If $\sup_n f_n$ is integrable, does equality hold in Fatou's Lemma? I am studying measure theory and I have a question regarding equality in Fatou's Lemma:
For any sequence of functions $\left\{ f_n \right\}$ that are in $L^+$,
$$\int \liminf_{n\rightarrow \infty} f_n d\mu \leq \liminf_{n\rightarrow \infty} \int f_n d\mu $$
I am wondering if there are any sufficient conditions for equality to hold.
I saw in this answer that if $\sup_n f_n$ is integrable, then equality holds where Davide quotes Dominated Converges Theorem for that.
https://math.stackexchange.com/a/545298/77767
However, DCT only works when the sequence of functions converges, I cannot see how that can be applied here. Is there any version of DCT that is stated using $\liminf$ instead of using $\lim$?
Any help will be appreciated!
 A: Equality rarely holds without convergence of $(f_n)$. Let $f_n=f$ for $n$ even and $g$ for $n$ odd. Then equality would mean $\int (f \wedge g)=(\int f )\wedge (\int g)$ which is not true in general.
Note: $\int (f \wedge g)=(\int f )\wedge (\int g)$ if and only if either $f \leq g$ a.e. or $g \leq f $ a.e. Proof: $(\int f )\wedge (\int g)$  is either $\int f$ of $\int g$. Suppose it is $\int f$. Then $\int (f-f\wedge g)=0$ and $f-f\wedge g$ is a non-negative measurable function. Hence, it is $0$ a.e. But $f=f\wedge g$ a.e. implies $f \leq g $ a.e. In  the case when $(\int f )\wedge (\int g) =\int g$ we get $g \leq f$ a.e.
Thus, you get a  counter-example by taking $f=\chi_A,g=\chi_B$ where $A$ and $B$ are any two disjoint sets with positive measure, say two disjoint open intervals.
A: Equality still need not hold. Consider $X=[0,2]$ with usual Lebesgue measure, and
\begin{align}
f_n=
\begin{cases}
\chi_{[0,1]}&\text{if $n$ odd}\\
\chi_{[1,2]}&\text{if $n$ even}
\end{cases}
\end{align}
Draw a sketch of these functions. So $f_n$ alternates between a 'left-step' and a 'right-step' function. Then we have $\liminf f_n = 0$, and $\int f_n = 1$ for all $n$, so we have a strict inequality in Fatou $0<1$.
What we can say is that if $\sup f_n$ is integrable, then
\begin{align}
\int \liminf f_n\leq \liminf \int f_n \leq \limsup\int f_n \leq \int \limsup f_n.
\end{align}
The first inequality is usual Fatou, second always holds (a basic property about numbers); the final step requires the domination assumption; see this answer
for further remarks. Note that this chain of inequalities actually implies DCT because if we assume further that $\lim f_n$ exists, then $\liminf f_n = \limsup f_n$ (again a basic property of real numbers applied pointwise: limit exists in $[-\infty,\infty]$ if and only if $\liminf$ and $\limsup$ are equal), and thus the first and last term of the inequality are the same thing, so all inequalities can be replaced by equalities.
Fatou's lemma is really just a general observation about integrals and limiting operations (and to be fully general we only consider $\liminf$). However, because of the weaker hypothesis (not-necessarily assuming convergence) the conclusions are weaker. Going one step further of making the domination assumption gives us the reverse Fatou inequality (the final inequality in the chain above). Going yet one step further and assuming pointwise a.e. convergence yields DCT.
