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Wikipedia article states that to obatin this lemma, the functions $g-f_n$ have to be considered (where $f_n \leq g$). However, the difference might not exist for some elements (e.g. $\infty - \infty$ or $-\infty - (-\infty)$). How is this problem circumvented?

Thanks, Phanindra

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In the Wikipedia article, $g$ is required to be integrable. This implies that the set $\{x \in S\colon g(x) = \pm \infty\}$ has measure zero.

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Thanks! I did not quite understand how integrability plays a role until I saw your answer. –  jpv Oct 29 '11 at 6:22
    
Supposing $g$ takes $\infty$ on a set of measure greater than zero. As $g$ is integrable the value of its integral is $\infty$. So, it seems that $g$ can take the extreme values on a set of positive measure. –  jpv Oct 29 '11 at 6:38
    
@jpv: No, a measurable function is defined to be integrable iff its integral exists and is finite. (Admittedly, it is kind of a strange convention...) –  Jesse Madnick Oct 29 '11 at 7:07
    
Thanks! The book I am following (books.google.com/books/about/…) defines a function to be integrable if either the integral or the negative or the positive part is finite. In general, this implies that the integral of the function can be +infinity or -infinity. However, the dominated convergence theorem is slightly different from the one in wikipedia. The authors specifically want finiteness probably to avoid the problem in the question. I believe I have interpreted integrability erroneously in the wikipedia page. –  jpv Oct 30 '11 at 1:59
    
Even if we impose finiteness (i.e. require that $\int_X|g| d\mu < \infty$) it is still weird that we can define integrals for functions $g-f_n$ which are undefined on a set of points in the domain. I am aware of the convention that $0*\infty = 0$ but nothing about $0 *(\infty - \infty)$. –  jpv Oct 30 '11 at 2:07

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