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 November 21st, 2013, 04:55 PM #1 Newbie   Joined: Sep 2013 Posts: 8 Thanks: 0 measure Regarding this statement: http://i.imgur.com/F9YyQKi.png I am having trouble incorporating $\lambda (A_{1}) \, < \, \infty$ in my proof. I understand why this must be true, but wouldn't the statement also still be true as long as $\lambda (A_{k}) \, < \, \infty$ for any k? Either way, I still don't know how to use it.. When trying to prove it Ive started by saying $A_{n} \subset A_{k} \; \forall k \, < \, n \; \rightarrow \; {\bigcap}_{k=1}^n A_{k}\, = \, A_{n} \rightarrow \lambda (A_{n}) \, = \, \lambda ({\bigcap}_{k=1}^n A_{k}) \rightarrow \lim_{n \to \infty} \lambda (A_{n}) \, = \, \lim_{n \to \infty} \lambda ({\bigcap}_{k=1}^n A_{k}) \, = \, \lambda ({\bigcap}_{k=1}^\infty A_{k})$ Obviously ive made some mistake along the way because my proof doesn't use $\lambda (A_{1}) \, < \, \infty$, can anyone point me in the right direction?
November 21st, 2013, 10:51 PM   #2
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 Originally Posted by DuncanThaw but wouldn't the statement also still be true as long as for any k?
- you are right, this is exactly what you need. If $meas(A_1)$ were infinite, then you could start from the $k$ for which $meas(A_k) < \infty$ because the intersection of A_m's from m = 1 to infinity and from m=k to infinity are the same. The reason you're told that $meas(A_1) < \infty$ is just so that you won't get a result like "$\infty= \infty$".

 November 22nd, 2013, 08:55 AM #3 Newbie   Joined: Sep 2013 Posts: 8 Thanks: 0 Re: measure Thanks for your response, can you point out the error in my proof? I did not use the assumption that $\lambda (A_{1})< \infty$ so it must be wrong in some way..
 November 22nd, 2013, 12:37 PM #4 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: measure What I suggest is that you justify the last step you took, where the limit has been passed into the measure $\lambda$ without explanation. You should justify doing this because some measures aren't exactly continuous functions on the measure space. Because $g_n := \lambda(\cap_{k=1}^n A_k)$ is a decreasing sequence of non-negative real numbers (by $\sigma$-additivity and non-negativity of all measures) then $lim_{n \rightarrow \infty} g_n$ exists and is finite. This is what permits us to pass the limit.
 November 23rd, 2013, 11:02 AM #5 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: measure DuncanThaw, alas I suggest that you write $\lambda(lim_{n\rightarrow \infty}A_n)$ instead of $lim_{n\rightarrow \infty}\lambda(A_n)$ to completely avoid passing the limit into the measure (I'm not at all comfortable with doing this). Still justify finiteness of the measure due to $\sigma$-additivity and non-negativeness of all measures.

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