Continuity of Lebesgue Stieltjes integral
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I am trying to prove that Lebesgue-Stieltjes integral defines a cadlag function (i.e. right continuous with left limits) when its integrator is a cadlag function.
Assume that $A(s)$, $sin mathbb{R}_+$ is a right-continuous function, has left limits and of finite variation. Also assume that function $H(s)$, $sin mathbb{R}_+$ is measurable function (i.e. Borel measurable on $mathbb{R}_+$).
It is well-known that $A$ defines a signed measure and thus we can define Lebesgue-Stieltjes integral $Y(t):=int_0^t H(s)dA(s)$ for any $t$ as long as $int_0^t |H(s)|d|A|(s)<infty$.
To prove that $Y(t)$ is right continuous, I choose a decreasing sequence $t_ndownarrow t$ and have $Y(t_n)=int_{mathbb{R}_+}1_{(0,t_n]}(s)H(s)dA(s)$. We can see that for every fixed $s$ the integrand $1_{(0,t_n]}(s)H(s)$ tends to $1_{(0,t]}(s)H(s)$.
As $int_0^u |H(s)|d|A|(s)<infty$ for all $u$, we can use dominated convergence and show that indeed $lim_{nto infty}Y(t_n)=Y(t)$. This proves that $Y(t)$ is right continuous.
Using exactly the same argumens (i.e. by choosing $t_nuparrow t$) we can show that $Y(t)$ is left continuous. However, my intuition says that when $A(s)$ has a jump discontinuity at point $t$, then $Y(t)$ should also have a jump of size $H(t)Delta A(t)$. I can't find a flaw in my proof above.
measure-theory lebesgue-integral stieltjes-integral signed-measures
asked Dec 1 at 13:00
Mushtandoid
9610
add a comment |
up vote
1
down vote
favorite
I am trying to prove that Lebesgue-Stieltjes integral defines a cadlag function (i.e. right continuous with left limits) when its integrator is a cadlag function.
Assume that $A(s)$, $sin mathbb{R}_+$ is a right-continuous function, has left limits and of finite variation. Also assume that function $H(s)$, $sin mathbb{R}_+$ is measurable function (i.e. Borel measurable on $mathbb{R}_+$).
It is well-known that $A$ defines a signed measure and thus we can define Lebesgue-Stieltjes integral $Y(t):=int_0^t H(s)dA(s)$ for any $t$ as long as $int_0^t |H(s)|d|A|(s)<infty$.
To prove that $Y(t)$ is right continuous, I choose a decreasing sequence $t_ndownarrow t$ and have $Y(t_n)=int_{mathbb{R}_+}1_{(0,t_n]}(s)H(s)dA(s)$. We can see that for every fixed $s$ the integrand $1_{(0,t_n]}(s)H(s)$ tends to $1_{(0,t]}(s)H(s)$.
As $int_0^u |H(s)|d|A|(s)<infty$ for all $u$, we can use dominated convergence and show that indeed $lim_{nto infty}Y(t_n)=Y(t)$. This proves that $Y(t)$ is right continuous.
Using exactly the same argumens (i.e. by choosing $t_nuparrow t$) we can show that $Y(t)$ is left continuous. However, my intuition says that when $A(s)$ has a jump discontinuity at point $t$, then $Y(t)$ should also have a jump of size $H(t)Delta A(t)$. I can't find a flaw in my proof above.
measure-theory lebesgue-integral stieltjes-integral signed-measures
asked Dec 1 at 13:00
Mushtandoid
9610
Is it possible that dominated convergence theorem for Lebesgue-Stieltjes integrals works only when $A(s)$ has no jumps?
– Mushtandoid
Dec 1 at 13:14
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to prove that Lebesgue-Stieltjes integral defines a cadlag function (i.e. right continuous with left limits) when its integrator is a cadlag function.
Assume that $A(s)$, $sin mathbb{R}_+$ is a right-continuous function, has left limits and of finite variation. Also assume that function $H(s)$, $sin mathbb{R}_+$ is measurable function (i.e. Borel measurable on $mathbb{R}_+$).
