Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / D / 4

The dominated convergence theorem states: if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le g$ almost everywhere for an integrable $g\ge 0$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$.

To derive it from the monotone convergence theorem, first prove Fatou's lemma. For nonnegative measurable $h_n$, the functions $k_N={\inf }_{n\ge N}{h}_n$ increase to $\mathrm{liminf}{h}_n$ and satisfy $\int k_N\le {\inf }_{n\ge N}\int h_n$. Monotone convergence yields $\int \mathrm{liminf}{h}_n\le \mathrm{liminf}\int h_n$.

Now $|f|\le g$ almost everywhere, so $f$ is integrable. Apply Fatou's lemma to the nonnegative functions $2g-|f_n-f|$, which converge almost everywhere to $2g$:Hence $\int |f_n-f|\to 0$, and $|\int f_n-\int f|\le \int |f_n-f|$ proves the assertion. Common exceptional null sets may be removed at the beginning since the sequence is countable.

For fixed $x>0$, $f_n(x)\to e^{-x}\sin x$; for $x\le 0$, the limit is zero. On $0<x<n$, the supplied inequality gives $(1-x/n{)}^n\le e^{-x}$, soThis is the given integrable dominating function. By the dominated convergence theorem,