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If a measurable set $E\subseteq [0,L]$ has measure $m(E)=L/2+\alpha$, then for $|t|<2\alpha$ the sets $E$ and $E+t$ lie in an interval of length $L+|t|$, soThus $(-2\alpha ,2\alpha )\subseteq E-E$. This quantitative overlap argument is a bounded form of the Steinhaus theorem.

Every Lebesgue-measurable additive function $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is continuous. Small-value preimages are measurable; one of their integer dilates has positive measure, so the Steinhaus theorem puts a neighbourhood of zero in their difference set. Additivity then makes $f$ small on that neighbourhood.

First intersect $E$ with a sufficiently large bounded interval to obtain a finite positive-measure set. By the Lebesgue density theorem, this set has a density-one point $x$. Hence, for every $\epsilon >0$, some sufficiently small interval $I=(a,b)$ centred at $x$ satisfieswhich is the high-density interval in a positive-measure subset of the real line.

To obtain the final claim, choose $\epsilon <1/2$, put $L=b-a$, and translate $A=E\cap (a,b)$ into $[0,L]$. WritePart (f), after rescaling the interval length, givesThus $E-E$ contains an open interval around zero, which is the Steinhaus theorem.

For $f\in L^{\infty }({\mathbb{R}}^n)$ and $g\in L^1({\mathbb{R}}^n)$, define their convolution byIt is bounded becauseIf ${\tau }_{h}g(z)=g(z+h)$, thenContinuity of translation in $L^1$ makes the right-hand side tend to zero with $h$, uniformly in $x$. Thus the convolution of L infinity and L1 functions is bounded and continuous.

Now let $f=1_A$ and $g(x)=1_A(-x)$. ThenAt zero this equals $|A|>0$. By continuity it remains positive throughout some open neighbourhood $U$ of zero. Positivity at $z$ means that some $y$ satisfies $y\in A$ and $y-z\in A$, so $z\in A-A$. HenceThis is the Steinhaus theorem.

The conclusion also holds when $|A|=\infty$. Since ${\mathbb{R}}^n$ is the union of bounded balls, some measurable subset $A\cap B(0,R)$ has finite positive measure. Its difference set lies inside $A-A$ and already contains a neighbourhood of zero.

For every positive integer $k$, additivity gives $f(kx)=kf(x)$. HenceConversely, if $z\in f^{-1}(B_{k\epsilon })$, write $z=k(z/k)$. Thenso $z/k\in f^{-1}(B_{\epsilon })$. Thus

To prove continuity, fix $\epsilon >0$ and putThese sets are measurable and ${\mathbb{R}}^n={\bigcup }_{k\ge 1}{E}_k$, so at least one $E_k$ has positive Lebesgue measure. Part (ii) gives $E_k=k{E}_1$, and therefore $E_1$ also has positive measure. By the Steinhaus theorem, $E_1-E_1$ contains an open neighbourhood $U$ of zero. Part (i) givesThis proves continuity at zero. Finally,as $h\to 0$, so $f$ is continuous at every $x$. Thus every such map is an instance of the theorem that a measurable additive function is continuous.