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How to calculate the Lebesgue measure $m(A+\lambda A)$ in terms of $m(A)$ and $\lambda$ where A is a convex set $\subset \mathbb{R}^d$, $\lambda$ is a constant, and addition is Minkowski sum?

My idea about solving it is to first consider the case for $A$ when it is a rectangle, then pass it to the case where $A$ is open set and write it as infinite sum of open sets, then compact set and finally the general case. But, that might be a long path and no guarantee. Any better idea? It should be simple.

Thanks,

Susan_Math123
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1 Answers1

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We assume that $\lambda >0$. You only need this:

$$A + \lambda A = (1+ \lambda )A = \{(1+ \lambda) a : a\in A\}$$

when $A$ is convex. Note that $\supset$ is obvious. On the other hand, let $y\in A + \lambda A$. Then by definition, $y = a_1 + \lambda a_2$, where $a_1, a_2 \in A$. Then

$$\frac{1}{1+\lambda} y = \frac{1}{1+ \lambda} a_1 + \frac{\lambda}{1+\lambda } a_2 $$

is in $A$, as the right hand side is a convex combination of $a_1, a_2$. Thus $y\in (1+ \lambda)A$.

As a result, we have that $$m(A+ \lambda A) = (1+\lambda)^d m(A),$$ where $d$ is the dimension.

Note that in general when $A$ is not convex, $A+A$ might have positive measure even if $A$ is of measure zero. Cantor set is one of the example.

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