Lebesgue Integral of $e^x$












1












$begingroup$


I want to show that that $f_{alpha}: ]0,infty[ to mathbb{R}$, where $x to e^{-alpha x}$ for any $alpha gt 0, alpha in mathbb{R}$ is Lebesgue integrable, i.e. $$int_{]0,infty[}abs(f_{alpha})dlambda ltinfty$$
Since I am not allowed to calculate the integral with Riemannian methods, I want to write out the integral expliclitly. For that I need a sequence of simple functions converging to $f_{alpha}$, something I cannot find. Otherwise, I could write $e^{-alpha x}$ as a composition of $e^x$ and $-alpha x$ and then use an according theorem, but there I would still need to calculate $e^x$. So this should look like $$int_{]0,infty[}f_{alpha}dlambda=sup_{kinmathbb{N}}int_{]0,infty[}f_{alpha,k}dlambda=sup_{kinmathbb{N}}left(sum_{i=0}^Nbeta_ilambda(A_i)right)_{alpha, k}$$with $lambda$ being the ordinary Borel-Lebesgue measure. I am now looking for $f_{alpha, k}=sum_{i=0}^Nbeta_i 1_{A_i}$. Any help greatly apppreciated!






real-analysis measure-theory







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  • $begingroup$
    $A_i$ being of course subsets of $]0,infty[$ with $bigcup_{iinmathbb{N}}A_i=]0,infty[$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:32


















1












$begingroup$


I want to show that that $f_{alpha}: ]0,infty[ to mathbb{R}$, where $x to e^{-alpha x}$ for any $alpha gt 0, alpha in mathbb{R}$ is Lebesgue integrable, i.e. $$int_{]0,infty[}abs(f_{alpha})dlambda ltinfty$$
Since I am not allowed to calculate the integral with Riemannian methods, I want to write out the integral expliclitly. For that I need a sequence of simple functions converging to $f_{alpha}$, something I cannot find. Otherwise, I could write $e^{-alpha x}$ as a composition of $e^x$ and $-alpha x$ and then use an according theorem, but there I would still need to calculate $e^x$. So this should look like $$int_{]0,infty[}f_{alpha}dlambda=sup_{kinmathbb{N}}int_{]0,infty[}f_{alpha,k}dlambda=sup_{kinmathbb{N}}left(sum_{i=0}^Nbeta_ilambda(A_i)right)_{alpha, k}$$with $lambda$ being the ordinary Borel-Lebesgue measure. I am now looking for $f_{alpha, k}=sum_{i=0}^Nbeta_i 1_{A_i}$. Any help greatly apppreciated!






real-analysis measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    $A_i$ being of course subsets of $]0,infty[$ with $bigcup_{iinmathbb{N}}A_i=]0,infty[$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:32
















1












1








1


2



$begingroup$


I want to show that that $f_{alpha}: ]0,infty[ to mathbb{R}$, where $x to e^{-alpha x}$ for any $alpha gt 0, alpha in mathbb{R}$ is Lebesgue integrable, i.e. $$int_{]0,infty[}abs(f_{alpha})dlambda ltinfty$$
Since I am not allowed to calculate the integral with Riemannian methods, I want to write out the integral expliclitly. For that I need a sequence of simple functions converging to $f_{alpha}$, something I cannot find. Otherwise, I could write $e^{-alpha x}$ as a composition of $e^x$ and $-alpha x$ and then use an according theorem, but there I would still need to calculate $e^x$. So this should look like $$int_{]0,infty[}f_{alpha}dlambda=sup_{kinmathbb{N}}int_{]0,infty[}f_{alpha,k}dlambda=sup_{kinmathbb{N}}left(sum_{i=0}^Nbeta_ilambda(A_i)right)_{alpha, k}$$with $lambda$ being the ordinary Borel-Lebesgue measure. I am now looking for $f_{alpha, k}=sum_{i=0}^Nbeta_i 1_{A_i}$. Any help greatly apppreciated!






real-analysis measure-theory







share|cite|improve this question









$endgroup$




I want to show that that $f_{alpha}: ]0,infty[ to mathbb{R}$, where $x to e^{-alpha x}$ for any $alpha gt 0, alpha in mathbb{R}$ is Lebesgue integrable, i.e. $$int_{]0,infty[}abs(f_{alpha})dlambda ltinfty$$
Since I am not allowed to calculate the integral with Riemannian methods, I want to write out the integral expliclitly. For that I need a sequence of simple functions converging to $f_{alpha}$, something I cannot find. Otherwise, I could write $e^{-alpha x}$ as a composition of $e^x$ and $-alpha x$ and then use an according theorem, but there I would still need to calculate $e^x$. So this should look like $$int_{]0,infty[}f_{alpha}dlambda=sup_{kinmathbb{N}}int_{]0,infty[}f_{alpha,k}dlambda=sup_{kinmathbb{N}}left(sum_{i=0}^Nbeta_ilambda(A_i)right)_{alpha, k}$$with $lambda$ being the ordinary Borel-Lebesgue measure. I am now looking for $f_{alpha, k}=sum_{i=0}^Nbeta_i 1_{A_i}$. Any help greatly apppreciated!






