Difference between spaces of integrable functions w.r.t Lebesgue measure and Borel measure - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-24T03:36:37Z http://mathoverflow.net/feeds/question/85447 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85447/difference-between-spaces-of-integrable-functions-w-r-t-lebesgue-measure-and-bore Difference between spaces of integrable functions w.r.t Lebesgue measure and Borel measure anonymous 2012-01-11T20:19:47Z 2012-01-11T20:31:25Z <p>Is there a difference between $L^p(\mathbb R,\mathfrak B,\beta)$ and $L^p(\mathbb R,\mathfrak L,\lambda)$ ? Here I denoted by $\lambda$ the Lebesgue measure, defined on the Lebesgue $\sigma$-algebra $\mathfrak L$ and by $\beta$ its restriction to the Borel $\sigma$-algebra $\beta$. Does the answer depend on wether I consider equivalence classes of functions or not?</p> http://mathoverflow.net/questions/85447/difference-between-spaces-of-integrable-functions-w-r-t-lebesgue-measure-and-bore/85448#85448 Answer by Kofi for Difference between spaces of integrable functions w.r.t Lebesgue measure and Borel measure Kofi 2012-01-11T20:31:25Z 2012-01-11T20:31:25Z <p>I don't exactly know what the Lebesgue sigma-algebra is, but I presume you mean the extension of - for example - the Borel algebra that gives a complete measure. I know this as Baire algebra, and it has a higher cardinality than the Borel algebra.</p> <p>The $L^p$ spaces however, constist both of equivalence classes of functions, and in fact the spaces are isomorphic via a natural embedding from the Borel one to the other. The difference is that the equivalence classes are bigger. You get more measurable functions in the $\mathcal{L}^p(\lambda)$ space since the sigma-algebra is bigger, but you factor out those you got more when descending to $L^p(\lambda)$.</p>