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I'm looking for two measurable sets in $\Bbb{R}^2$ st their sum is not measurable.

I found the example , let $A\subset \Bbb{R}$ be a non-measurable set in $\Bbb{R}$ and consider the sets $A\times \{0\}$ and $\{0\}\times \mathbb R$, these both sets have measure zero in $\Bbb{R}^2$ but their sum $A\times \mathbb R$ is non measurable in $\Bbb{R}^2$. My question is why is $A\times \mathbb R$ non measurable in $\Bbb{R}^2$?

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Mathronaut
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    What is "sum" in this context? $X+Y={ x+y : x \in X,y \in Y }$? – Ian Mar 01 '15 at 20:46
  • @Ian yes that's correct – Mathronaut Mar 01 '15 at 20:47
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    I think your example is harder to handle than it has to be. Try taking $A$ to be (for concreteness) the Vitali set, and replace your ${ 0 } \times \mathbb{R}$ with ${ 0 } \times [0,1]$. This will ensure that the sum is a subset of a set which has finite measure (namely $[0,1] \times [0,1]$), which will help you get a contradiction (just as it did with the Vitali set). – Ian Mar 01 '15 at 20:50

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Without loss of generality we can assume that $l_*(A)=0$ and $l^*(A)=a>0$, where $l_*$ and $l^*$ denote inner and outer measures in $R$ produced by one-dimensional Lebesgue measure $l$ in $R$. Notice that $(A \times \{0\})+(\{0\}\times R)=A \times R$. If we assume that $A \times R$ is measurable then the intersection $(A \times R) \times (R \times [0,1])=A\times [0,1]$ also will be measurable. Now by using Fubini theorem it is obvious to check that $ (l\times l)_*(A\times[0,1])=0$ and $(l\times l)^*(A\times[0,1]) \ge a $, where $ (l\times l)_*$ and $ (l\times l)^*$ denote inner and outer measures in $R^2$ produced by two-dimensional Lebesgue measure $l\times l$ in $R^2$. The latter relation means that $A \times [0,1]$ is not $l\times l$-measurable. This mens that our assumption that $A \times R$ is measurable is false. This ends the proof of the fact that $A \times R$ is non-measurable.

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