Finding $F$ from a Lebesgue-Stieltjes Measure












1












$begingroup$



Exercise: Define a function $F:mathbb{R}tomathbb{R}$ such that the associated Lebesgue-Stieltjes measure verifies conditions: $mu_F({1})=frac{1}{4}$,$mu_F((-infty,1))=0$,$mu_F((1,frac{3}{2}))=1$ and $mu_F([2,x))=frac{x}{2}$,$forall x>2$




$mu_F({1})=F(1)=frac{1}{4}$



$mu_F((1,frac{3}{2}))=F(frac{3}{2})-F(1)=1implies F(frac{3}{2})-frac{1}{4}=1implies F(frac{3}{2})=frac{5}{4} $



However I have no idea on how to deal with $mu_F([2,x))=frac{x}{2}$,$forall x>2$
. I am not understanding the aim of the exercise I tried to draw the function and I can see a straight line from minus infinity to one at the height of $frac{1}{4}$ and then a point at $1.5$ below the $X$ axis $frac{3}{4}$.



Question:



Since the function is not assumed to be either continuous or trivial how should I find the aforementioned function $F$?



Thanks in advance!






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you have $mu_F((1,3/2))$ reversed - it should be $F(3/2) - F(1)$.
    $endgroup$
    – Math1000
    Dec 21 '18 at 18:03










  • $begingroup$
    @Math1000 Thanks you are right!
    $endgroup$
    – Pedro Gomes
    Dec 21 '18 at 19:46


















1












$begingroup$



Exercise: Define a function $F:mathbb{R}tomathbb{R}$ such that the associated Lebesgue-Stieltjes measure verifies conditions: $mu_F({1})=frac{1}{4}$,$mu_F((-infty,1))=0$,$mu_F((1,frac{3}{2}))=1$ and $mu_F([2,x))=frac{x}{2}$,$forall x>2$




$mu_F({1})=F(1)=frac{1}{4}$



$mu_F((1,frac{3}{2}))=F(frac{3}{2})-F(1)=1implies F(frac{3}{2})-frac{1}{4}=1implies F(frac{3}{2})=frac{5}{4} $



However I have no idea on how to deal with $mu_F([2,x))=frac{x}{2}$,$forall x>2$
. I am not understanding the aim of the exercise I tried to draw the function and I can see a straight line from minus infinity to one at the height of $frac{1}{4}$ and then a point at $1.5$ below the $X$ axis $frac{3}{4}$.



Question:



Since the function is not assumed to be either continuous or trivial how should I find the aforementioned function $F$?



Thanks in advance!






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you have $mu_F((1,3/2))$ reversed - it should be $F(3/2) - F(1)$.
    $endgroup$
    – Math1000
    Dec 21 '18 at 18:03










  • $begingroup$
    @Math1000 Thanks you are right!
    $endgroup$
    – Pedro Gomes
    Dec 21 '18 at 19:46
















1












1








1





$begingroup$



Exercise: Define a function $F:mathbb{R}tomathbb{R}$ such that the associated Lebesgue-Stieltjes measure verifies conditions: $mu_F({1})=frac{1}{4}$,$mu_F((-infty,1))=0$,$mu_F((1,frac{3}{2}))=1$ and $mu_F([2,x))=frac{x}{2}$,$forall x>2$




$mu_F({1})=F(1)=frac{1}{4}$



$mu_F((1,frac{3}{2}))=F(frac{3}{2})-F(1)=1implies F(frac{3}{2})-frac{1}{4}=1implies F(frac{3}{2})=frac{5}{4} $



However I have no idea on how to deal with $mu_F([2,x))=frac{x}{2}$,$forall x>2$
. I am not understanding the aim of the exercise I tried to draw the function and I can see a straight line from minus infinity to one at the height of $frac{1}{4}$ and then a point at $1.5$ below the $X$ axis $frac{3}{4}$.



Question:



Since the function is not assumed to be either continuous or trivial how should I find the aforementioned function $F$?



Thanks in advance!






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$





Exercise: Define a function $F:mathbb{R}tomathbb{R}$ such that the associated Lebesgue-Stieltjes measure verifies conditions: $mu_F({1})=frac{1}{4}$,$mu_F((-infty,1))=0$,$mu_F((1,frac{3}{2}))=1$ and $mu_F([2,x))=frac{x}{2}$,$forall x>2$




$mu_F({1})=F(1)=frac{1}{4}$



$mu_F((1,frac{3}{2}))=F(frac{3}{2})-F(1)=1implies F(frac{3}{2})-frac{1}{4}=1implies F(frac{3}{2})=frac{5}{4} $



However I have no idea on how to deal with $mu_F([2,x))=frac{x}{2}$,$forall x>2$
. I am not understanding the aim of the exercise I tried to draw the function and I can see a straight line from minus infinity to one at the height of $frac{1}{4}$ and then a point at $1.5$ below the $X$ axis $frac{3}{4}$.



