Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 25K (Probability and Measure)

Fatou's lemma states that for nonnegative measurable $h_n$,Indeed $g_n={\inf }_{k\ge n}{h}_k$ is measurable, increases to $\mathrm{liminf}{h}_n$, and satisfies $g_n\le h_k$ for every $k\ge n$. The monotone convergence theorem gives $\int \mathrm{liminf}{h}_n={\lim }_n\int g_n\le {\lim }_n{\inf }_{k\ge n}\int h_k$, proving the claim, including infinite values.

The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ almost everywhere for one integrable nonnegative $g$, then $h$ is integrable and $\int h_n\to \int h$; in fact $\int |h_n-h|\to 0$. Integrability follows from $|h|\le g$. For real functions, Fatou applied to $g+h_n$ gives $\int h\le \mathrm{liminf}\int h_n$, and applied to $g-h_n$ gives the reverse inequality against the upper limit. Thus the integrals converge. To obtain the stronger conclusion, Fatou applied to $2g-|h_n-h|\ge 0$ gives $2\int g\le 2\int g-\mathrm{limsup}\int |h_n-h|$, forcing the latter limit to be zero. Complex functions may be treated through their real and imaginary parts or the same absolute-difference argument.

Write $A_n=\int f_n^{+}$, $B_n=\int f_n^{-}$, $A=\int f^{+}$ and $B=\int f^{-}$. Almost-everywhere convergence gives $f_n^{\pm }\to f^{\pm }$, so Fatou's lemma gives $A\le \mathrm{liminf}{A}_n$ and $B\le \mathrm{liminf}{B}_n$. The given norm condition is $A_n+B_n\to A+B$. Consequentlyand the symmetric argument bounds $\mathrm{limsup}{B}_n\le B$. Combining upper and lower limits provesNo common integrable dominating function for $f_n$ has been assumed.

Now $f\ge 0$ almost everywhere. The functions $\min (f_n,f)$ converge almost everywhere to $f$ and are dominated by the integrable $f$. The dominated convergence theorem gives $\int \min (f_n,f)\to \int f$. Using $|f_n-f|=f_n+f-2\min (f_n,f)$ and the assumed convergence of the nonnegative integrals yieldsThis is the nonnegative form of Scheffé's lemma: almost-everywhere convergence together with convergence of total mass implies convergence in the L1 norm.