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Is the following scenario possible? Provide an example or argue why not.

Let $\{f_n\}_{n=1}^{\infty}$ be measurable non-negative functions on $[0,1]$ converging to $f(x)$ pointwise Lebesgue-almsot everywhere on $[0,1]$ and $\displaystyle \lim_{n \to \infty} \int_{0}^{1} f_n =2$ and $\displaystyle \int_{0}^{1} f =1$.

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Are you sure this is what you want to ask? If $\int_0^1 f_n=1$ then the sequence $\{\int_0^1 f_n\}$ is constant and its limit is $1$. –  Julián Aguirre Jan 31 '13 at 16:59
    
I have now fixed the problem. –  user44069 Jan 31 '13 at 17:01
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1 Answer

up vote 2 down vote accepted

Take the functions from the example here and shift them up by $1$. More precisely, let $f \equiv 1$ and $$ f_n = f+n\cdot1_{\left(0,\frac1n\right]} $$

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I fixed it. Apologies –  user44069 Jan 31 '13 at 17:00
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