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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?

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closed as too localized by Anthony Quas, Dmitri Pavlov, Alain Valette, Bill Johnson, Gerald Edgar Jan 12 '12 at 3:35

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Sounds like a homework assignment... See FAQ –  Anthony Quas Jan 11 '12 at 20:28

1 Answer 1

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.

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 $L^p(\lambda)$.

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I thought the Baire algebra was the sigma-algebra generated by the zero sets. $\;$ –  Ricky Demer Jan 11 '12 at 20:39
It's not helpful to post solutions to homework assignments on the forum: it just encourages other inappropriate postings –  Anthony Quas Jan 11 '12 at 22:47