$G(t)$ is a decreasing function and $0\leq G(t) \leq 1$, for all $t \in [0, \infty).$ I have calculate $$\lim_{\theta\to \infty}\int_{0}^{\infty}[G(t)]^\theta \,dt. $$ The answer to the question involve evaluating the integral and then taking the limit to prove the result; but I was wondering if it would be valid to move the integral inside the limit, that is $$\int_{0}^{\infty}\lim_{\theta\to \infty}[G(t)]^\theta \,dt=0.$$ Can you help me?
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Relelvant: https://math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral – Theo Bendit Nov 10 '17 at 05:14
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I remember there are some regularity conditions to do so. So in general, the answer is negative. You may add some more conditions on the function. – ZhuShY Nov 10 '17 at 05:15
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Is $G$ strictly decreasing? Or strictly less than $1$? Otherwise, $G(t) = 1$ would be a counterexample. – Theo Bendit Nov 10 '17 at 05:15
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G is strictly decreasing. Is it true? – ALPHA Nov 10 '17 at 05:31
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If you assume also that the integral is convergent for $\theta = 1$, then yes, the result is true. I do have a way of proving it, but I fear it's a little inelegant. I'd be willing to write up an answer if you want to see it, but I think it'd be better if someone else wrote out a proof. – Theo Bendit Nov 10 '17 at 05:38
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G is survival function for non-negative continuous random variable – ALPHA Nov 10 '17 at 05:52