Mixing
Properties of Counting Measures Restricted to a Particular Sigma Algebra
Define $mathcal{F}={Asubseteqmathbb{R} vert 0in A^{mathrm{o}} or 0 in (A^c)^{mathrm{o}} }$, where $A^{mathrm{o}}$ is the interior of $A$. It can be shown quite easily that $mathcal{F}$ is an algebra of sets, and that $sigma(mathcal{F})$ contains the singletons of $mathbb{R}$.
If $gamma$ is the counting measure on $(mathbb{R},mathcal{P}(mathbb{R}))$, let $alpha(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} })$ and $beta(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} } cup {0})$. If we define $mu := left.alpha right|_{sigma(mathcal{F})}$ and $nu := left.beta right|_{sigma(mathcal{F})}$ (the restrictions of $mu$ and $nu$ to
$sigma(mathcal{F})$) then it turns out that:
a) both $mu$ and $nu$ are $sigma$-finite.
b) $mu$ and $nu$ coincide on $mathcal{F}$ (that is, $mu(A)=nu(A)$, $forall A in mathcal{F}$).
c) $mu neq nu(A)$ on $sigma(mathcal{F})$.
I am having trouble in proving these 3 facts, and any help/hints to help push me towards the right direction will be much appreciated.
In particular, I'm having difficulty in working with any general set $Ain sigma(mathcal{F})$; I currently have no clue on how to construct a sequence $(A_n)_{n=1}^{infty}$ in $sigma(mathcal{F})$ such that $bigcuplimits_{n=1}^{infty}A_n = mathbb{R}$ with $mu(A_n)<infty$ and $nu(A_n)<infty$.
EDIT: I've actually made some progress with this since asking: I have solutions to b) and c), and have partial work for a). The struggle in part a) remains with constructing such a sequence of sets. Best I've come up with is $(A_k)_{k=1}^{infty}=(-k,-frac{1}{k}] cup [frac{1}{k},k)$, and in this case, $0 in (A_k^c)^{mathrm{o}}$ with $mu(A_k)<infty$ and $nu(A_k)<infty$. Problem here is that the countable union isn't the whole real line (it misses ${0}$!). Similarly, I've also considered the sequence $(A_k)_{k=1}^{infty}=(-k,0] cup [frac{1}{k},k)$, whose countable union is all of $mathbb{R}$, $mu$ and $nu$ are finite $forall k$, but this sequence is not in $mathcal{F}$.
real-analysis analysis measure-theory
asked Nov 18 '18 at 19:33
Buttfor
806
add a comment |
Define $mathcal{F}={Asubseteqmathbb{R} vert 0in A^{mathrm{o}} or 0 in (A^c)^{mathrm{o}} }$, where $A^{mathrm{o}}$ is the interior of $A$. It can be shown quite easily that $mathcal{F}$ is an algebra of sets, and that $sigma(mathcal{F})$ contains the singletons of $mathbb{R}$.
If $gamma$ is the counting measure on $(mathbb{R},mathcal{P}(mathbb{R}))$, let $alpha(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} })$ and $beta(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} } cup {0})$. If we define $mu := left.alpha right|_{sigma(mathcal{F})}$ and $nu := left.beta right|_{sigma(mathcal{F})}$ (the restrictions of $mu$ and $nu$ to
$sigma(mathcal{F})$) then it turns out that:
a) both $mu$ and $nu$ are $sigma$-finite.
b) $mu$ and $nu$ coincide on $mathcal{F}$ (that is, $mu(A)=nu(A)$, $forall A in mathcal{F}$).
c) $mu neq nu(A)$ on $sigma(mathcal{F})$.
I am having trouble in proving these 3 facts, and any help/hints to help push me towards the right direction will be much appreciated.
In particular, I'm having difficulty in working with any general set $Ain sigma(mathcal{F})$; I currently have no clue on how to construct a sequence $(A_n)_{n=1}^{infty}$ in $sigma(mathcal{F})$ such that $bigcuplimits_{n=1}^{infty}A_n = mathbb{R}$ with $mu(A_n)<infty$ and $nu(A_n)<infty$.
