Limit of measures on set intersection
$begingroup$
While studying measures I was wondering about the following:
Let $(X,mathcal{A},mu)$ be a measure space and assume we have a sequence of sets ${A_n}_{ninmathbb{N}} subset mathcal{A}$ with $A_{n+1} subseteq A_{n}$ such that $mu(A_n) = 1$ for all $nin mathbb{N}$, can we conclude that the following statement is always true:
$mu(bigcap_{ninmathbb{N}} A_n) =lim_{n to infty}mu(A_n) = 1 $ ?
I would say that it is, since the function $f(n):= mu(A_n) = 1$ is constant over $mathbb{N}$ but I'm not sure. Some help would be highly appreciated
I'm re-editing this question because I think I actually came up with a proof:
We first define $I_n:= A_0setminus A_n$. Then we have that:
begin{equation}
begin{split}
mu(bigcup_{nin mathbb{N}} I_n) = mu(bigcup_{nin mathbb{N}}( A_0setminus A_n ) &quad \= mu(A_0setminusbigcap_{nin mathbb{N}}A_n) = mu(A_0) - mu(bigcap_{nin mathbb{N}}A_n) = mu(A_0) - lim_{nto infty}mu(A_n) \ leq sum_{ninmathbb{N}}mu( A_0setminus A_n ) &quad \= sum_{ninmathbb{N}} mu( A_0) - ( A_n ) &quad \= sum_{ninmathbb{N}} (1-1) &quad \= 0 .
end{split}
end{equation}
Finally $mu(A_0) - lim_{nto infty}mu(A_n) = 0$ implies that $lim_{nto infty}mu(A_n) =mu(A_0) = 1$
Does that make sense?
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 17:44
jffijffi
878
$endgroup$
add a comment |
$begingroup$
While studying measures I was wondering about the following:
Let $(X,mathcal{A},mu)$ be a measure space and assume we have a sequence of sets ${A_n}_{ninmathbb{N}} subset mathcal{A}$ with $A_{n+1} subseteq A_{n}$ such that $mu(A_n) = 1$ for all $nin mathbb{N}$, can we conclude that the following statement is always true:
$mu(bigcap_{ninmathbb{N}} A_n) =lim_{n to infty}mu(A_n) = 1 $ ?
I would say that it is, since the function $f(n):= mu(A_n) = 1$ is constant over $mathbb{N}$ but I'm not sure. Some help would be highly appreciated
I'm re-editing this question because I think I actually came up with a proof:
We first define $I_n:= A_0setminus A_n$. Then we have that:
begin{equation}
begin{split}
mu(bigcup_{nin mathbb{N}} I_n) = mu(bigcup_{nin mathbb{N}}( A_0setminus A_n ) &quad \= mu(A_0setminusbigcap_{nin mathbb{N}}A_n) = mu(A_0) - mu(bigcap_{nin mathbb{N}}A_n) = mu(A_0) - lim_{nto infty}mu(A_n) \ leq sum_{ninmathbb{N}}mu( A_0setminus A_n ) &quad \= sum_{ninmathbb{N}} mu( A_0) - ( A_n ) &quad \= sum_{ninmathbb{N}} (1-1) &quad \= 0 .
end{split}
end{equation}
Finally $mu(A_0) - lim_{nto infty}mu(A_n) = 0$ implies that $lim_{nto infty}mu(A_n) =mu(A_0) = 1$
Does that make sense?
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 17:44
jffijffi
878
$endgroup$
add a comment |
$begingroup$
While studying measures I was wondering about the following:
Let $(X,mathcal{A},mu)$ be a measure space and assume we have a sequence of sets ${A_n}_{ninmathbb{N}} subset mathcal{A}$ with $A_{n+1} subseteq A_{n}$ such that $mu(A_n) = 1$ for all $nin mathbb{N}$, can we conclude that the following statement is always true:
$mu(bigcap_{ninmathbb{N}} A_n) =lim_{n to infty}mu(A_n) = 1 $ ?
I would say that it is, since the function $f(n):= mu(A_n) = 1$ is constant over $mathbb{N}$ but I'm not sure. Some help would be highly appreciated
I'm re-editing this question because I think I actually came up with a proof:
We first define $I_n:= A_0setminus A_n$. Then we have that:
begin{equation}
begin{split}
mu(bigcup_{nin mathbb{N}} I_n) = mu(bigcup_{nin mathbb{N}}( A_0setminus A_n ) &quad \= mu(A_0setminusbigcap_{nin mathbb{N}}A_n) = mu(A_0) - mu(bigcap_{nin mathbb{N}}A_n) = mu(A_0) - lim_{nto infty}mu(A_n) \ leq sum_{ninmathbb{N}}mu( A_0setminus A_n ) &quad \= sum_{ninmathbb{N}} mu( A_0) - ( A_n ) &quad \= sum_{ninmathbb{N}} (1-1) &quad \= 0 .
end{split}
end{equation}
Finally $mu(A_0) - lim_{nto infty}mu(A_n) = 0$ implies that $lim_{nto infty}mu(A_n) =mu(A_0) = 1$
Does that make sense?