real-analysis measure-theory




real-analysis measure-theory






share|cite|improve this question













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asked Jan 15 at 8:30









Michael MaierMichael Maier

859




859












  • $begingroup$
    $A_i$ being of course subsets of $]0,infty[$ with $bigcup_{iinmathbb{N}}A_i=]0,infty[$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:32




















  • $begingroup$
    $A_i$ being of course subsets of $]0,infty[$ with $bigcup_{iinmathbb{N}}A_i=]0,infty[$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:32


















$begingroup$
$A_i$ being of course subsets of $]0,infty[$ with $bigcup_{iinmathbb{N}}A_i=]0,infty[$
$endgroup$
– Michael Maier
Jan 15 at 8:32






$begingroup$
$A_i$ being of course subsets of $]0,infty[$ with $bigcup_{iinmathbb{N}}A_i=]0,infty[$
$endgroup$
– Michael Maier
Jan 15 at 8:32












1 Answer
1






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$begingroup$

Take $f_{alpha,k} =sumlimits_{j=1}^{k} frac 1 {j^{2}} I_{A_j}$ where $A_j=(frac {2 ln , j} {alpha},frac {2 ln , (j+1)} {alpha})$. Note that if $x in A_j$ then $e^{-alpha, x}$ lies between $frac 1 {(j+1)^{2}}$ and $frac 1 {j^{2}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    One question: If I want $f_{alpha, k} to f_{alpha}$, following your proposal, let $ktoinfty$, hence $f_{alpha, k} to sum_{j=1}^infty frac{1}{j^2}1_{A_j}$. I don't quite see how this approaches $e^{-alpha x} = sum_{j=1}^infty frac{1}{j!}(-alpha x)^j$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:57








  • 1




    $begingroup$
    @MichaelMaier Any $x$ lies in one of the sets $A_j$. For that $j$, $|e^{-alpha, x} -f_{alpha,k} (x)| leq frac 1 {j^{2}} -frac 1 {(j+1)^{2}}$ because if to numbers are in an interval $(a,b)$ then their difference is less than or equal to $b-a$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 9:02













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1 Answer
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1 Answer
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active

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2












$begingroup$

Take $f_{alpha,k} =sumlimits_{j=1}^{k} frac 1 {j^{2}} I_{A_j}$ where $A_j=(frac {2 ln , j} {alpha},frac {2 ln , (j+1)} {alpha})$. Note that if $x in A_j$ then $e^{-alpha, x}$ lies between $frac 1 {(j+1)^{2}}$ and $frac 1 {j^{2}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    One question: If I want $f_{alpha, k} to f_{alpha}$, following your proposal, let $ktoinfty$, hence $f_{alpha, k} to sum_{j=1}^infty frac{1}{j^2}1_{A_j}$. I don't quite see how this approaches $e^{-alpha x} = sum_{j=1}^infty frac{1}{j!}(-alpha x)^j$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:57








  • 1




    $begingroup$
    @MichaelMaier Any $x$ lies in one of the sets $A_j$. For that $j$, $|e^{-alpha, x} -f_{alpha,k} (x)| leq frac 1 {j^{2}} -frac 1 {(j+1)^{2}}$ because if to numbers are in an interval $(a,b)$ then their difference is less than or equal to $b-a$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 9:02


















2












$begingroup$

Take $f_{alpha,k} =sumlimits_{j=1}^{k} frac 1 {j^{2}} I_{A_j}$ where $A_j=(frac {2 ln , j} {alpha},frac {2 ln , (j+1)} {alpha})$. Note that if $x in A_j$ then $e^{-alpha, x}$ lies between $frac 1 {(j+1)^{2}}$ and $frac 1 {j^{2}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    One question: If I want $f_{alpha, k} to f_{alpha}$, following your proposal, let $ktoinfty$, hence $f_{alpha, k} to sum_{j=1}^infty frac{1}{j^2}1_{A_j}$. I don't quite see how this approaches $e^{-alpha x} = sum_{j=1}^infty frac{1}{j!}(-alpha x)^j$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:57








