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I fail to see where the finite measure assumption is used in Klenke's characterization of uniform integrability (the part: Uniform Integrability $\implies$ (ii) in the very last paragraph). As i see it, since $0\leq\tilde{g}_{\epsilon/2}\in\mathcal{L}^1\left(\mu\right)$, $B\mapsto\int_B \tilde{g}_{\epsilon/2} d\mu$ defines a finite measure on $\mathcal{B}$ whether or not $\mu\left(\Omega\right)<\infty$. Hence it is possible to find $a_\epsilon$ such that $\int_{\{\tilde{g}_{\epsilon/2}>a_\epsilon\}} \tilde{g}_{\epsilon/2} d\mu<\frac{\epsilon}{2}$ and the rest of the proof doesn't make use of $\mu$'s finiteness.

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Please make your posts self-contained, rather than relying on off-site images. – Did Aug 8 '12 at 21:36
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The phrasing is indeed ambiguous since uniform integrability implies condition (i)-(ii) with no restriction on the finiteness of $\mu$ while condition (i)-(ii) implies uniform integrability only when $\mu$ is finite. When $\mu$ is infinite, condition (i)-(ii) holds for every uniformly bounded family of functions, although nonzero constant functions, for example, are not even integrable in this case hence they cannot belong to any uniformly integrable family. (The trouble is visible when the author mentions the smaller set of constant functions, thus, implicitly considering that constant functions are integrable.)

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I see. Great. Thanks. – Evan Aad Aug 8 '12 at 22:53

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