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 17:44
jffijffi
878
$endgroup$
While studying measures I was wondering about the following:
Let $(X,mathcal{A},mu)$ be a measure space and assume we have a sequence of sets ${A_n}_{ninmathbb{N}} subset mathcal{A}$ with $A_{n+1} subseteq A_{n}$ such that $mu(A_n) = 1$ for all $nin mathbb{N}$, can we conclude that the following statement is always true:
$mu(bigcap_{ninmathbb{N}} A_n) =lim_{n to infty}mu(A_n) = 1 $ ?
I would say that it is, since the function $f(n):= mu(A_n) = 1$ is constant over $mathbb{N}$ but I'm not sure. Some help would be highly appreciated
I'm re-editing this question because I think I actually came up with a proof:
We first define $I_n:= A_0setminus A_n$. Then we have that:
begin{equation}
begin{split}
mu(bigcup_{nin mathbb{N}} I_n) = mu(bigcup_{nin mathbb{N}}( A_0setminus A_n ) &quad \= mu(A_0setminusbigcap_{nin mathbb{N}}A_n) = mu(A_0) - mu(bigcap_{nin mathbb{N}}A_n) = mu(A_0) - lim_{nto infty}mu(A_n) \ leq sum_{ninmathbb{N}}mu( A_0setminus A_n ) &quad \= sum_{ninmathbb{N}} mu( A_0) - ( A_n ) &quad \= sum_{ninmathbb{N}} (1-1) &quad \= 0 .
end{split}
end{equation}
Finally $mu(A_0) - lim_{nto infty}mu(A_n) = 0$ implies that $lim_{nto infty}mu(A_n) =mu(A_0) = 1$
Does that make sense?
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Jan 7 at 17:44
jffijffi
878
asked Jan 7 at 17:44
jffijffi
878
edited Jan 7 at 18:23
jffi
asked Jan 7 at 17:44
jffijffi
878
asked Jan 7 at 17:44
jffijffi
878
asked Jan 7 at 17:44
jffijffi
878
878
add a comment |
add a comment |
1 Answer
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$begingroup$
Under your monotonicity assumption ($A_{n+1} subset A_n$ for every $n$) and the assumption that $mu(A_N) < +infty$ for some $N$ (and hence for every $ngeq N$), you always have that
$$
muleft(bigcap_n A_nright) = lim_n mu(A_n).
$$
(See for example here for the analogous result for increasing sequences of sets.)
answered Jan 7 at 18:06
RigelRigel
11.2k11320
$endgroup$
$begingroup$
So $lim mu (A_n)$ = 1?
$endgroup$
– jffi
Jan 7 at 18:11
add a comment |
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$begingroup$
Under your monotonicity assumption ($A_{n+1} subset A_n$ for every $n$) and the assumption that $mu(A_N) < +infty$ for some $N$ (and hence for every $ngeq N$), you always have that
$$
muleft(bigcap_n A_nright) = lim_n mu(A_n).
$$
(See for example here for the analogous result for increasing sequences of sets.)
answered Jan 7 at 18:06
RigelRigel
11.2k11320
$endgroup$
$begingroup$
So $lim mu (A_n)$ = 1?
$endgroup$
– jffi
Jan 7 at 18:11
add a comment |
$begingroup$
Under your monotonicity assumption ($A_{n+1} subset A_n$ for every $n$) and the assumption that $mu(A_N) < +infty$ for some $N$ (and hence for every $ngeq N$), you always have that
$$
muleft(bigcap_n A_nright) = lim_n mu(A_n).
$$
(See for example here for the analogous result for increasing sequences of sets.)
answered Jan 7 at 18:06
RigelRigel
11.2k11320
$endgroup$
$begingroup$
So $lim mu (A_n)$ = 1?
$endgroup$
– jffi
Jan 7 at 18:11
add a comment |
$begingroup$
Under your monotonicity assumption ($A_{n+1} subset A_n$ for every $n$) and the assumption that $mu(A_N) < +infty$ for some $N$ (and hence for every $ngeq N$), you always have that
$$
muleft(bigcap_n A_nright) = lim_n mu(A_n).
$$
(See for example here for the analogous result for increasing sequences of sets.)
answered Jan 7 at 18:06
RigelRigel
11.2k11320
$endgroup$
Under your monotonicity assumption ($A_{n+1} subset A_n$ for every $n$) and the assumption that $mu(A_N) < +infty$ for some $N$ (and hence for every $ngeq N$), you always have that
$$
muleft(bigcap_n A_nright) = lim_n mu(A_n).
$$
(See for example here for the analogous result for increasing sequences of sets.)
answered Jan 7 at 18:06
RigelRigel
11.2k11320
answered Jan 7 at 18:06
RigelRigel
11.2k11320
answered Jan 7 at 18:06
RigelRigel
11.2k11320
answered Jan 7 at 18:06
RigelRigel
11.2k11320
11.2k11320
$begingroup$
So $lim mu (A_n)$ = 1?
$endgroup$
– jffi
Jan 7 at 18:11
add a comment |
$begingroup$
So $lim mu (A_n)$ = 1?
$endgroup$
– jffi
Jan 7 at 18:11
$begingroup$
So $lim mu (A_n)$ = 1?
$endgroup$
– jffi
Jan 7 at 18:11
$begingroup$
So $lim mu (A_n)$ = 1?
$endgroup$
– jffi
Jan 7 at 18:11
add a comment |
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