EDIT: I've actually made some progress with this since asking: I have solutions to b) and c), and have partial work for a). The struggle in part a) remains with constructing such a sequence of sets. Best I've come up with is $(A_k)_{k=1}^{infty}=(-k,-frac{1}{k}] cup [frac{1}{k},k)$, and in this case, $0 in (A_k^c)^{mathrm{o}}$ with $mu(A_k)<infty$ and $nu(A_k)<infty$. Problem here is that the countable union isn't the whole real line (it misses ${0}$!). Similarly, I've also considered the sequence $(A_k)_{k=1}^{infty}=(-k,0] cup [frac{1}{k},k)$, whose countable union is all of $mathbb{R}$, $mu$ and $nu$ are finite $forall k$, but this sequence is not in $mathcal{F}$.
real-analysis analysis measure-theory
asked Nov 18 '18 at 19:33
Buttfor
806
you don't have to construct a sequence which the union is equal $sigma(mathcal{F})$, but $X$ instead. As $Xin sigma(mathcal{F})$ the sigma-algebra is a family of sets, not a kind of 'measure subspace'
– Robson
Nov 18 '18 at 22:33
Thank you, I'll make the edit!
– Buttfor
Nov 18 '18 at 22:39
Why do you want $A_k$ to be in $mathcal{F}$? For sigma-finite you only need $A_kin sigma(mathcal{F})$.
– Severin Schraven
Nov 25 '18 at 18:50
In fact, your proposed $A_k = (-k, 0] cup [frac{1}{k}, k)$ works just fine.
– Severin Schraven
Nov 25 '18 at 19:17
add a comment |
Define $mathcal{F}={Asubseteqmathbb{R} vert 0in A^{mathrm{o}} or 0 in (A^c)^{mathrm{o}} }$, where $A^{mathrm{o}}$ is the interior of $A$. It can be shown quite easily that $mathcal{F}$ is an algebra of sets, and that $sigma(mathcal{F})$ contains the singletons of $mathbb{R}$.
If $gamma$ is the counting measure on $(mathbb{R},mathcal{P}(mathbb{R}))$, let $alpha(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} })$ and $beta(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} } cup {0})$. If we define $mu := left.alpha right|_{sigma(mathcal{F})}$ and $nu := left.beta right|_{sigma(mathcal{F})}$ (the restrictions of $mu$ and $nu$ to
$sigma(mathcal{F})$) then it turns out that:
a) both $mu$ and $nu$ are $sigma$-finite.
b) $mu$ and $nu$ coincide on $mathcal{F}$ (that is, $mu(A)=nu(A)$, $forall A in mathcal{F}$).
c) $mu neq nu(A)$ on $sigma(mathcal{F})$.
I am having trouble in proving these 3 facts, and any help/hints to help push me towards the right direction will be much appreciated.
In particular, I'm having difficulty in working with any general set $Ain sigma(mathcal{F})$; I currently have no clue on how to construct a sequence $(A_n)_{n=1}^{infty}$ in $sigma(mathcal{F})$ such that $bigcuplimits_{n=1}^{infty}A_n = mathbb{R}$ with $mu(A_n)<infty$ and $nu(A_n)<infty$.
EDIT: I've actually made some progress with this since asking: I have solutions to b) and c), and have partial work for a). The struggle in part a) remains with constructing such a sequence of sets. Best I've come up with is $(A_k)_{k=1}^{infty}=(-k,-frac{1}{k}] cup [frac{1}{k},k)$, and in this case, $0 in (A_k^c)^{mathrm{o}}$ with $mu(A_k)<infty$ and $nu(A_k)<infty$. Problem here is that the countable union isn't the whole real line (it misses ${0}$!). Similarly, I've also considered the sequence $(A_k)_{k=1}^{infty}=(-k,0] cup [frac{1}{k},k)$, whose countable union is all of $mathbb{R}$, $mu$ and $nu$ are finite $forall k$, but this sequence is not in $mathcal{F}$.
real-analysis analysis measure-theory
asked Nov 18 '18 at 19:33
Buttfor
806
Define $mathcal{F}={Asubseteqmathbb{R} vert 0in A^{mathrm{o}} or 0 in (A^c)^{mathrm{o}} }$, where $A^{mathrm{o}}$ is the interior of $A$. It can be shown quite easily that $mathcal{F}$ is an algebra of sets, and that $sigma(mathcal{F})$ contains the singletons of $mathbb{R}$.
