Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 25J (Probability and Measure)

A measure space consists of a set $\Omega$, a sigma-algebra $\mathcal{A}$ of subsets, and a countably additive map $\mu :\mathcal{A}\to [0,\infty ]$ with $\mu (\varnothing )=0$. A sigma-algebra contains $\Omega$ and is closed under complements and countable unions. Countable additivity means $\mu ({\bigcup }_n{A}_n)={\sum }_n\mu (A_n)$ for pairwise disjoint measurable sets. Such sets are called measurable; a real function is measurable when the inverse image of every Borel set belongs to $\mathcal{A}$.

The original PDF has height $1/n$ on the interval $[0,e^n]$. For finite $p$,so the sequence cannot converge in $L^p$, since a norm-convergent sequence is bounded. For $p=\infty$, $\Vert f_n{\Vert }_{\infty }=1/n\to 0$, so it converges to zero in $L^{\infty }$. It also converges pointwise, and hence almost everywhere, to zero. Finally for each $ϵ>0$, once $n>1/ϵ$ the set $\{|f_n|>ϵ\}$ is empty. Thus it converges in measure to zero, even on this infinite-measure space.

Convergence to a finite real limit is equivalent to the Cauchy property. The convergence set is thereforeEach set in braces is measurable, and sigma-algebras are closed under all the displayed countable operations. Completeness of $\mathbb{R}$ ensures that the Cauchy property gives a finite limit, proving the claim.

Set $Y_n={\sup }_{m\ge n}|f_m|$. It is an extended-real measurable function and decreases to $Y={\mathrm{limsup}}_m|f_m|$. If $f_n\to 0$ almost surely, then $Y_n\to 0$ almost surely. For any $ϵ>0$, the decreasing sets $\{Y_n>ϵ\}$ have intersection contained, up to a null set, in the exceptional set for convergence. Continuity from above of the probability measure gives $P(Y_n>ϵ)\to 0$.

Conversely, if $Y_n\to 0$ in probability, then $P(Y>ϵ)\le P(Y_n>ϵ)\to 0$ for every positive $ϵ$. Taking rational $ϵ>0$ gives $Y=0$ almost surely, so $|f_n|\to 0$ almost surely. This proves both directions, even when some individual $Y_n$ take the value infinity.