Proof of Fatou-Lebesgue Theorem

Question

Good evening everyone,

how can I prove the following inequality?

Let $f,f_n\in L(X,\mu)$ on a measure space with $0\leq f_n(x)\leq f(x)$. If $f,f_n$ are $\mu$-almost everywhere in $X$, then is

$$\int_X\liminf_{n\to\infty} f_n d\mu \leq \liminf_{n\to\infty}\int_X f_n d\mu\leq\limsup_{n\to\infty}\int_X f_n d\mu\leq \int_X\limsup_{n\to\infty} f_nd\mu.$$

How can I prove it?

Firstable, I know that

$$\liminf_{n\to\infty} \int_X f_n d\mu \leq \limsup_{n\to\infty}\int_X f_n d\mu.$$

But I don't know how I can prove the inequality. Could someone help me please?

Answer 1

Okay, I've tried to prove the "big inequality", so:

We have the property that $f,f_n$ are measurable and $0\leq f_n(x)\leq f(x)$. Let be

$$a_n:=f+f_n,\qquad b_n := f-f_n.$$ Then, $a_n,b_n$ are real and integrable because of $f$ and $f_n$. So, $$0\leq a_n\leq 2\cdot f,\ 0\leq b_n\leq a_n\leq 2\cdot f.$$ Now we consider $\liminf$ of the sequences $a_n,b_n$: $$\begin{align} \liminf_{n\to\infty} a_n &= f+\liminf_{n\to\infty} f_n,\\ \liminf_{n\to\infty} b_n &= f+\liminf_{n\to\infty} (-f_n) = f-\limsup_{n\to\infty} f_n. \end{align}$$ By Fatou's lemma is $$\int_X\liminf_{n\to\infty} f_n d\mu \leq \liminf_{n\to\infty} \int_X f_n d\mu.$$ So: $$\begin{align} \int_X\liminf_{n\to\infty} a_n d\mu &= \int_X(f+\liminf_{n\to\infty} f_n)d\mu\\ &=\int_X fd\mu+\int_X\liminf_{n\to\infty} f_n d\mu\\ \Leftrightarrow\int_X\liminf_{n\to\infty} f_n d\mu &=\int_X\liminf_{n\to\infty} a_n d\mu-\int_X fd\mu\\ &\leq\liminf_{n\to\infty}\int_X a_nd\mu -\int_X fd\mu\\ &=\liminf_{n\to\infty}\int_X (f+f_n) d\mu-\int_X fd\mu\\ &=\int_X fd\mu-\int_X fd\mu+\liminf_{n\to\infty}\int_X f_nd\mu\\ &=\liminf_{n\to\infty}\int_X f_nd\mu. \end{align}$$

Now for $b_n$:

$$\begin{align} \int_X\liminf_{n\to\infty} b_n d\mu &= \int_X(f-\limsup_{n\to\infty} f_n)d\mu\\ &=\int_X fd\mu-\int_X\limsup_{n\to\infty} f_n d\mu\\ \Leftrightarrow\int_X\limsup_{n\to\infty} f_n d\mu &=-\int_X\liminf_{n\to\infty} b_n d\mu+\int_X fd\mu\\ &\geq\int_X fd\mu-\liminf_{n\to\infty}\int_X b_nd\mu\\ &=\int_X fd\mu-\liminf_{n\to\infty}\int_X (f-f_n)d\mu\\ &=\int_X f d\mu-\int_X f d\mu - \liminf_{n\to\infty}\int_X (-f_n)d\mu\\ &=-\left(-\limsup_{n\to\infty}\int_X f_n d\mu\right)\\ &=\limsup_{n\to\infty}\int_X f_n d\mu. \end{align}$$

Could anyone help me, how I can prove the following inequality? $$\liminf_{n\to\infty}\int_X f_n d\mu \leq \limsup_{n\to\infty}\int_X f_n d\mu$$

Answer 2

The inequality you're asking about has nothing to do with sequences of integrals (as pointed out in the first comment). In general $$\liminf_{n} x_{n} \leq \limsup_{n} x_{n}$$ for any sequence $\{x_{n}\}$.

If you really need to prove the inequality, you should do it for a sequence of real numbers (or extended real numbers). It's a matter of checking the definitions of $\liminf$ and $\limsup$.