If $gamma$ is the counting measure on $(mathbb{R},mathcal{P}(mathbb{R}))$, let $alpha(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} })$ and $beta(A)=gamma(A cap {frac{1}{n} vert n in mathbb{N} } cup {0})$. If we define $mu := left.alpha right|_{sigma(mathcal{F})}$ and $nu := left.beta right|_{sigma(mathcal{F})}$ (the restrictions of $mu$ and $nu$ to
$sigma(mathcal{F})$) then it turns out that:
a) both $mu$ and $nu$ are $sigma$-finite.
b) $mu$ and $nu$ coincide on $mathcal{F}$ (that is, $mu(A)=nu(A)$, $forall A in mathcal{F}$).
c) $mu neq nu(A)$ on $sigma(mathcal{F})$.
I am having trouble in proving these 3 facts, and any help/hints to help push me towards the right direction will be much appreciated.
In particular, I'm having difficulty in working with any general set $Ain sigma(mathcal{F})$; I currently have no clue on how to construct a sequence $(A_n)_{n=1}^{infty}$ in $sigma(mathcal{F})$ such that $bigcuplimits_{n=1}^{infty}A_n = mathbb{R}$ with $mu(A_n)<infty$ and $nu(A_n)<infty$.
EDIT: I've actually made some progress with this since asking: I have solutions to b) and c), and have partial work for a). The struggle in part a) remains with constructing such a sequence of sets. Best I've come up with is $(A_k)_{k=1}^{infty}=(-k,-frac{1}{k}] cup [frac{1}{k},k)$, and in this case, $0 in (A_k^c)^{mathrm{o}}$ with $mu(A_k)<infty$ and $nu(A_k)<infty$. Problem here is that the countable union isn't the whole real line (it misses ${0}$!). Similarly, I've also considered the sequence $(A_k)_{k=1}^{infty}=(-k,0] cup [frac{1}{k},k)$, whose countable union is all of $mathbb{R}$, $mu$ and $nu$ are finite $forall k$, but this sequence is not in $mathcal{F}$.
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Nov 18 '18 at 19:33
Buttfor
806
asked Nov 18 '18 at 19:33
Buttfor
806
edited Nov 21 '18 at 2:56
asked Nov 18 '18 at 19:33
Buttfor
806
asked Nov 18 '18 at 19:33
Buttfor
806
asked Nov 18 '18 at 19:33
Buttfor
806
806
you don't have to construct a sequence which the union is equal $sigma(mathcal{F})$, but $X$ instead. As $Xin sigma(mathcal{F})$ the sigma-algebra is a family of sets, not a kind of 'measure subspace'
– Robson
Nov 18 '18 at 22:33
Thank you, I'll make the edit!
– Buttfor
Nov 18 '18 at 22:39
Why do you want $A_k$ to be in $mathcal{F}$? For sigma-finite you only need $A_kin sigma(mathcal{F})$.
– Severin Schraven
Nov 25 '18 at 18:50
In fact, your proposed $A_k = (-k, 0] cup [frac{1}{k}, k)$ works just fine.
– Severin Schraven
Nov 25 '18 at 19:17
add a comment |
you don't have to construct a sequence which the union is equal $sigma(mathcal{F})$, but $X$ instead. As $Xin sigma(mathcal{F})$ the sigma-algebra is a family of sets, not a kind of 'measure subspace'
– Robson
Nov 18 '18 at 22:33
Thank you, I'll make the edit!
– Buttfor
Nov 18 '18 at 22:39
Why do you want $A_k$ to be in $mathcal{F}$? For sigma-finite you only need $A_kin sigma(mathcal{F})$.
– Severin Schraven
Nov 25 '18 at 18:50
In fact, your proposed $A_k = (-k, 0] cup [frac{1}{k}, k)$ works just fine.
