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Using Lebesgue measurable set which is uncountable, one can show that the cardinality of the set of all Lebesgue measurable functions is $2^\mathbb{R}$

I know that Borel $\sigma$-algebra on $\mathbb{R}$ is of cardinality $\mathbb{R}$ (even if I haven't read proof). Then how can I show that the cardinality of the set of all borel measurable functions is $\mathbb{R}$?

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Since all constant functions are Borel measurable, the cardinality of the set of all real, Borel-measurable functions on $\Bbb R$ must be at least $|{\Bbb R}|=2^{\aleph_0}$. Going the other way, a real-valued function $f$ is determined by the sequence of sets $(f^{-1}(r,\infty)\mid r\in {\Bbb Q})$. If $f$ is Borel measurable, then each of these sets is Borel measurable, so there are only $2^{\aleph_0}$ possible choices for each $f^{-1}(r,\infty)$. Therefore, since there are only $\aleph_0$ rational numbers, the cardinality of the set of real, Borel-measurable functions on $\Bbb R$ is at most $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$.

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  • $\begingroup$ Oh, I was just going to edit my answer to add this argument... but now there's no point! $\endgroup$ – Asaf Karagila Apr 23 '13 at 0:16
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One method is to show that every Borel measurable function can be constructed as an iteration of pointwise convergence starting from continuous functions, the construction is analogous to the construction of the Borel $\sigma$-algebra internally from the open sets by taking intersections and complements repeatedly $\aleph_1$ times.

Then the proof of cardinality works here as well, there are $2^{\aleph_0}$ continuous functions; and therefore only $2^{\aleph_0}$ sequences of such, so at each stage we add at most $2^{\aleph_0}$ new functions. There are $\aleph_1$ stages in the construction so overall we have at least $2^{\aleph_0}$ and at most $2^{\aleph_0}\cdot\aleph_1=2^{\aleph_0}$ Borel measurable functions.

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