanks!
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
asked Jan 6 at 20:24
Math LadyMath Lady
1196
asked Jan 6 at 20:24
Math LadyMath Lady
1196
asked Jan 6 at 20:24
Math LadyMath Lady
1196
asked Jan 6 at 20:24
Math LadyMath Lady
1196
asked Jan 6 at 20:24
Math LadyMath Lady
1196
1196
add a comment |
add a comment |
2 Answers
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$begingroup$
We can write
$$begin{eqnarray}
int_{E_beta}(phi(f(x))-phi(beta))dmu(x) &=&int_{E_beta}left(int_beta^{f(x)}phi'(t)dtright)dmu(x) \
&=&int_{E_beta}int_{beta<t<f(x)}phi'(t)dtdmu(x) \
&=&int_{t>beta}int_{E_betabigcap {t<f(x)}}phi'(t)dmu(x)dt\
&=&int_{t>beta}left(int_{E_t} 1dmu(x)right)phi'(t)dt\
&=&int_beta^infty phi'(t)lambda(t)dt,
end{eqnarray}$$ by Tonelli's theorem.
answered Jan 6 at 21:04
SongSong
10.4k627
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add a comment |
$begingroup$
$f(x) >beta$ and $beta <t<f(x)$ iff $beta <t<infty$ and $f(x) >t$. Hence $xin E_{beta}$ and $beta <t <f(x)$ iff $t >beta$ and $x in E_t$.
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
We can write
$$begin{eqnarray}
int_{E_beta}(phi(f(x))-phi(beta))dmu(x) &=&int_{E_beta}left(int_beta^{f(x)}phi'(t)dtright)dmu(x) \
&=&int_{E_beta}int_{beta<t<f(x)}phi'(t)dtdmu(x) \
&=&int_{t>beta}int_{E_betabigcap {t<f(x)}}phi'(t)dmu(x)dt\
&=&int_{t>beta}left(int_{E_t} 1dmu(x)right)phi'(t)dt\
&=&int_beta^infty phi'(t)lambda(t)dt,
end{eqnarray}$$ by Tonelli's theorem.
answered Jan 6 at 21:04
SongSong
10.4k627
$endgroup$
add a comment |
$begingroup$
We can write
$$begin{eqnarray}
int_{E_beta}(phi(f(x))-phi(beta))dmu(x) &=&int_{E_beta}left(int_beta^{f(x)}phi'(t)dtright)dmu(x) \
&=&int_{E_beta}int_{beta<t<f(x)}phi'(t)dtdmu(x) \
&=&int_{t>beta}int_{E_betabigcap {t<f(x)}}phi'(t)dmu(x)dt\
&=&int_{t>beta}left(int_{E_t} 1dmu(x)right)phi'(t)dt\
&=&int_beta^infty phi'(t)lambda(t)dt,
end{eqnarray}$$ by Tonelli's theorem.
answered Jan 6 at 21:04
SongSong
10.4k627
$endgroup$
add a comment |
$begingroup$
We can write
$$begin{eqnarray}
int_{E_beta}(phi(f(x))-phi(beta))dmu(x) &=&int_{E_beta}left(int_beta^{f(x)}phi'(t)dtright)dmu(x) \
&=&int_{E_beta}int_{beta<t<f(x)}phi'(t)dtdmu(x) \
&=&int_{t>beta}int_{E_betabigcap {t<f(x)}}phi'(t)dmu(x)dt\
&=&int_{t>beta}left(int_{E_t} 1dmu(x)right)phi'(t)dt\
&=&int_beta^infty phi'(t)lambda(t)dt,
end{eqnarray}$$ by Tonelli's theorem.
answered Jan 6 at 21:04
SongSong
10.4k627
$endgroup$
We can write
$$begin{eqnarray}
int_{E_beta}(phi(f(x))-phi(beta))dmu(x) &=&int_{E_beta}left(int_beta^{f(x)}phi'(t)dtright)dmu(x) \
&=&int_{E_beta}int_{beta<t<f(x)}phi'(t)dtdmu(x) \
&=&int_{t>beta}int_{E_betabigcap {t<f(x)}}phi'(t)dmu(x)dt\
&=&int_{t>beta}left(int_{E_t} 1dmu(x)right)phi'(t)dt\
&=&int_beta^infty phi'(t)lambda(t)dt,
end{eqnarray}$$ by Tonelli's theorem.
answered Jan 6 at 21:04
SongSong
10.4k627
answered Jan 6 at 21:04
SongSong
10.4k627
answered Jan 6 at 21:04
SongSong
10.4k627
answered Jan 6 at 21:04
SongSong
10.4k627
10.4k627
add a comment |
add a comment |
$begingroup$
$f(x) >beta$ and $beta <t<f(x)$ iff $beta <t<infty$ and $f(x) >t$. Hence $xin E_{beta}$ and $beta <t <f(x)$ iff $t >beta$ and $x in E_t$.
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
$f(x) >beta$ and $beta <t<f(x)$ iff $beta <t<infty$ and $f(x) >t$. Hence $xin E_{beta}$ and $beta <t <f(x)$ iff $t >beta$ and $x in E_t$.
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
$f(x) >beta$ and $beta <t<f(x)$ iff $beta <t<infty$ and $f(x) >t$. Hence $xin E_{beta}$ and $beta <t <f(x)$ iff $t >beta$ and $x in E_t$.
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$f(x) >beta$ and $beta <t<f(x)$ iff $beta <t<infty$ and $f(x) >t$. Hence $xin E_{beta}$ and $beta <t <f(x)$ iff $t >beta$ and $x in E_t$.
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 6 at 23:52
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
add a comment |
add a comment |
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