> [!theoremb] Theorem > > Let $\seq{f_n} \subset L^+$ be a [[Sequence|sequence]] of [[Measurable Function|measurable functions]], then the [[Limit Superior and Inferior|limit inferior]] can be interchanged with the [[Integral|integral]] under an inequality > $ > \int \liminf f_n \le \liminf \int f_n > $ > *Proof*. See [[Upper and Lower Approximations]]. Let $\phi_n \in \ul{\Sigma}_{\inf_{k \ge n}f_k} \subseteq \ul{\Sigma}_{\liminf f_n}$, then since $f_n \ge \inf_{k \ge n}f_k$, there exists $\psi_n \in \ul{\Sigma}_{f_n}$ such that $\int \phi_n \le \int \psi_n \le \int f_n$. This gives > $ > \begin{align*} > \limv{n}\int \phi_n \le \liminf_{n \to \infty} \int \psi_n &\le \liminf_{n \to \infty} \int f_n \\ > \int \liminf_{n \to \infty}f_n &\le \liminf_{n \to \infty}\int f_n > \end{align*} > $ > [!theoremb]- Why is it an inequality? > > To see why this is an inequality I think I can use the good old sine trick like this. Let $X = [0, 2\pi]$ and $f_j(x) = \sin(x + j) + 1$, then > > ![[fatou_strict.png|400]] > > $ > 0 = \int \liminf f_n \le \liminf \int f_n = 2\pi > $ > > The right-hand side says that the *any limit volume represented by the integral is at least this much*, but it doesn't say how it is distributed. While the $\liminf f_n$ says that *"at any given point", the limiting volume is this much*. Integrating it yields the result of minimising the volume at each point, and then combining them into one minimum volume. Any distribution that minimises mass at each point must also minimise the total volume. > > We can also cut the space into two pieces $X = A \cup B$, then > $ > \liminf\int_A f_n + \liminf\int_B f_n \le \liminf \int f_n > $ > Since there may be limit distributions where all the mass is in $A$, or limit distributions where all the mass is in $B$. The total mass from these two pieces cannot exceed the lower limit, because there are only so much mass to distribute. > [!theorem] > > Let $\seq{f_n} \subset L^+$, $f \in L^+$, and $f_n \to f$ [[Pointwise Convergence|pointwise]] [[Almost Everywhere|a.e.]], then > $ > \int f \le \liminf \int f_n > $ > *Proof*. If $f_n \to f$ everywhere, then this is exactly Fatou's Lemma. Let > $ > N \supset \bracs{x \in X: \limv{n}f_n(x) \ne f(x)} \quad \mu(n) = 0 > $ > then setting > $ > f'_n(x) = \begin{cases} > f_n(x) &x \not\in N \\ > 0 &x = 0 > \end{cases} > $ > and $f' = \limv{n}f_n'(x)$. This gives $f - f' = 0$ a.e., $\int f = \int f'$ and $\int f_n = \int f'_n$ a.e. We can apply Fatou's lemma on $f'$ and $f'_n$ which extends the inequality > $ > \int f = \int f' \le \liminf \int f'_n = \liminf \int f_n > $