Does anyone know the cardinality of $$ \{f\colon \mathbb{R}\to\mathbb{R} \text{ Lebesgue measurable}\}/\sim $$ where $$ f\sim g\Leftrightarrow \text{f=g Lebesgue a.e.}? $$

  • $\begingroup$ Just wondering: is this question out of pure curiosity o is it context-related? $\endgroup$ – b00n heT Sep 6 '16 at 14:50
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    $\begingroup$ @boon I read that an easy proof that not all functions can be expressed as Fourier series (as Fourier has allegedly claimed) is an cardinality argument: The set of all functions has cardinality $|\mathbb{R}^\mathbb{R}|$, whereas the Fourier coefficients have cardinality $|\mathbb{R}^{\mathbb{N}}|=|\mathbb{R}|$. This argument does not work anymore (as per answer below) if we identify a.e. identical functions $\endgroup$ – Fleshman Sep 6 '16 at 17:13

Yes: it's $2^{\aleph_0}$. A Lebesgue-measurable function is a.e. identical to a Borel-measurable function. The set of Borel-measurable functions has cardinality $2^{\aleph_0}$. Of course, since all constants are Lebesgue-measurable, there are at least $2^{\aleph_0}$ pairwise non-almost-everywhere-identical measurable functions.

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  • $\begingroup$ There is the related question for sets, (which I find easier to think about), each Lebesgue set is a.e. equal to a borel set. Can OP's question be reduced to this case ? Maybe look at the graph of $f$ ? $\endgroup$ – Rene Schipperus Sep 6 '16 at 15:08
  • $\begingroup$ @Rene Schipperus: This also follows from the fact that the $L^{p}$ spaces are metric spaces that are complete and separable and contain no isolated points. See also this 11 January 2007 sci.math post. $\endgroup$ – Dave L. Renfro Sep 6 '16 at 15:19
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    $\begingroup$ @Rene Yes, your lemma is the reason why a L.-measurable function is L.-a.e. equal to a B.-measurable function (paired with the fact that a set is L.-negligible iff it is contained in some B.-negligible set). Write $f$ L.-measurable as $\lim_{n\to\infty}\sum_{k=1}^{h_n}a_n^k1_{E^n_k}$ for L.-sets $E^n_k$. There are B.-sets $A^n_k$ and $N^n_k$ such that $E^n_k\setminus A^n_k\subseteq N^n_k$, $A^n_k\cap N^n_k=\emptyset$ and $N^n_k$ is B.-negligible. The set $N=\bigcup_{n,k}N^n_k$ is B.-negligible. The function$$\tilde f=1_N+\lim_{n\to\infty}\sum_{k=1}^{h_n}a_n^k1_{A_n^k\setminus N}$$is your pal. $\endgroup$ – user228113 Sep 6 '16 at 15:38
  • $\begingroup$ @G.Sassatelli Thanks for that. I was thinking more along the lines of approximating its graph by a Borel set, but thats probably nonsense. $\endgroup$ – Rene Schipperus Sep 6 '16 at 15:48
  • $\begingroup$ (of course, in my above statement I should have added $A^n_k\subseteq E^n_k$ as well) $\endgroup$ – user228113 Sep 6 '16 at 16:02

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