Past exam of the options II course of the University of Oxford / 2007 / D / D4

The dominated convergence theorem says that if measurable $f_n\to f$ almost everywhere and $|f_n|\le g$ for an integrable $g$, then $f$ is integrable, $\int |f_n-f|\to 0$, and consequently $\int f_n\to \int f$.

First derive Fatou's lemma from monotone convergence. For nonnegative $u_n$, the measurable functions $v_n={\inf }_{j\ge n}{u}_j$ increase to $\mathrm{liminf}{u}_n$, and $\int v_n\le {\inf }_{j\ge n}\int u_j$. Monotone convergence therefore yields $\int \mathrm{liminf}{u}_n\le \mathrm{liminf}\int u_n$.

Now $f$ is measurable and $|f|\le g$ almost everywhere, so is integrable. Apply Fatou's lemma to $u_n=2g-|f_n-f|\ge 0$, which converges almost everywhere to $2g$:The last integrals are nonnegative, so they tend to zero. Finally $|\int f_n-\int f|\le \int |f_n-f|\to 0$. This argument also works for complex-valued functions.

At each $x>0$, eventually $x<n$, and the stipulated power limit gives $f_n(x)\to e^{-x}\sin x$. At nonpositive $x$ the functions vanish. On their positive support,The supplied integrable exponential therefore dominates the sequence. Dominated convergence givesThe final improper integral is absolutely convergent, so agrees with its Lebesgue integral.