- For this problem, consider just Lebesgue measurable functions \(f : [ 0 , 1 ] \to \mathbb { R }\) . together with the Lebesgue measure.
State Fatou’s lemma (no proof required).
State and prove the Dominated Convergence Theorem.
Give an example where \(f_n(x) \to 0\) a.e. on \([0,1]\), but \(\int_0^1 f_n(x)\,dx \to 1\).
Solution
Throughout, \((X,\mu)\) is \([0,1]\) with Lebesgue measure.
1. (a) Fatou’s lemma: if \(f_n\ge0\) are measurable, then \(\int\liminf_n f_n\le\liminf_n\int f_n\).
This is the requested statement.
2. (b) If \(f_n\to f\) a.e., and \(\abs{f_n}\le g\) a.e. for all \(n\) with \(g\in L^1\), then \(f\in L^1\) and \(\int f_n\to\int f\).
2.1. \(f\in L^1\).
Letting \(n\to\infty\) in \(\abs{f_n}\le g\) gives \(\abs f\le g\) a.e., and \(g\) is integrable.
2.2. \(\int f\le\liminf_n\int f_n\).
Fatou’s lemma applied to \(g+f_n\ge0\) gives \(\int(g+f)\le\int g+\liminf_n\int f_n\), and \(\int g\) is finite.
2.3. \(\limsup_n\int f_n\le\int f\).
Fatou’s lemma applied to \(g-f_n\ge0\) gives \(\int(g-f)\le\int g-\limsup_n\int f_n\), and \(\int g\) is finite.
2.4. Q.E.D.
Steps 2 and 3 give \(\limsup_n\int f_n\le\int f\le\liminf_n\int f_n\), so \(\int f_n\to\int f\); step 1 gives \(f\in L^1\).
3. (c) \(\boxed{f_n=n\,\mathbf 1_{(0,1/n)}}\) satisfies \(f_n\to0\) everywhere on \([0,1]\) and \(\int_0^1f_n=1\) for all \(n\).
\(f_n(0)=0\), and for \(x\in(0,1]\), \(f_n(x)=0\) once \(n>1/x\). Also \(\int_0^1f_n=n\cdot\frac1n=1\).
4. Q.E.D.
Steps 1, 2 and 3 answer parts (a), (b) and (c).