Past exam of the options II course of the University of Oxford / 2005 / D / 3

The monotone convergence theorem says that if $0\le f_n\uparrow f$ almost everywhere and the functions are measurable on a measure space, thenwith infinite integrals permitted. The limit is measurable after the usual modification on any exceptional null set.

The dominated convergence theorem says that if measurable functions $f_n$ converge almost everywhere to $f$ and satisfy $|f_n|\le g$ almost everywhere for an integrable nonnegative function $g$, then $f$ is integrable andThe theorem applies equally to real or complex functions. In particular domination must be by one fixed Lebesgue integrable function, rather than by a different bound for each $n$.

Put $G_N={\sum }_{n=1}^N|f_n|$. These are nonnegative measurable functions increasing to $G={\sum }_{n=1}^{\infty }|f_n|$, possibly initially with value infinity. Monotone convergence theorem givesAn infinite value on a set of positive measure would force an infinite integral. Thus $G$ is finite almost everywhere and, after setting it to zero on its null exceptional set if desired, is integrable.

Where $G$ is finite, the real series ${\sum }_n{f}_n$ is absolutely convergent. Define its sum to be $h$ there and zero on the exceptional null set. Measurability follows as an almost-everywhere limit of partial sums, and $|h|\le G$ almost everywhere gives integrability. Every partial sum is dominated by $G$, so dominated convergence theorem yieldsIn fact the convergence is also in the integrable-function norm:

Use $|\sin (\alpha x)|\le |\alpha |x$ for $x\ge 0$. For the source's functions,The last integral follows by the substitution $u=nx$ and integration by parts. Since ${\sum }_n{n}^{-2}$ converges, (ii) permits summing the integrals and identifies an integrable almost-everywhere sum.

For every $x>0$, the geometric series givesThe functions vanish on $x\le 0$, and the single endpoint does not change the integral. Use the supplied integral of $f_n$ to obtainThe estimates include negative $\alpha$ through its absolute value, and the case $\alpha =0$ gives zero on both sides.