Translation Invariance as an Integration Axiom

Adding this axiom allows simplification of the previous axioms

This note is part of the story of sketching out an axiomatic understanding of integration. The previous note in this series considers a collection of four axioms strictly stronger than the original three:

• Normalization: The function $1$ is integrable on any interval $I$, and returns the length of the interval: $\int_I 1 = |I|$

• Nonnegativity: If $f$ is integrable on $I$ and $f\geq 0$ then $\int_I f\geq 0$.

• Subdivision: If $[a,b]$ is an interval and $c\in[a,b]$, then $f$ is integrable on $[a,b]$ if and only if its integrable on $[a,c]$ and $[c,b]$. Furthermore, in this case $\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f$

• Linearity: If $f,g$ are integrable on $I$ then so is $cf+kg$ for any $c,k\in\mathbb{R}$ and $\int_{I}(cf+kg)=c\int_I f+k\int_I g$

A previous note showed that the property of translation invariance, coupled with linearity, is a promising way forward (we can prove the specific function $\chi_\QQ$ is no longer a counterexample).

This note proposes adding this as an additional axiom to give a new set:

• The Original 3 axioms
• Linearity
• Translation Invariance

A natural question that this post does not resolve is whether or not this is actually stronger

QUESTION: Is translation invariance implied by the previous axioms?

However, its good to note that without the first normalization axiom setting $\int_{[a,b]}1=b-a$ one cannot hope to prove translation invariance, as there are integrals satisfying all the other axioms which are not translation invariant (for example, the Lebesgue integral against a measure $m(x)dx$ for $m$ a positive nonconstant function).

Like when adding linearity to the axiom list, the power of of translation invariance renders some assumptions of the other axioms redundant. The main simplification possible here is to the normalization axiom, where we can get away by fixing the value of only a single integral:

The main content of this post is to prove that this simplified set of axioms is equivalent to the original three + linearity + translation invariance.

Proving Equivalence

Given an integral satisfying the above axioms, we want to prove the original normalization axiom: that $1$ is integrable on any interval, and that the returned value is the interval’s length. We follow steps familiar from many introductory real analysis arguments:

• The function 1 is integrable on all intervals of the form $[0,x]$ for $x>0$.
• It suffices to prove that $\int_{[0,x]}1=x$.
• The claimed equality is true for $x$ of the form $x=1/n$.
• The claimed equality is true for rational $x$.
• The function $F(x)=\int_{[0,x]} 1$ is continuous.
• Thus agreement on the rationals implies agreement everywhere.