Mixing
Monotone Convergence Theorem in calculating Lebesgue Integral
$begingroup$
Given the Lebesgue measure $m_2$ on $(mathbb{R}^2,mathbb{B}_2)$, I have the following integral:
$$mu(mathbb{R}^2)=...=int 1_{[1,infty )}frac{1}{x^2}dm(x)$$
My professor insist on using the Monotone Convergence Theorem:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=lim_{ntoinfty}int 1_{[1,n]}frac{1}{x^2}dm(x)=lim_{ntoinfty}int_{1}^nfrac{1}{x^2}dx$$
$$=lim_{ntoinfty}left(-frac{1}{n}-left(-frac{1}{1}right)right)=1$$
Why can I not just make it into the Riemann-integral from first step as:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=int_{1}^inftyfrac{1}{x^2}dx=...,$$
instead of using the Monotone Convergence Theorem? Because the requirements, that $frac{1}{x^2}$ should be positive and continuous are fulfilled.
Any help would be greatly appreciated, as I cannot se why this step is needed.
integration lebesgue-integral lebesgue-measure
asked Jan 12 at 13:11
FrederikFrederik
1009
$endgroup$
add a comment |
$begingroup$
Given the Lebesgue measure $m_2$ on $(mathbb{R}^2,mathbb{B}_2)$, I have the following integral:
$$mu(mathbb{R}^2)=...=int 1_{[1,infty )}frac{1}{x^2}dm(x)$$
My professor insist on using the Monotone Convergence Theorem:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=lim_{ntoinfty}int 1_{[1,n]}frac{1}{x^2}dm(x)=lim_{ntoinfty}int_{1}^nfrac{1}{x^2}dx$$
$$=lim_{ntoinfty}left(-frac{1}{n}-left(-frac{1}{1}right)right)=1$$
Why can I not just make it into the Riemann-integral from first step as:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=int_{1}^inftyfrac{1}{x^2}dx=...,$$
instead of using the Monotone Convergence Theorem? Because the requirements, that $frac{1}{x^2}$ should be positive and continuous are fulfilled.
Any help would be greatly appreciated, as I cannot se why this step is needed.
integration lebesgue-integral lebesgue-measure
asked Jan 12 at 13:11
FrederikFrederik
1009
$endgroup$
add a comment |
$begingroup$
Given the Lebesgue measure $m_2$ on $(mathbb{R}^2,mathbb{B}_2)$, I have the following integral:
$$mu(mathbb{R}^2)=...=int 1_{[1,infty )}frac{1}{x^2}dm(x)$$
My professor insist on using the Monotone Convergence Theorem:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=lim_{ntoinfty}int 1_{[1,n]}frac{1}{x^2}dm(x)=lim_{ntoinfty}int_{1}^nfrac{1}{x^2}dx$$
$$=lim_{ntoinfty}left(-frac{1}{n}-left(-frac{1}{1}right)right)=1$$
Why can I not just make it into the Riemann-integral from first step as:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=int_{1}^inftyfrac{1}{x^2}dx=...,$$
instead of using the Monotone Convergence Theorem? Because the requirements, that $frac{1}{x^2}$ should be positive and continuous are fulfilled.
Any help would be greatly appreciated, as I cannot se why this step is needed.
integration lebesgue-integral lebesgue-measure
asked Jan 12 at 13:11
FrederikFrederik
1009
$endgroup$
Given the Lebesgue measure $m_2$ on $(mathbb{R}^2,mathbb{B}_2)$, I have the following integral:
$$mu(mathbb{R}^2)=...=int 1_{[1,infty )}frac{1}{x^2}dm(x)$$
My professor insist on using the Monotone Convergence Theorem:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=lim_{ntoinfty}int 1_{[1,n]}frac{1}{x^2}dm(x)=lim_{ntoinfty}int_{1}^nfrac{1}{x^2}dx$$
$$=lim_{ntoinfty}left(-frac{1}{n}-left(-frac{1}{1}right)right)=1$$
Why can I not just make it into the Riemann-integral from first step as:
$$mu(mathbb{R}^2)=int 1_{[1,infty )}frac{1}{x^2}dm(x)=int_{1}^inftyfrac{1}{x^2}dx=...,$$
instead of using the Monotone Convergence Theorem? Because the requirements, that $frac{1}{x^2}$ should be positive and continuous are fulfilled.
Any help would be greatly appreciated, as I cannot se why this step is needed.
integration lebesgue-integral lebesgue-measure
integration lebesgue-integral lebesgue-measure
asked Jan 12 at 13:11
FrederikFrederik
1009
asked Jan 12 at 13:11
FrederikFrederik
1009
asked Jan 12 at 13:11
FrederikFrederik
1009
asked Jan 12 at 13:11
FrederikFrederik
1009
asked Jan 12 at 13:11
FrederikFrederik
1009
1009
add a comment |
add a comment |
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