Mixing
Verification of argument in Corollary 3.3 Stein and Shakarchi Real analysis
$begingroup$
I thoght Proof of Corollary 3.3 (ii) is not complete.
As E is not shown to be measurable there.
I thought following argument . As the complement of a measurable set is measurable then $E_1^c, E_2^c....$ are measurable.
this sequence is increasing like (i)
Using that $m(E^c)=lim_{Ntoinfty}m(E_N^c)$
and $E^c=cup E^c_N$ so it is measurable.
Complement is also measurable
SO E is measurable.
Please See my argument. If there is any other argument please suggest.
measure-theory proof-verification lebesgue-measure
edited Feb 3 at 14:04
Fozoro
1266
asked Feb 3 at 12:50
SRJSRJ
1,8981620
$endgroup$
add a comment |
$begingroup$
I thoght Proof of Corollary 3.3 (ii) is not complete.
As E is not shown to be measurable there.
I thought following argument . As the complement of a measurable set is measurable then $E_1^c, E_2^c....$ are measurable.
this sequence is increasing like (i)
Using that $m(E^c)=lim_{Ntoinfty}m(E_N^c)$
and $E^c=cup E^c_N$ so it is measurable.
Complement is also measurable
SO E is measurable.
Please See my argument. If there is any other argument please suggest.
measure-theory proof-verification lebesgue-measure
edited Feb 3 at 14:04
Fozoro
1266
asked Feb 3 at 12:50
SRJSRJ
1,8981620
$endgroup$
$begingroup$
$m^*(E) = m(E)$ only if $E$ is measurable.
$endgroup$
– Dbchatto67
Feb 3 at 12:56
add a comment |
$begingroup$
I thoght Proof of Corollary 3.3 (ii) is not complete.
As E is not shown to be measurable there.
I thought following argument . As the complement of a measurable set is measurable then $E_1^c, E_2^c....$ are measurable.
this sequence is increasing like (i)
Using that $m(E^c)=lim_{Ntoinfty}m(E_N^c)$
and $E^c=cup E^c_N$ so it is measurable.
Complement is also measurable
SO E is measurable.
Please See my argument. If there is any other argument please suggest.
measure-theory proof-verification lebesgue-measure
edited Feb 3 at 14:04
Fozoro
1266
asked Feb 3 at 12:50
SRJSRJ
1,8981620
$endgroup$
I thoght Proof of Corollary 3.3 (ii) is not complete.
As E is not shown to be measurable there.
I thought following argument . As the complement of a measurable set is measurable then $E_1^c, E_2^c....$ are measurable.
this sequence is increasing like (i)
Using that $m(E^c)=lim_{Ntoinfty}m(E_N^c)$
and $E^c=cup E^c_N$ so it is measurable.
Complement is also measurable
SO E is measurable.
Please See my argument. If there is any other argument please suggest.
measure-theory proof-verification lebesgue-measure
measure-theory proof-verification lebesgue-measure
edited Feb 3 at 14:04
Fozoro
1266
asked Feb 3 at 12:50
SRJSRJ
1,8981620
edited Feb 3 at 14:04
Fozoro
1266
asked Feb 3 at 12:50
SRJSRJ
1,8981620
edited Feb 3 at 14:04
Fozoro
1266
edited Feb 3 at 14:04
Fozoro
1266
edited Feb 3 at 14:04
Fozoro
1266
1266
asked Feb 3 at 12:50
SRJSRJ
1,8981620
asked Feb 3 at 12:50
SRJSRJ
1,8981620
asked Feb 3 at 12:50
SRJSRJ
1,8981620
1,8981620
$begingroup$
$m^*(E) = m(E)$ only if $E$ is measurable.
$endgroup$
– Dbchatto67
Feb 3 at 12:56
add a comment |
$begingroup$
$m^*(E) = m(E)$ only if $E$ is measurable.
$endgroup$
– Dbchatto67
Feb 3 at 12:56
$begingroup$
$m^*(E) = m(E)$ only if $E$ is measurable.
$endgroup$
– Dbchatto67
Feb 3 at 12:56
$begingroup$
$m^*(E) = m(E)$ only if $E$ is measurable.
$endgroup$
– Dbchatto67
Feb 3 at 12:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$E$ is shown to be a measurable set here; from the marked identity:
$$E = E_1 setminus bigcup_{k=1}^infty G_i$$
$E_1$ is measurable and so are the $G_i$ as differences of measurable sets so $E$ is measurable. I don't see any incompleteness. The author could have made it more explicit perhaps. But from the start $E=bigcap E_n$ is clearly measurable when all $E_n$ are. (That's what $E_n searrow E$ means.)
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
$begingroup$
Sir (ii) part . $E_1=Ecup cup G_k$
$endgroup$
– SRJ
Feb 3 at 12:55
$begingroup$
@SRJ see my edit.
$endgroup$
– Henno Brandsma
Feb 3 at 12:56
$begingroup$
Sorry Sir. I understood now. $E=E_1cap (cup G_k)^c$
$endgroup$
– SRJ
Feb 3 at 12:58
$begingroup$
@SRJ also $E=bigcap_n E_n$, which is easier still.
$endgroup$
– Henno Brandsma
Feb 3 at 12:59
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$E$ is shown to be a measurable set here; from the marked identity:
$$E = E_1 setminus bigcup_{k=1}^infty G_i$$
$E_1$ is measurable and so are the $G_i$ as differences of measurable sets so $E$ is measurable. I don't see any incompleteness. The author could have made it more explicit perhaps. But from the start $E=bigcap E_n$ is clearly measurable when all $E_n$ are. (That's what $E_n searrow E$ means.)
