Fatou’s Lemma Suppose \(\{f_n\}\) is a sequence of measurable functions with \(f_n \geq 0\). If \(\lim\limits_{n\rightarrow\infty}f_n(x) = f(x)\,\, a.e.\), then \[\int f \leq \lim \inf \int f_n\]
Intuition : The RHS of Fatou’s Lemma has \(\lim \inf\) because we don’t always have \(f_n\leq f\) even though \(f_n\rightarrow f\).
By the definition of the integral of non-negative function: \[\int f = \sup \int g, \,\,\text{ where } g \leq f, \text{ bounded and supported in a finite set}\]
It’s not hard to construct a proof by bounded convergence theorem, that if we add a condition \(f_n\leq f\) to Fatou’s Lemma, the result will become \[\int f = \lim \int f_n\]
Fatou’s Lemma is actually one direction of the above equality.
Proof Sketch: Suppose \(0\leq g\leq f\), where \(g\) is bounded and supported on a finite set \(E\). Let \(g_n = \min(f_n, g)\), then \(g_n\rightarrow g\) is also bounded and supported on a finite set. By BCT \[\int g = \lim \int g_n \leq \int f_n\] Thus \[\int f = \sup \int g \leq \lim\inf \int f_n\]
Here is the end of the proof of Fatou’s Lemma.
And since \(f_n \leq f\), the other direction \[\lim \sup\int f_n \leq \int f\]
Q.E.D.
An example of \(f_n\rightarrow f, f_n\geq f\): \[f_n = \begin{cases} 1/n,\quad x\in[-1/n, 1/n] \\ 0,\quad otherwise \end{cases}\] \[f (x)= 0\] Then \[\int f = 0, \quad \quad \lim \int f_n = 2\]
The Fatou’s Lemma holds but \(\int f < \lim \int f_n\).