Problem 3. Page 240. Barry Simon. (Associated Lebesgue-Stieltjes measure )
In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
($M$ is a algebra).
Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.
In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$
Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
For any Baire set $A.$
I have this:
$mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$
Why this sum is $0$ when $1/2not in A?$
real-analysis measure-theory lebesgue-measure stieltjes-integral
asked Nov 25 at 20:57
eraldcoil
358111
add a comment |
In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
($M$ is a algebra).
Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.
In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$
Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
For any Baire set $A.$
I have this:
$mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$
Why this sum is $0$ when $1/2not in A?$
real-analysis measure-theory lebesgue-measure stieltjes-integral
asked Nov 25 at 20:57
eraldcoil
358111
add a comment |
In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
($M$ is a algebra).
Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.
In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$
Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
For any Baire set $A.$
I have this:
$mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$
Why this sum is $0$ when $1/2not in A?$
real-analysis measure-theory lebesgue-measure stieltjes-integral
asked Nov 25 at 20:57
eraldcoil
358111
In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
($M$ is a algebra).
Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.
In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$
Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
For any Baire set $A.$
I have this:
$mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$
Why this sum is $0$ when $1/2not in A?$
real-analysis measure-theory lebesgue-measure stieltjes-integral
real-analysis measure-theory lebesgue-measure stieltjes-integral
asked Nov 25 at 20:57
eraldcoil
358111
asked Nov 25 at 20:57
eraldcoil
358111
asked Nov 25 at 20:57
eraldcoil
358111
asked Nov 25 at 20:57
eraldcoil
358111
asked Nov 25 at 20:57
eraldcoil
358111
358111
add a comment |
add a comment |
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