State and prove Fatou’s Lemma on a general measurable space.
Solution
Fatou’s lemma. Let \((X, \mathcal M, \mu)\) be a measure space and \(\{f_n\}\) a sequence of measurable functions with \(f_n \geq 0\) for all \(n\). Then \[\int_X \liminf_{n \to \infty} f_n ~d\mu \leq \liminf_{n \to \infty} \int_X f_n ~d\mu,\] where both sides may be \(+\infty\).
1. Proof of Fatou’s lemma.
1.1. Define \(g_k \definedas \inf_{n \geq k} f_n\); then \(g_k\) is measurable, \(0 \leq g_k \leq f_n\) for all \(n \geq k\), and \(g_k\) increases to \(\liminf_n f_n\) as \(k \to \infty\).
Each \(g_k\) is a countable infimum of measurable functions, hence measurable; \(g_k \leq f_n\) for \(n \geq k\) by definition; the sequence \((g_k)\) is nondecreasing; and \(\lim_k g_k = \lim_k \inf_{n \geq k} f_n = \liminf_n f_n\) by definition of \(\liminf\).
1.2. Monotone convergence: \(\int_X g_k ~d\mu \uparrow \int_X \liminf_n f_n ~d\mu\).
\(g_k \uparrow \liminf f_n\) pointwise (<2>1) and the \(g_k\) are nonnegative, so the monotone convergence theorem applies.
1.3. For each fixed \(k\), \(\int_X g_k ~d\mu \leq \int_X f_n ~d\mu\) for every \(n \geq k\), hence \(\int_X g_k ~d\mu \leq \inf_{n \geq k} \int_X f_n ~d\mu\).
\(g_k \leq f_n\) pointwise for \(n \geq k\) (<2>1) and both are nonnegative, so monotonicity of the integral gives the inequality; taking the infimum over \(n \geq k\) gives the second.
1.4. Take the limit as \(k \to \infty\): \[\int_X \liminf_n f_n ~d\mu = \lim_k \int_X g_k ~d\mu \leq \lim_k \inf_{n \geq k} \int_X f_n ~d\mu = \liminf_n \int_X f_n ~d\mu.\] Proof: Left equality by 2; inequality by 3 (passing to the limit); right equality by definition of \(\liminf\) applied to the sequence \(\int_X f_n\).
1.5. Q.E.D.
The inequality of step 4 is the conclusion of Fatou’s lemma.