V venturozzaccio May 2010 7 1 May 29, 2010 #1 Hi! I want to ask if is \(\displaystyle R^2\) is Lebesgue-measurable? If yes why? This is true also for R, R^3 and so on? Are they Peano-Jordan measurable? I thought they were because their complementar is empty. Sorry for my english. Thank you.