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Let $f_{n}$ be positive measurable functions that converge pointwise to $f$ and such that $ \int f = \lim \int f_{n} < \infty$ Prove that for any measurable set $E$, we have $ \int f = \lim \int f_{n}$ if both integrals are taken over the set $E$.

Any ideas? thanks

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    Hint: http://en.wikipedia.org/wiki/Dominated_convergence_theorem2011-11-13
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    No, there's no assumption that these are dominated by an integrable function.2011-11-13
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    @Martin: The dominated convergence theorem is not good enough, as Robert points out. You need to generalize it slightly using Fatou since you don't have a dominating function.2011-11-13
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    @t.b. I thought that I can take $f$ as the dominating function in this special case. (The post contains and assumption $\int_X f = \lim_X \int f_{n} < \infty$, which means that $f$ is integrable; at least that how I understood it.)2011-11-13
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    @Martin: but you don't know that $f_n \leq f$.2011-11-13
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    @t.b. Ok, I understand. (I should not have missed that.)2011-11-13

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Hint: try using Fatou's lemma.

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    got that thank you. but what is the relevance of the less than infinity? is there a counter example where the integral is equal to infinity?2011-11-13
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    @alice: Take your favourite example of positive $f_n$ converging pointwise to $f$ on, say, $[0,1]$ with $\lim \int_0^1 f_n(x)\ dx \ne\int_0^1 f(x)\ dx$, and take $f_n(x) = f(x) = 1$ on $[1,\infty)$. Let $E = [0,1]$.2011-11-13