It is well-known that $A$ defines a signed measure and thus we can define Lebesgue-Stieltjes integral $Y(t):=int_0^t H(s)dA(s)$ for any $t$ as long as $int_0^t |H(s)|d|A|(s)<infty$.
To prove that $Y(t)$ is right continuous, I choose a decreasing sequence $t_ndownarrow t$ and have $Y(t_n)=int_{mathbb{R}_+}1_{(0,t_n]}(s)H(s)dA(s)$. We can see that for every fixed $s$ the integrand $1_{(0,t_n]}(s)H(s)$ tends to $1_{(0,t]}(s)H(s)$.
As $int_0^u |H(s)|d|A|(s)<infty$ for all $u$, we can use dominated convergence and show that indeed $lim_{nto infty}Y(t_n)=Y(t)$. This proves that $Y(t)$ is right continuous.
Using exactly the same argumens (i.e. by choosing $t_nuparrow t$) we can show that $Y(t)$ is left continuous. However, my intuition says that when $A(s)$ has a jump discontinuity at point $t$, then $Y(t)$ should also have a jump of size $H(t)Delta A(t)$. I can't find a flaw in my proof above.
measure-theory lebesgue-integral stieltjes-integral signed-measures
asked Dec 1 at 13:00
Mushtandoid
9610
I am trying to prove that Lebesgue-Stieltjes integral defines a cadlag function (i.e. right continuous with left limits) when its integrator is a cadlag function.
Assume that $A(s)$, $sin mathbb{R}_+$ is a right-continuous function, has left limits and of finite variation. Also assume that function $H(s)$, $sin mathbb{R}_+$ is measurable function (i.e. Borel measurable on $mathbb{R}_+$).
It is well-known that $A$ defines a signed measure and thus we can define Lebesgue-Stieltjes integral $Y(t):=int_0^t H(s)dA(s)$ for any $t$ as long as $int_0^t |H(s)|d|A|(s)<infty$.
To prove that $Y(t)$ is right continuous, I choose a decreasing sequence $t_ndownarrow t$ and have $Y(t_n)=int_{mathbb{R}_+}1_{(0,t_n]}(s)H(s)dA(s)$. We can see that for every fixed $s$ the integrand $1_{(0,t_n]}(s)H(s)$ tends to $1_{(0,t]}(s)H(s)$.
As $int_0^u |H(s)|d|A|(s)<infty$ for all $u$, we can use dominated convergence and show that indeed $lim_{nto infty}Y(t_n)=Y(t)$. This proves that $Y(t)$ is right continuous.
Using exactly the same argumens (i.e. by choosing $t_nuparrow t$) we can show that $Y(t)$ is left continuous. However, my intuition says that when $A(s)$ has a jump discontinuity at point $t$, then $Y(t)$ should also have a jump of size $H(t)Delta A(t)$. I can't find a flaw in my proof above.
measure-theory lebesgue-integral stieltjes-integral signed-measures
measure-theory lebesgue-integral stieltjes-integral signed-measures
asked Dec 1 at 13:00
Mushtandoid
9610
asked Dec 1 at 13:00
Mushtandoid
9610
asked Dec 1 at 13:00
Mushtandoid
9610
asked Dec 1 at 13:00
Mushtandoid
9610
asked Dec 1 at 13:00
Mushtandoid
9610
9610
Is it possible that dominated convergence theorem for Lebesgue-Stieltjes integrals works only when $A(s)$ has no jumps?
– Mushtandoid
Dec 1 at 13:14
add a comment |
Is it possible that dominated convergence theorem for Lebesgue-Stieltjes integrals works only when $A(s)$ has no jumps?
– Mushtandoid
Dec 1 at 13:14
Is it possible that dominated convergence theorem for Lebesgue-Stieltjes integrals works only when $A(s)$ has no jumps?
– Mushtandoid
Dec 1 at 13:14
Is it possible that dominated convergence theorem for Lebesgue-Stieltjes integrals works only when $A(s)$ has no jumps?
– Mushtandoid
Dec 1 at 13:14
add a comment |
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Is it possible that dominated convergence theorem for Lebesgue-Stieltjes integrals works only when $A(s)$ has no jumps?
– Mushtandoid
Dec 1 at 13:14