An Alternative Proof of Steinhaus Theorem

TL;DR

This paper presents a straightforward, elementary proof of Steinhaus theorem, a fundamental result in measure theory concerning the difference of two positive measure sets.

Contribution

It offers a simpler, more accessible proof of Steinhaus theorem, improving understanding and potentially broadening its applicability.

Findings

Provided an elementary proof of Steinhaus theorem

Simplified the understanding of the theorem's core result

Potentially facilitates teaching and further research in measure theory

Abstract

In measure theory, Steinhaus theorem is a result that deals with a property of the difference between two sets of positive measure. We give a simple elementary proof of the result.

Full text

An Alternative proof of Steinhaus Theorem

Arpan Sadhukhan, Tata Institute of Fudamental Research, Bangalore, India

Steinhaus’s Theorem states that if $A$ is a lebesgue measurable set on the real line such that the lebesgue measure of $A$ is not zero then the difference set $A-A=\{a-b|a,b\in A\}$ contains an open neighbourhood of origin[1]. Since there is a compact set of positive measure inside any set of positive measure. It is enough to prove the result for compact sets.

Theorem For any compact set $C\subset\mathbb{R}$ of positive measure, the difference set $D=C-C=\{a-b|a,b\in C\}$ contains an interval containing [math].

Proof: Suppose the set $D$ does not contain an interval around origin, then $\exists$ a sequence $x_{n}\rightarrow 0$ such that $x_{n}\notin D$, $\forall n\in\mathbb{N}$.

We proceed by induction to create arbitrarily large number of mutually disjoint sets $A_{1},A_{2},A_{3}\ldots$such that $A_{i}$ is of the form $x_{n_{i}}+C$ for all $i\geq 2$, $i\in\mathbb{N}$.

Let $A_{1}=x_{1}+C$, clearly $A_{1}$ satisfies the above property. Now suppose $A_{1},A_{2}\ldots A_{k}$ is created in such a manner. We will now create $A_{k+1}$.

Suppose for every $n\in\mathbb{N}$, $\exists$ an $i\leq k$ such that the set $x_{n}+C$ intersects $A_{i}$, then there exists a $j\leq k$ such that $x_{n}+C$ intersects $A_{j}$ for infinitely values of $n\in\mathbb{N}$, so $\exists$ a subsequence $\{y_{m}\}$ of $\{x_{n}\}$ such that $x_{n_{j}}-y_{m}\in D$ for all $m\in\mathbb{N}$. As $y_{m}\rightarrow 0$ and $D$ is compact, $x_{n_{j}}$ belongs to $D$ (a contradiction). So there exists an $N\in\mathbb{N}$ such that the set $x_{N}+C$ does not intersect $A_{i}$ for any $i\leq k$. Define $A_{k+1}=x_{N}+C$. So we have our desired set.

Now for any $n\in\mathbb{N}$, the set $x_{n}+C$ lies in some bounded set $[-R,R]$ for some $R\in\mathbb{N}$ and for any $p,q\in\mathbb{N}$ measure of $A_{p}>0$, $A_{q}>0$ and measure of $A_{p}$ equals measure of $A_{q}$. So we have an arbitrary large number collection of mutually disjoint sets of the same positive measure in $[-R,R]$ (a contradiction). Hence the result follows.

1 References

$[1]$. T.Tao, An Introduction to Measure Theory, $1$st edition, American Mathematical Society.

(Accepted American Mathematical Monthly(16/3/19)