Borel-Cantelli-like inequality
Clash Royale CLAN TAG#URR8PPP
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IâÂÂm wondering if the following is true:
If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.
Here $m$ is the Lebesgue measure.
measure-theory
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up vote
0
down vote
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IâÂÂm wondering if the following is true:
If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.
Here $m$ is the Lebesgue measure.
measure-theory
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
IâÂÂm wondering if the following is true:
If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.
Here $m$ is the Lebesgue measure.
measure-theory
IâÂÂm wondering if the following is true:
If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.
Here $m$ is the Lebesgue measure.
measure-theory
asked Aug 17 at 2:28
Seh-kai
1027
1027
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1 Answer
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Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$
add a comment |Â
up vote
3
down vote
accepted
Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$
Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$
answered Aug 17 at 3:51
spaceisdarkgreen
28k21548
28k21548
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