– Severin Schraven
Nov 25 '18 at 19:17
you don't have to construct a sequence which the union is equal $sigma(mathcal{F})$, but $X$ instead. As $Xin sigma(mathcal{F})$ the sigma-algebra is a family of sets, not a kind of 'measure subspace'
– Robson
Nov 18 '18 at 22:33
you don't have to construct a sequence which the union is equal $sigma(mathcal{F})$, but $X$ instead. As $Xin sigma(mathcal{F})$ the sigma-algebra is a family of sets, not a kind of 'measure subspace'
– Robson
Nov 18 '18 at 22:33
Thank you, I'll make the edit!
– Buttfor
Nov 18 '18 at 22:39
Thank you, I'll make the edit!
– Buttfor
Nov 18 '18 at 22:39
Why do you want $A_k$ to be in $mathcal{F}$? For sigma-finite you only need $A_kin sigma(mathcal{F})$.
– Severin Schraven
Nov 25 '18 at 18:50
Why do you want $A_k$ to be in $mathcal{F}$? For sigma-finite you only need $A_kin sigma(mathcal{F})$.
– Severin Schraven
Nov 25 '18 at 18:50
In fact, your proposed $A_k = (-k, 0] cup [frac{1}{k}, k)$ works just fine.
– Severin Schraven
Nov 25 '18 at 19:17
In fact, your proposed $A_k = (-k, 0] cup [frac{1}{k}, k)$ works just fine.
– Severin Schraven
Nov 25 '18 at 19:17
add a comment |
1 Answer
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1.) We have
$$ (-infty, 0] = bigcap_{ngeq 1} (-infty, frac{1}{n}) in sigma(mathcal{F}) $$
as well as $ [frac{1}{k}, k) $ for all $kin mathbb{N}$. We define
$$ A_k := (-infty, 0] cup [frac{1}{k}, k) in sigma(mathcal{F}).$$
We note that
$$ mathbb{R} = bigcup_{kgeq 1} A_k.$$
If we can show that those $A_k$ have finite measure for both $mu$ and $nu$, then we have shown that $mu$ and $nu$ are sigma-finite. Indeed, we have
$$ mu(A_k) = gamma ({ nin mathbb{N} : frac{1}{n} in A_k }) = gamma({1, dots, k}) = k. $$
Similarly, we have
$$ nu(A_k) = k+1 $$
(the plus one comes from $0in A_k$). We conclude that $mu, nu$ are sigma-finite.
2.) Pick $Ain mathcal{F}$. By the very definition of $mathcal{F}$ we have two cases:
Firstly, we consider the case $0in A^o$. This means, that there exists $1>varepsilon>0$ such that $(-varepsilon, varepsilon)subseteq A$. However, then we have
$$ mu(A) = infty = nu(A) $$
as there exists $N>1$ such that for all $ngeq N$ we have $0<frac{1}{n}<varepsilon <frac{1}{N}$ and hence
$$ mu(A) geq mu((-varepsilon, varepsilon)) = gamma({ nin mathbb{N} : n>N })= infty. $$
Similarly one shows that $nu(A)=infty$.
In the second case we have that $0in (mathbb{R}setminus A)^o$. Then we have that $0notin A$ and thus, we have $mu(A)=nu(A)$ as we have by definition
$$ nu(A) = mu(A) + gamma(Acap {0}). $$
c.) For this we recall that the singelons are in the sigma-algebra and hence, we have ${0}in sigma(mathcal{F})$ and
$$ nu({0})= 1 neq 0 = mu({0}).$$
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
add a comment |
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1 Answer
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1 Answer
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1.) We have
$$ (-infty, 0] = bigcap_{ngeq 1} (-infty, frac{1}{n}) in sigma(mathcal{F}) $$
as well as $ [frac{1}{k}, k) $ for all $kin mathbb{N}$. We define
$$ A_k := (-infty, 0] cup [frac{1}{k}, k) in sigma(mathcal{F}).$$
We note that
$$ mathbb{R} = bigcup_{kgeq 1} A_k.$$
If we can show that those $A_k$ have finite measure for both $mu$ and $nu$, then we have shown that $mu$ and $nu$ are sigma-finite. Indeed, we have
$$ mu(A_k) = gamma ({ nin mathbb{N} : frac{1}{n} in A_k }) = gamma({1, dots, k}) = k. $$
Similarly, we have
$$ nu(A_k) = k+1 $$
(the plus one comes from $0in A_k$). We conclude that $mu, nu$ are sigma-finite.