Question:



Since the function is not assumed to be either continuous or trivial how should I find the aforementioned function $F$?



Thanks in advance!






measure-theory lebesgue-measure




measure-theory lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 19:49







Pedro Gomes

















asked Dec 21 '18 at 15:20









Pedro GomesPedro Gomes

1,8892721




1,8892721












  • $begingroup$
    I think you have $mu_F((1,3/2))$ reversed - it should be $F(3/2) - F(1)$.
    $endgroup$
    – Math1000
    Dec 21 '18 at 18:03










  • $begingroup$
    @Math1000 Thanks you are right!
    $endgroup$
    – Pedro Gomes
    Dec 21 '18 at 19:46




















  • $begingroup$
    I think you have $mu_F((1,3/2))$ reversed - it should be $F(3/2) - F(1)$.
    $endgroup$
    – Math1000
    Dec 21 '18 at 18:03










  • $begingroup$
    @Math1000 Thanks you are right!
    $endgroup$
    – Pedro Gomes
    Dec 21 '18 at 19:46


















$begingroup$
I think you have $mu_F((1,3/2))$ reversed - it should be $F(3/2) - F(1)$.
$endgroup$
– Math1000
Dec 21 '18 at 18:03




$begingroup$
I think you have $mu_F((1,3/2))$ reversed - it should be $F(3/2) - F(1)$.
$endgroup$
– Math1000
Dec 21 '18 at 18:03












$begingroup$
@Math1000 Thanks you are right!
$endgroup$
– Pedro Gomes
Dec 21 '18 at 19:46






$begingroup$
@Math1000 Thanks you are right!
$endgroup$
– Pedro Gomes
Dec 21 '18 at 19:46












1 Answer
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oldest

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1












$begingroup$

Note that we cannot determine $F=mu_F((-infty,x])$ uniquely using only given information. Specifically, we cannot determine $F$ on $[3/2,2]$. Notice that
$$
lim_{xto 2}mu_F([2,x)) =1=mu_F(2) = F(2)-F(2^{-}).
$$ Here, $F(x^{-})$ denotes left limit of $F$ at $x$. This means that $F$ has a jump discontinuity of size $1$ at $x=2$. Also, note that
$$
F(x^{-})=mu_F((-infty, x)),quadforall xinmathbb{R}.
$$ Hence $mu_F([2,x))=frac{x}{2}$ means that $F(x^{-})-F(2^{-}) = F(x^{-})-F(2)+1=frac{x}{2}.$ Since $F(x^{-}) = frac{x}{2} +F(2) -1$ is continuous on $x>2$, we have
$$
F(x)=F(x^{-}) = frac{x}{2} +F(2) -1, quadforall x>2.$$



To conclude, $F$ has the following form
$$
F(x) = begin{cases}0 ,quad x<1\1/4,quad x=1\ 5/4,quad x=(frac{3}{2})^{-}\ 5/4+delta ,quad x=2^{-}\9/4+delta,quad x=2\5/4+delta +frac{x}{2},quad x>2,
end{cases}
$$where $0leqdelta<infty.$






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  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 22 '18 at 22:10











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1 Answer
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1












$begingroup$

Note that we cannot determine $F=mu_F((-infty,x])$ uniquely using only given information. Specifically, we cannot determine $F$ on $[3/2,2]$. Notice that
$$
lim_{xto 2}mu_F([2,x)) =1=mu_F(2) = F(2)-F(2^{-}).
$$ Here, $F(x^{-})$ denotes left limit of $F$ at $x$. This means that $F$ has a jump discontinuity of size $1$ at $x=2$. Also, note that
$$
F(x^{-})=mu_F((-infty, x)),quadforall xinmathbb{R}.
$$ Hence $mu_F([2,x))=frac{x}{2}$ means that $F(x^{-})-F(2^{-}) = F(x^{-})-F(2)+1=frac{x}{2}.$ Since $F(x^{-}) = frac{x}{2} +F(2) -1$ is continuous on $x>2$, we have
$$
F(x)=F(x^{-}) = frac{x}{2} +F(2) -1, quadforall x>2.$$



