Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 24J (Probability and Measure)

Working with measurable functions modulo equality almost everywhere, the sequence is Cauchy in Lp space if for every $ϵ>0$ there exists $N$ such that $\Vert f_n-f_m{\Vert }_p<ϵ$ for all $m,n\ge N$. This is a statement about integral norm distance, not pointwise convergence. Without quotienting null functions the displayed quantity is a seminorm, so the conventional $L^p$ normed space uses these equivalence classes.

Completeness means that every norm-Cauchy sequence has a limit belonging to the same Lp space, with $\Vert f_n-f{\Vert }_p\to 0$. To prove it, choose a subsequence $f_{n_j}$ with $\Vert f_{n_{j+1}}-f_{n_j}{\Vert }_p\le 2^{-j}$. The functionshave $p$-norm bounded by $\Vert f_{n_1}{\Vert }_p+1$, by Minkowski inequality. They increase to $H$, and Fatou's lemma gives $\int H^p\le (\Vert f_{n_1}{\Vert }_p+1{)}^p$. Thus $H$ is finite almost everywhere, so the subsequence telescopes absolutely there to a measurable limit $f$ with $|f|\le H$ and $f\in L^p$.

The same argument on the tails gives $\Vert f-f_{n_j}{\Vert }_p\le {\sum }_{k\ge j}{2}^{-k}\to 0$. Given $ϵ$, choose such a subsequence term beyond the Cauchy threshold and use $\Vert f_n-f{\Vert }_p\le \Vert f_n-f_{n_j}{\Vert }_p+\Vert f_{n_j}-f{\Vert }_p$. This proves convergence of the full sequence and completeness for every $1\le p<\infty$, on an arbitrary measure space.

For $m>n$, the $L^1$ difference is ${\int }_n^m{x}^{-1}dx=\log (m/n)$. Choosing $m=2n$ makes it $\log 2$, so the sequence is not Cauchy in $L^1$. Its squared $L^2$ difference is ${\int }_n^m{x}^{-2}dx\le 1/n$, which tends uniformly to zero for $m>n$. Hence it is Cauchy in $L^2$, with norm limit $x^{-1}$ on $(1,\infty )$.

Here the $L^1$ difference is ${\int }_n^m{x}^{-2}dx\le 1/n$, and the squared $L^2$ difference is ${\int }_n^m{x}^{-4}dx\le 1/(3n^3)$. Both vanish uniformly for $m>n$, so the sequence is Cauchy in both spaces, with limit $x^{-2}$ in each.