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?

closed as too localized by Anthony Quas, Dmitri Pavlov, Alain Valette, Bill Johnson, Gerald Edgar Jan 12 '12 at 3:35
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
I don't exactly know what the Lebesgue sigmaalgebra 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. 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 sigmaalgebra is bigger, but you factor out those you got more when descending to $L^p(\lambda)$. 