To conclude, $F$ has the following form
$$
F(x) = begin{cases}0 ,quad x<1\1/4,quad x=1\ 5/4,quad x=(frac{3}{2})^{-}\ 5/4+delta ,quad x=2^{-}\9/4+delta,quad x=2\5/4+delta +frac{x}{2},quad x>2,
end{cases}
$$where $0leqdelta<infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 22 '18 at 22:10
















1












$begingroup$

Note that we cannot determine $F=mu_F((-infty,x])$ uniquely using only given information. Specifically, we cannot determine $F$ on $[3/2,2]$. Notice that
$$
lim_{xto 2}mu_F([2,x)) =1=mu_F(2) = F(2)-F(2^{-}).
$$ Here, $F(x^{-})$ denotes left limit of $F$ at $x$. This means that $F$ has a jump discontinuity of size $1$ at $x=2$. Also, note that
$$
F(x^{-})=mu_F((-infty, x)),quadforall xinmathbb{R}.
$$ Hence $mu_F([2,x))=frac{x}{2}$ means that $F(x^{-})-F(2^{-}) = F(x^{-})-F(2)+1=frac{x}{2}.$ Since $F(x^{-}) = frac{x}{2} +F(2) -1$ is continuous on $x>2$, we have
$$
F(x)=F(x^{-}) = frac{x}{2} +F(2) -1, quadforall x>2.$$



To conclude, $F$ has the following form
$$
F(x) = begin{cases}0 ,quad x<1\1/4,quad x=1\ 5/4,quad x=(frac{3}{2})^{-}\ 5/4+delta ,quad x=2^{-}\9/4+delta,quad x=2\5/4+delta +frac{x}{2},quad x>2,
end{cases}
$$where $0leqdelta<infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 22 '18 at 22:10














1












1








1





$begingroup$

Note that we cannot determine $F=mu_F((-infty,x])$ uniquely using only given information. Specifically, we cannot determine $F$ on $[3/2,2]$. Notice that
$$
lim_{xto 2}mu_F([2,x)) =1=mu_F(2) = F(2)-F(2^{-}).
$$ Here, $F(x^{-})$ denotes left limit of $F$ at $x$. This means that $F$ has a jump discontinuity of size $1$ at $x=2$. Also, note that
$$
F(x^{-})=mu_F((-infty, x)),quadforall xinmathbb{R}.
$$ Hence $mu_F([2,x))=frac{x}{2}$ means that $F(x^{-})-F(2^{-}) = F(x^{-})-F(2)+1=frac{x}{2}.$ Since $F(x^{-}) = frac{x}{2} +F(2) -1$ is continuous on $x>2$, we have
$$
F(x)=F(x^{-}) = frac{x}{2} +F(2) -1, quadforall x>2.$$



To conclude, $F$ has the following form
$$
F(x) = begin{cases}0 ,quad x<1\1/4,quad x=1\ 5/4,quad x=(frac{3}{2})^{-}\ 5/4+delta ,quad x=2^{-}\9/4+delta,quad x=2\5/4+delta +frac{x}{2},quad x>2,
end{cases}
$$where $0leqdelta<infty.$






share|cite|improve this answer









$endgroup$



Note that we cannot determine $F=mu_F((-infty,x])$ uniquely using only given information. Specifically, we cannot determine $F$ on $[3/2,2]$. Notice that
$$
lim_{xto 2}mu_F([2,x)) =1=mu_F(2) = F(2)-F(2^{-}).
$$ Here, $F(x^{-})$ denotes left limit of $F$ at $x$. This means that $F$ has a jump discontinuity of size $1$ at $x=2$. Also, note that
$$
F(x^{-})=mu_F((-infty, x)),quadforall xinmathbb{R}.
$$ Hence $mu_F([2,x))=frac{x}{2}$ means that $F(x^{-})-F(2^{-}) = F(x^{-})-F(2)+1=frac{x}{2}.$ Since $F(x^{-}) = frac{x}{2} +F(2) -1$ is continuous on $x>2$, we have
$$
F(x)=F(x^{-}) = frac{x}{2} +F(2) -1, quadforall x>2.$$



To conclude, $F$ has the following form
$$
F(x) = begin{cases}0 ,quad x<1\1/4,quad x=1\ 5/4,quad x=(frac{3}{2})^{-}\ 5/4+delta ,quad x=2^{-}\9/4+delta,quad x=2\5/4+delta +frac{x}{2},quad x>2,
end{cases}
$$where $0leqdelta<infty.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 20:43









SongSong

16.2k1739




16.2k1739












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 22 '18 at 22:10


















  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 22 '18 at 22:10
















$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 22 '18 at 22:10




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 22 '18 at 22:10


















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