 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/23h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Equip

N

#

. For a nonnegative function

f : N \to [0, \infty]

n \geq 1 \sum f (n) := \int_{N} f d # = N \geq 1 sup n = 1 \sum N f (n) .

This lies in

[0, \infty]

For

f : N \to C

\sum_{n \geq 1} f (n) = \int_{N} f d #

when

f

is integrable, which here means

n \geq 1 \sum ∣ f (n) ∣ < \infty .

Thus the complex sum is defined under absolute convergence and can be obtained by integrating the real and imaginary parts.

N

is a function with finite range, equivalently

s = \sum_{j = 1}^{m} a_{j} 1_{A_{j}}

for a finite measurable partition by subsets

A_{j} \subseteq 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.

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