TITLE: Unnecessary condition of Lebesgue's monotone convergence theorem?
QUESTION [6 upvotes]: Rudin RCA p.21
Let $(X,\Sigma,\mu)$ be a measure space and $\{f_n\}$ be a sequence of measurable functions on $X$ and suppose that
(a) $0\leq f_1\leq f_2\leq\cdots\leq\infty$ for every $x\in X,$
(b) $f_n\rightarrow f$ pointwise on $X$.
Then $\int_X f_n d\mu \rightarrow \int_X f d\mu$.
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I followed the proof, but cannot see why the condition (b) is essential. Every monotonic sequence in extended real system has a limit as a supremum or infimum, so I think the condition (a) implies the condition (b). Am I wrong? How?

REPLY [6 votes]: If you omitted condition (b), then nothing in the hypothesis tells you what $f$ is.   
Remember that a theorem should be correct no matter what particular values you give its variables; as long as the hypotheses are true, the conclusion must be true also.  Now suppose you chose some reasonable functions as your $f_n$'s, satisfying hypothesis (a), so they converge, but you chose some totally different function as your $f$, not the limit to which the $f_n$'s converge.  For this choice of $f_n$'s and $f$, the conclusion would probably be false (unless you happened to choose a particularly lucky $f$), but all the hypotheses except (b) are true.  
Therefore, if you omit (b), the theorem becomes incorrect.  There are choices of $f_n$'s and $f$ that make the surviving hypotheses true but make the conclusion false.
It is possible to omit hypothesis (b) and compensate for the omission so as to keep the theorem correct. For example, by writing the last integral in the conclusion as $$\int_X\lim_{n\to\infty}f_n\,d\mu.$$