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Let $\mu$ finite measure then $A_n\rightarrow A\implies \mu(A_n)\rightarrow \mu(A)$.

My Attempt:

$$ \mu(A)=\mu(\lim_{n\rightarrow\infty}\sup A_n) =\mu\left(\lim_{n\rightarrow\infty}\bigcup^\infty_{k=n} A_k\right) =\lim_{n\rightarrow\infty}\mu\left(\bigcup^\infty_{k=n} A_k\right) =\lim_{n\rightarrow\infty}\sum^\infty_{k=n} \mu(A_k) =\lim_{n\rightarrow\infty} \mu(A_n) $$

Is this correct?

VIVID
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gbd
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  • https://math.stackexchange.com/questions/2439848/show-that-if-a-n-to-a-then-pa-n-to-pa/2439884#2439884 –  Oct 21 '21 at 16:18
  • @d.k.o. I think this only works for a monotone sequence. – gbd Oct 21 '21 at 16:20
  • What only works for monotone sequences? You have $A_n\to A$ which means that $\limsup_n A_n=A=\liminf_n A_n$. –  Oct 21 '21 at 18:22
  • @d.k.o. The link you provided is for an increasing sequence. But I have to prove for a general convergent squence. – gbd Oct 21 '21 at 18:26
  • Nope. The question is about general converging sequences. –  Oct 21 '21 at 18:35

1 Answers1

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It might be wise to provide more concrete assumptions if there are any. Based on your 2nd equality where you write that the measure of the limsup is equal to the measure of the limit of unions, you may be missing some conditions on $A_k$. I say this because the definition of the lim sup for sets is defined as $\displaystyle\limsup_{n\rightarrow\infty} A_n=\bigcap_{n\geq 1}\bigcup_{k\geq n} A_k$.

Furthermore, it seems that you are attempting to use continuity from below. DO you have the conditions for this?

JJ Hoo
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  • I dont have the conditions. This is just my attempt to the question. – gbd Oct 21 '21 at 15:54
  • If there are no conditions, this will not be true in general. The measure of all sets $A_n$ must be finite. If they are, then I recommend trying to use the Dominated Convergence Theorem to get the result. – JJ Hoo Oct 21 '21 at 16:00
  • @JJHoo Did you read the first four words in the statement of the problem??? – David C. Ullrich Oct 21 '21 at 16:33