Theorem of Steinhaus

The Steinhaus theorem says that if a set $A \subset \mathbb R^n$ is of positive inner Lebesgue measure then $\operatorname{int}{(A+A)} \neq \emptyset$. Is it true that also $\operatorname{int}{(tA+(1-t)A)} \neq \emptyset$ for $t \in(0,1)$? It is clear for $t=\frac{1}{2}$? But in general? Thanks.

Solution 1:

Yes.

If $A$ has positive inner measure then it contains a measurable set $B$ of positive finite measure.

Now for $0 \lt t \lt 1$ both $tB$ and $(1-t)B$ have positive measure, hence $tB + (1-t)B$ contains an open set by the result mentioned below. Therefore the interior of $tA + (1-t)A \supset tB + (1-t)B$ is non-empty.

In Corollary 20.17 on page 296 of Hewitt–Ross, Abstract Harmonic Analysis, I the following general result is shown:

Let $X$ and $Y$ be measurable sets of (finite) positive measure in a locally compact group with left Haar measure $\mu$. Then $XY$ contains an open set.

The proof relies on showing that the convolution $[X]\ast[Y](z)$ of the characteristic functions $[X]$ and $[Y]$ is equal to the function $z \mapsto \mu(X \cap zY^{-1})$; it is continuous; it vanishes outside $XY$; and it is non-zero because $\int [X]\ast[Y]=\mu(X) \mu(Y) \gt 0$, hence $XY$ must contain an interior point.

Some background, applications and further links are contained in this thread here.

If you read French you may want to have a look at Hugo Steinhaus's original article Sur les distances des points des ensembles de mesure positive, Fund. Math. 1 (1920), 93–104.

In fact, the entire first issue of Fundamenta Mathematicae is packed with gems of this sort. Fourteen (!) articles are by Wacław Sierpiński.