limit of Lebesgue integrable functions

Given two sequences of integrable functions $\{f_{n}\}, \{g_{n}\}$ with limits $f$ and $g$ both also integrable. Does this always hold

$$\lim_{n}(f_{n}-g_{n})=\lim_{n}f_{n}-\lim_{n}g_{n}=f-g$$

I mean what if for some point x, $f(x)=\infty$ and $g(x)=\infty$ that would make $f-g=\infty-\infty$. Then what happened at that point? or in order for a sequence of, in this case, integrable functions one should have that the limit is different to $\infty$ at every point. thanks for the answers beforehand

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What kind of limit are you taking? –  Qiaochu Yuan May 16 '11 at 6:18

In the Lebesgue theory, one generally deals with equivalence classes of functions, rather than with individual functions; maybe a better way of saying this is that when one appears to be referring to a function $f$, one is really referring to the class of functions equivalent to $f$. Functions that differ at just one point are equivalent (indeed, functions that differ at a set of measure zero are equivalent) so it doesn't matter if there's one point $x$ where $f(x)=g(x)=\infty$.
Alternatively, if you insist on dealing with functions, not equivalence classes, then you can't say $f$ is integrable if there's an $x$ with $f(x)=\infty$, because that means $f$ isn't even defined on its putative domain.
I disagree with your second paragraph. It makes perfect sense to look at extended real-valued functions (at the cost of dealing with a vector space before identifying functions if they agree a.e.). Since the integral is defined as a limit of simple functions, we can define the integral even if $f$ takes on the values $\pm \infty$ (as long as the set where these values are attained are negligible) and $f$ agrees a.e. with an integrable function. –  t.b. May 16 '11 at 6:30