Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 23H (Analysis of Functions)

For each $n\in \mathbb{N}$, the open intervalmeets $\mathbb{N}$ only in $\{n\}$. Thus every singleton is open in the subspace topology on $\mathbb{N}$. Every subset $A\subseteq \mathbb{N}$ is the union $A={\bigcup }_{n\in A}\{n\}$ and is therefore open. Hence the induced topology is the discrete topology:

Because every subset of $\mathbb{N}$ is open, the Borel sigma-algebra of a discrete space isIf $f:\mathbb{N}\to \mathbb{R}$ or $f:\mathbb{N}\to [0,\infty ]$ and $B$ is a Borel subset of the codomain, then $f^{-1}(B)$ is some subset of $\mathbb{N}$, hence is Borel. Thus every such function is a Borel measurable function.

Equip $\mathbb{N}$ with counting measure $\#$. For a nonnegative function $f:\mathbb{N}\to [0,\infty ]$, defineThis lies in $[0,\infty ]$ and agrees with the series as a Lebesgue integral against counting measure.

For $f:\mathbb{N}\to \mathbb{C}$, definewhen $f$ is integrable, which here meansThus the complex sum is defined under absolute convergence and can be obtained by integrating the real and imaginary parts.

A simple function on $\mathbb{N}$ is a function with finite range, equivalentlyfor a finite measurable partition by subsets $A_j\subseteq \mathbb{N}$. Its nonnegative sum is ${\sum }_j{a}_j\#A_j$. It is finite exactly when every level set with $a_j>0$ is finite, equivalently when $s$ has finite support after its zero level is discarded.

The Beppo Levi theorem, or monotone convergence theorem, states that if $(f_n)$ is a sequence of nonnegative measurable functions on a measure space andpointwise almost everywhere, then $f$ is measurable andwhere either side may be infinite.

Measurability of $f={\sup }_n{f}_n$ follows from measurability of countable suprema. Since $f_n\le f$, monotonicity of the Lebesgue integral givesFor the reverse inequality, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. DefineThen $E_n\uparrow X$ up to the null set on which convergence fails. Since $f_n\ge c,s{1}_{E_n}$,by continuity from below of a measure, applied to the finitely many level sets of $s$. Hence ${\lim }_n\int f_n\ge c\int s$. Letting $c\uparrow 1$ and taking the supremum over all simple $s\le f$, as in the definition of the Lebesgue integral, gives the reverse inequality and proves the theorem.

For integer $k\ge 1$, define $f_k(n)=f(k,n)$. The assumed monotonicity in $t$ givesfor every $n$. Apply the monotone convergence theorem on $\mathbb{N}$ with counting measure:The function $t\mapsto {\sum }_{n}f(t,n)$ is itself nondecreasing, and the positive integers are cofinal in $\mathbb{R}$ as $t\to \infty$, so its limit over real $t$ equals its limit over integer $k$. Therefore

Without monotonicity, takeFor each fixed $n$, $f(t,n)\to 0$, so the sum of the pointwise limits is zero. At every $t\ge 1$, however, exactly one term is one, so ${\sum }_{n}f(t,n)=1$ and its limit is one. This moving-spike obstruction to interchanging a limit and an infinite sum gives the required counterexample.