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If $(f_n)$ is a pointwise convergent, uniformly bounded sequence of measurable functions on $I:[0,1]$, then its limit $f$ is also measurable.

I think I will use the property that since it is pointwise convergent, we have $\lim \sup a_n = \lim \inf a_n$. But how can I prove the sup/inf is measurable? Is there any proposition could help?

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  • $\begingroup$ Uniform boundedness is not needed. $\endgroup$ – Did Mar 16 '15 at 0:34
  • $\begingroup$ Does the boundedness indicates the existence of sup and inf? $\endgroup$ – user3124 Mar 16 '15 at 3:40
  • $\begingroup$ No, supremum and infimum do not have to be finite (and neither do measurable functions). $\endgroup$ – Did Mar 16 '15 at 8:52
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Let $f = \sup_{n\geq 1} f_n$. Notice that $$f^{-1}((-\infty,a]) = \{x : f_n(x) \leq a, n \geq 1\} = \cap_{n=1}^\infty f_n^{-1}((-\infty,a]),$$ which is a countable sequence of measurable sets, hence measurable.

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