Editing Monotone convergence theorem (section)

==Proof based on Fatou's lemma==
The proof can also be based on Fatou's lemma instead of a direct proof as above, because Fatou's lemma can be proved independent of the monotone convergence theorem.  
However the monotone convergence theorem is in some ways more primitive than Fatou's lemma. It easily follows from the monotone convergence theorem and proof of Fatou's lemma is similar and arguably slightly less natural than the proof above.

As before, measurability follows from the fact that <math display="inline">f = \sup_k f_k = \lim_{k \to \infty} f_k  = \liminf_{k \to \infty}f_k</math> almost everywhere.  The interchange of limits and integrals is then an easy consequence of Fatou's lemma.  One has <math display="block">\int_X f\,d\mu = \int_X \liminf_k f_k\,d\mu \le \liminf \int_X f_k\,d\mu</math> by Fatou's lemma, and then, since  <math>\int f_k \,d\mu \le \int f_{k + 1} \,d\mu \le \int f d\mu</math> (monotonicity), 
<math display="block">\liminf \int_X f_k\,d\mu \le \limsup_k \int_X f_k\,d\mu = \sup_k \int_X f_k\,d\mu \le \int_X f\,d\mu.</math> Therefore 
<math display="block">\int_X f \, d\mu = \liminf_{k \to\infty} \int_X f_k\,d\mu = \limsup_{k \to\infty} \int_X f_k\,d\mu = \lim_{k \to\infty} \int_X f_k \, d\mu = \sup_k \int_X f_k\,d\mu.</math>