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
$begingroup$
Sir (ii) part . $E_1=Ecup cup G_k$
$endgroup$
– SRJ
Feb 3 at 12:55
$begingroup$
@SRJ see my edit.
$endgroup$
– Henno Brandsma
Feb 3 at 12:56
$begingroup$
Sorry Sir. I understood now. $E=E_1cap (cup G_k)^c$
$endgroup$
– SRJ
Feb 3 at 12:58
$begingroup$
@SRJ also $E=bigcap_n E_n$, which is easier still.
$endgroup$
– Henno Brandsma
Feb 3 at 12:59
add a comment |
$begingroup$
$E$ is shown to be a measurable set here; from the marked identity:
$$E = E_1 setminus bigcup_{k=1}^infty G_i$$
$E_1$ is measurable and so are the $G_i$ as differences of measurable sets so $E$ is measurable. I don't see any incompleteness. The author could have made it more explicit perhaps. But from the start $E=bigcap E_n$ is clearly measurable when all $E_n$ are. (That's what $E_n searrow E$ means.)
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
$begingroup$
Sir (ii) part . $E_1=Ecup cup G_k$
$endgroup$
– SRJ
Feb 3 at 12:55
$begingroup$
@SRJ see my edit.
$endgroup$
– Henno Brandsma
Feb 3 at 12:56
$begingroup$
Sorry Sir. I understood now. $E=E_1cap (cup G_k)^c$
$endgroup$
– SRJ
Feb 3 at 12:58
$begingroup$
@SRJ also $E=bigcap_n E_n$, which is easier still.
$endgroup$
– Henno Brandsma
Feb 3 at 12:59
add a comment |
$begingroup$
$E$ is shown to be a measurable set here; from the marked identity:
$$E = E_1 setminus bigcup_{k=1}^infty G_i$$
$E_1$ is measurable and so are the $G_i$ as differences of measurable sets so $E$ is measurable. I don't see any incompleteness. The author could have made it more explicit perhaps. But from the start $E=bigcap E_n$ is clearly measurable when all $E_n$ are. (That's what $E_n searrow E$ means.)
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
$E$ is shown to be a measurable set here; from the marked identity:
$$E = E_1 setminus bigcup_{k=1}^infty G_i$$
$E_1$ is measurable and so are the $G_i$ as differences of measurable sets so $E$ is measurable. I don't see any incompleteness. The author could have made it more explicit perhaps. But from the start $E=bigcap E_n$ is clearly measurable when all $E_n$ are. (That's what $E_n searrow E$ means.)
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
edited Feb 3 at 12:58
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
answered Feb 3 at 12:53
Henno BrandsmaHenno Brandsma
117k349127
117k349127
$begingroup$
Sir (ii) part . $E_1=Ecup cup G_k$
$endgroup$
– SRJ
Feb 3 at 12:55
$begingroup$
@SRJ see my edit.
$endgroup$
– Henno Brandsma
Feb 3 at 12:56
$begingroup$
Sorry Sir. I understood now. $E=E_1cap (cup G_k)^c$
$endgroup$
– SRJ
Feb 3 at 12:58
$begingroup$
@SRJ also $E=bigcap_n E_n$, which is easier still.
$endgroup$
– Henno Brandsma
Feb 3 at 12:59
add a comment |
$begingroup$
Sir (ii) part . $E_1=Ecup cup G_k$
$endgroup$
– SRJ
Feb 3 at 12:55
$begingroup$
@SRJ see my edit.
$endgroup$
– Henno Brandsma
Feb 3 at 12:56
$begingroup$
Sorry Sir. I understood now. $E=E_1cap (cup G_k)^c$
$endgroup$
– SRJ
Feb 3 at 12:58
$begingroup$
@SRJ also $E=bigcap_n E_n$, which is easier still.
$endgroup$
– Henno Brandsma
Feb 3 at 12:59
$begingroup$
Sir (ii) part . $E_1=Ecup cup G_k$
$endgroup$
– SRJ
Feb 3 at 12:55
$begingroup$
Sir (ii) part . $E_1=Ecup cup G_k$
$endgroup$
– SRJ
Feb 3 at 12:55
$begingroup$
@SRJ see my edit.
$endgroup$
– Henno Brandsma
Feb 3 at 12:56
$begingroup$
@SRJ see my edit.
$endgroup$
– Henno Brandsma
Feb 3 at 12:56
$begingroup$
Sorry Sir. I understood now. $E=E_1cap (cup G_k)^c$
$endgroup$
– SRJ
Feb 3 at 12:58
$begingroup$
Sorry Sir. I understood now. $E=E_1cap (cup G_k)^c$
$endgroup$
– SRJ
Feb 3 at 12:58
$begingroup$
@SRJ also $E=bigcap_n E_n$, which is easier still.
$endgroup$
– Henno Brandsma
Feb 3 at 12:59
$begingroup$
@SRJ also $E=bigcap_n E_n$, which is easier still.
$endgroup$
– Henno Brandsma
Feb 3 at 12:59
add a comment |
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$begingroup$
$m^*(E) = m(E)$ only if $E$ is measurable.
$endgroup$
– Dbchatto67
Feb 3 at 12:56