# Two questions on Fatou's Lemma

While reading the following paragraph from Real Analysis by Stein (I hope this does not breach any copyright; if so, I have to type it out), two questions occurred to me.

1. In the proof of Fauto's lemma, it reads $\int g_n \leq \int f_n$, for all $n$, and therefore, $$\int g = \lim_n \int g_n \leq \liminf_n \int f_n.(+)$$ I understand that the reason why $\lim_n g_n$ exists is that $g_n$ is bounded and measurable on a finite support. And in addition, if $\lim_n g_n$ exists, then one has $$\lim_n g_n = \liminf_n g_n = \limsup_n g_n. (*)$$ However, I do NOT understand why one can take limit on the left hand side of $(+)$ and take liminf on the right hand side of $(+)$. I do NOT think one can use $(*)$ to adjustify $(+)$. If so, one might replace $\liminf_n f_n$ by $\limsup_n f_n$. That is, the theorem becomes $$\int g = \lim_n \int g_n \leq \limsup_n \int f_n. (++)$$ Is $(++)$ still true? Could anyone clarify this, please?
2. Corollary $1.8$ seems to me stronger than the usual Monotone Convergence Theorem (DCT) since all it needs is to have $f_n \leq f$ in this corollary (plus pointwise convergence of course). There is no requirement on how $f_n$ should converge to $f$. To be more specific, $f_n$ certainly does NOT need to increasing at all. My question is then why one always works with DCT. Isn't it natural to work with a theorem which imposes looser constraint, please? Thank you! • 1. If $a_n\leq b_n$, then $\liminf a_n \leq \liminf b_n$ 2. It's a (version of) dominated convergence theorem, not MCT. Different theorem. In my (admittedly, limited) experience, indeed more easily applicable and used more often. Jun 18 '14 at 8:05

1. For two real sequences $a_n \leq b_n$, there always results $\liminf_n a_n \leq \liminf_n b_n$. In your case, $\liminf_n a_n = \lim_n a_n$.
• As for 1, if what you said is true, then can I write the theorem as $\int g = \limsup \int g_n \leq \limsup \int f_n$, please? Jun 18 '14 at 8:46
• @20824 Note that $$\liminf \int f_n \leq \limsup \int f_n$$ holds in any case.
• Since $\inf \leq \sup$, it is wiser to use the best inequality, since $\sup = +\infty$ is allowed. And I can't understand question 2, again. Jun 18 '14 at 9:05