2.) Pick $Ain mathcal{F}$. By the very definition of $mathcal{F}$ we have two cases:
Firstly, we consider the case $0in A^o$. This means, that there exists $1>varepsilon>0$ such that $(-varepsilon, varepsilon)subseteq A$. However, then we have
$$ mu(A) = infty = nu(A) $$
as there exists $N>1$ such that for all $ngeq N$ we have $0<frac{1}{n}<varepsilon <frac{1}{N}$ and hence
$$ mu(A) geq mu((-varepsilon, varepsilon)) = gamma({ nin mathbb{N} : n>N })= infty. $$
Similarly one shows that $nu(A)=infty$.
In the second case we have that $0in (mathbb{R}setminus A)^o$. Then we have that $0notin A$ and thus, we have $mu(A)=nu(A)$ as we have by definition
$$ nu(A) = mu(A) + gamma(Acap {0}). $$
c.) For this we recall that the singelons are in the sigma-algebra and hence, we have ${0}in sigma(mathcal{F})$ and
$$ nu({0})= 1 neq 0 = mu({0}).$$
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
add a comment |
1.) We have
$$ (-infty, 0] = bigcap_{ngeq 1} (-infty, frac{1}{n}) in sigma(mathcal{F}) $$
as well as $ [frac{1}{k}, k) $ for all $kin mathbb{N}$. We define
$$ A_k := (-infty, 0] cup [frac{1}{k}, k) in sigma(mathcal{F}).$$
We note that
$$ mathbb{R} = bigcup_{kgeq 1} A_k.$$
If we can show that those $A_k$ have finite measure for both $mu$ and $nu$, then we have shown that $mu$ and $nu$ are sigma-finite. Indeed, we have
$$ mu(A_k) = gamma ({ nin mathbb{N} : frac{1}{n} in A_k }) = gamma({1, dots, k}) = k. $$
Similarly, we have
$$ nu(A_k) = k+1 $$
(the plus one comes from $0in A_k$). We conclude that $mu, nu$ are sigma-finite.
2.) Pick $Ain mathcal{F}$. By the very definition of $mathcal{F}$ we have two cases:
Firstly, we consider the case $0in A^o$. This means, that there exists $1>varepsilon>0$ such that $(-varepsilon, varepsilon)subseteq A$. However, then we have
$$ mu(A) = infty = nu(A) $$
as there exists $N>1$ such that for all $ngeq N$ we have $0<frac{1}{n}<varepsilon <frac{1}{N}$ and hence
$$ mu(A) geq mu((-varepsilon, varepsilon)) = gamma({ nin mathbb{N} : n>N })= infty. $$
Similarly one shows that $nu(A)=infty$.
In the second case we have that $0in (mathbb{R}setminus A)^o$. Then we have that $0notin A$ and thus, we have $mu(A)=nu(A)$ as we have by definition
$$ nu(A) = mu(A) + gamma(Acap {0}). $$
c.) For this we recall that the singelons are in the sigma-algebra and hence, we have ${0}in sigma(mathcal{F})$ and
$$ nu({0})= 1 neq 0 = mu({0}).$$
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
add a comment |
1.) We have
$$ (-infty, 0] = bigcap_{ngeq 1} (-infty, frac{1}{n}) in sigma(mathcal{F}) $$
as well as $ [frac{1}{k}, k) $ for all $kin mathbb{N}$. We define
$$ A_k := (-infty, 0] cup [frac{1}{k}, k) in sigma(mathcal{F}).$$
We note that
$$ mathbb{R} = bigcup_{kgeq 1} A_k.$$
If we can show that those $A_k$ have finite measure for both $mu$ and $nu$, then we have shown that $mu$ and $nu$ are sigma-finite. Indeed, we have
$$ mu(A_k) = gamma ({ nin mathbb{N} : frac{1}{n} in A_k }) = gamma({1, dots, k}) = k. $$
Similarly, we have
$$ nu(A_k) = k+1 $$
(the plus one comes from $0in A_k$). We conclude that $mu, nu$ are sigma-finite.
