- #1
Tedjn
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Rudin Theorem 1.14: If fn: X → [-∞,∞] is measurable, for n = 1,2,3,..., and
then g and h are measurable. (I can post more details if necessary.)
I find it amazing that measurable functions handle these limiting processes so well; of course, that's what gives the theory so much power. Logically, the proof follows from the properties of sigma algebras. However, I am wondering if anyone has a more intuitive way of understanding why this theorem is true.
Put another way, it seems to me very non-obvious that the motivating definitions behind sigma algebras would lead to this result. Am I missing a better way of understanding this concept? Is this just the culminating work after lots of dead ends that somehow led to this? Is this something that I will just have to learn to live with and eventually incorporate from frequent exposure?
By the way, this could be the start of many questions, as I am planning on slowly working through Rudin. Thanks for any comments, whatever they may be.
[tex]g = \sup_{n \geq 1} f_n, \qquad h = \limsup_{n \rightarrow \infty} f_n,[/tex]
then g and h are measurable. (I can post more details if necessary.)
I find it amazing that measurable functions handle these limiting processes so well; of course, that's what gives the theory so much power. Logically, the proof follows from the properties of sigma algebras. However, I am wondering if anyone has a more intuitive way of understanding why this theorem is true.
Put another way, it seems to me very non-obvious that the motivating definitions behind sigma algebras would lead to this result. Am I missing a better way of understanding this concept? Is this just the culminating work after lots of dead ends that somehow led to this? Is this something that I will just have to learn to live with and eventually incorporate from frequent exposure?
By the way, this could be the start of many questions, as I am planning on slowly working through Rudin. Thanks for any comments, whatever they may be.