$\int\liminf f_n\le\liminf\int f_n$; if $f_n\le g$ with $g\in L^1$, then $\limsup\int f_n\le\int\limsup f_n$.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space and let \(f_n\colon X\to[0,\infty]\) be measurable for \(n\geq1\).
(Fatou’s lemma [@Fol13]) \[ \liminf_{n\to\infty} \int_X f_n\dmu \geq \int_X \liminf_{n\to\infty} f_n\dmu . \]
(Reverse Fatou’s lemma) If moreover there is an integrable \(g\colon X\to[0,\infty)\) with \(f_n\leq g\) for all \(n\), then \[ \limsup_{n\to\infty} \int_X f_n\dmu \leq \int_X \limsup_{n\to\infty} f_n\dmu . \]
The domination hypothesis in (2) cannot be dropped. On \(\RR\) with Lebesgue measure, \(f_n\coloneqq\chi_{[n,n+1]}\) satisfies \(\int f_n=1\) for all \(n\), while \(\limsup_n f_n=0\) pointwise, so \(\limsup_n\int f_n=1>0=\int\limsup_n f_n\).