2.) Pick $Ain mathcal{F}$. By the very definition of $mathcal{F}$ we have two cases:
Firstly, we consider the case $0in A^o$. This means, that there exists $1>varepsilon>0$ such that $(-varepsilon, varepsilon)subseteq A$. However, then we have
$$ mu(A) = infty = nu(A) $$
as there exists $N>1$ such that for all $ngeq N$ we have $0<frac{1}{n}<varepsilon <frac{1}{N}$ and hence
$$ mu(A) geq mu((-varepsilon, varepsilon)) = gamma({ nin mathbb{N} : n>N })= infty. $$
Similarly one shows that $nu(A)=infty$.
In the second case we have that $0in (mathbb{R}setminus A)^o$. Then we have that $0notin A$ and thus, we have $mu(A)=nu(A)$ as we have by definition
$$ nu(A) = mu(A) + gamma(Acap {0}). $$
c.) For this we recall that the singelons are in the sigma-algebra and hence, we have ${0}in sigma(mathcal{F})$ and
$$ nu({0})= 1 neq 0 = mu({0}).$$
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
1.) We have
$$ (-infty, 0] = bigcap_{ngeq 1} (-infty, frac{1}{n}) in sigma(mathcal{F}) $$
as well as $ [frac{1}{k}, k) $ for all $kin mathbb{N}$. We define
$$ A_k := (-infty, 0] cup [frac{1}{k}, k) in sigma(mathcal{F}).$$
We note that
$$ mathbb{R} = bigcup_{kgeq 1} A_k.$$
If we can show that those $A_k$ have finite measure for both $mu$ and $nu$, then we have shown that $mu$ and $nu$ are sigma-finite. Indeed, we have
$$ mu(A_k) = gamma ({ nin mathbb{N} : frac{1}{n} in A_k }) = gamma({1, dots, k}) = k. $$
Similarly, we have
$$ nu(A_k) = k+1 $$
(the plus one comes from $0in A_k$). We conclude that $mu, nu$ are sigma-finite.
2.) Pick $Ain mathcal{F}$. By the very definition of $mathcal{F}$ we have two cases:
Firstly, we consider the case $0in A^o$. This means, that there exists $1>varepsilon>0$ such that $(-varepsilon, varepsilon)subseteq A$. However, then we have
$$ mu(A) = infty = nu(A) $$
as there exists $N>1$ such that for all $ngeq N$ we have $0<frac{1}{n}<varepsilon <frac{1}{N}$ and hence
$$ mu(A) geq mu((-varepsilon, varepsilon)) = gamma({ nin mathbb{N} : n>N })= infty. $$
Similarly one shows that $nu(A)=infty$.
In the second case we have that $0in (mathbb{R}setminus A)^o$. Then we have that $0notin A$ and thus, we have $mu(A)=nu(A)$ as we have by definition
$$ nu(A) = mu(A) + gamma(Acap {0}). $$
c.) For this we recall that the singelons are in the sigma-algebra and hence, we have ${0}in sigma(mathcal{F})$ and
$$ nu({0})= 1 neq 0 = mu({0}).$$
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
answered Nov 25 '18 at 21:15
Severin Schraven
5,8281934
5,8281934
add a comment |
add a comment |
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you don't have to construct a sequence which the union is equal $sigma(mathcal{F})$, but $X$ instead. As $Xin sigma(mathcal{F})$ the sigma-algebra is a family of sets, not a kind of 'measure subspace'
– Robson
Nov 18 '18 at 22:33
Thank you, I'll make the edit!
– Buttfor
Nov 18 '18 at 22:39
Why do you want $A_k$ to be in $mathcal{F}$? For sigma-finite you only need $A_kin sigma(mathcal{F})$.
– Severin Schraven
Nov 25 '18 at 18:50
In fact, your proposed $A_k = (-k, 0] cup [frac{1}{k}, k)$ works just fine.
– Severin Schraven
Nov 25 '18 at 19:17