  • 1




    $begingroup$
    @MichaelMaier Any $x$ lies in one of the sets $A_j$. For that $j$, $|e^{-alpha, x} -f_{alpha,k} (x)| leq frac 1 {j^{2}} -frac 1 {(j+1)^{2}}$ because if to numbers are in an interval $(a,b)$ then their difference is less than or equal to $b-a$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 9:02
















2












2








2





$begingroup$

Take $f_{alpha,k} =sumlimits_{j=1}^{k} frac 1 {j^{2}} I_{A_j}$ where $A_j=(frac {2 ln , j} {alpha},frac {2 ln , (j+1)} {alpha})$. Note that if $x in A_j$ then $e^{-alpha, x}$ lies between $frac 1 {(j+1)^{2}}$ and $frac 1 {j^{2}}$.






share|cite|improve this answer









$endgroup$



Take $f_{alpha,k} =sumlimits_{j=1}^{k} frac 1 {j^{2}} I_{A_j}$ where $A_j=(frac {2 ln , j} {alpha},frac {2 ln , (j+1)} {alpha})$. Note that if $x in A_j$ then $e^{-alpha, x}$ lies between $frac 1 {(j+1)^{2}}$ and $frac 1 {j^{2}}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 8:38









Kavi Rama MurthyKavi Rama Murthy

57.8k42160




57.8k42160












  • $begingroup$
    One question: If I want $f_{alpha, k} to f_{alpha}$, following your proposal, let $ktoinfty$, hence $f_{alpha, k} to sum_{j=1}^infty frac{1}{j^2}1_{A_j}$. I don't quite see how this approaches $e^{-alpha x} = sum_{j=1}^infty frac{1}{j!}(-alpha x)^j$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:57








  • 1




    $begingroup$
    @MichaelMaier Any $x$ lies in one of the sets $A_j$. For that $j$, $|e^{-alpha, x} -f_{alpha,k} (x)| leq frac 1 {j^{2}} -frac 1 {(j+1)^{2}}$ because if to numbers are in an interval $(a,b)$ then their difference is less than or equal to $b-a$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 9:02




















  • $begingroup$
    One question: If I want $f_{alpha, k} to f_{alpha}$, following your proposal, let $ktoinfty$, hence $f_{alpha, k} to sum_{j=1}^infty frac{1}{j^2}1_{A_j}$. I don't quite see how this approaches $e^{-alpha x} = sum_{j=1}^infty frac{1}{j!}(-alpha x)^j$
    $endgroup$
    – Michael Maier
    Jan 15 at 8:57








  • 1




    $begingroup$
    @MichaelMaier Any $x$ lies in one of the sets $A_j$. For that $j$, $|e^{-alpha, x} -f_{alpha,k} (x)| leq frac 1 {j^{2}} -frac 1 {(j+1)^{2}}$ because if to numbers are in an interval $(a,b)$ then their difference is less than or equal to $b-a$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 9:02


















$begingroup$
One question: If I want $f_{alpha, k} to f_{alpha}$, following your proposal, let $ktoinfty$, hence $f_{alpha, k} to sum_{j=1}^infty frac{1}{j^2}1_{A_j}$. I don't quite see how this approaches $e^{-alpha x} = sum_{j=1}^infty frac{1}{j!}(-alpha x)^j$
$endgroup$
– Michael Maier
Jan 15 at 8:57






$begingroup$
One question: If I want $f_{alpha, k} to f_{alpha}$, following your proposal, let $ktoinfty$, hence $f_{alpha, k} to sum_{j=1}^infty frac{1}{j^2}1_{A_j}$. I don't quite see how this approaches $e^{-alpha x} = sum_{j=1}^infty frac{1}{j!}(-alpha x)^j$
$endgroup$
– Michael Maier
Jan 15 at 8:57






1




1




$begingroup$
@MichaelMaier Any $x$ lies in one of the sets $A_j$. For that $j$, $|e^{-alpha, x} -f_{alpha,k} (x)| leq frac 1 {j^{2}} -frac 1 {(j+1)^{2}}$ because if to numbers are in an interval $(a,b)$ then their difference is less than or equal to $b-a$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 9:02






$begingroup$
@MichaelMaier Any $x$ lies in one of the sets $A_j$. For that $j$, $|e^{-alpha, x} -f_{alpha,k} (x)| leq frac 1 {j^{2}} -frac 1 {(j+1)^{2}}$ because if to numbers are in an interval $(a,b)$ then their difference is less than or equal to $b-a$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 9:02




















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