# Cardinality of Lebesgue measurable functions under a.e.-identification

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

• Just wondering: is this question out of pure curiosity o is it context-related? – b00n heT Sep 6 '16 at 14:50
• @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 – 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.
• 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 $f$ ? – Rene Schipperus Sep 6 '16 at 15:08
• @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. – Dave L. Renfro Sep 6 '16 at 15:19
• @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. – user228113 Sep 6 '16 at 15:38
• (of course, in my above statement I should have added $A^n_k\subseteq E^n_k$ as well) – user228113 Sep 6 '16 at 16:02