If ${\{f_i\}}_{i\in I}$ is a family of Lebesgue measurable functions from $\mathbb{R}$ to $\mathbb{R}$, then the initial sigma algebra of this family is the sigma algebra generated by all sets of the form $\{x\in\mathbb{R}:f_i(x)<a\}$. This is always a sub-sigma algebra of the Lebesgue sigma algebra. But my question is, what is the smallest family of Lebesgue measurable functions whose initial sigma algebra is the Lebesgue sigma algebra itself?

Would a finite family suffice, or a countable family, or what?

  • $\begingroup$ $\{f\}$ with $f(x)=x$ for example. $\endgroup$
    – amsmath
    Commented Oct 13, 2019 at 22:48
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    $\begingroup$ @amsmath that would be the Borel sigma algebra, not the Lebesgue $\endgroup$
    – s.harp
    Commented Oct 13, 2019 at 23:00
  • $\begingroup$ @s.harp Correct. Hmm... $\endgroup$
    – amsmath
    Commented Oct 13, 2019 at 23:04
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    $\begingroup$ The extra sets you get from Lebesgue algebra are too crazy for me to think about, I don't think there is going to be a simple answer. $\endgroup$
    – s.harp
    Commented Oct 13, 2019 at 23:06
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    $\begingroup$ I don't see why a smallest family should exist. $\endgroup$ Commented Oct 13, 2019 at 23:29

1 Answer 1


A $\sigma$-algebra generated by $\kappa$ sets (for $\kappa>1$) has at most $\kappa^{\aleph_0}$ elements (see cardinality of the Borel $\sigma$-algebra of a second countable space). It follows that the same is true for functions as long as $\kappa$ is infinite: a $\sigma$-algebra which is initial for by $\kappa$ functions to $\mathbb{R}$ (in the sense you describe) has at most $\kappa^{\aleph_0}$ elements (since the each function contributes countably many generating sets, namely $\{x\in\mathbb{R}:f_i(x)<a\}$ for $a\in\mathbb{Q}$).

The Lebesgue $\sigma$-algebra has $2^{2^{\aleph_0}}$ elements, since for instance it contains all subsets of the Cantor set. It follows that it cannot be initial for any family of $\kappa$ functions where $\kappa^{\aleph_0}<2^{2^{\aleph_0}}$ (in particular, this includes $\kappa\leq 2^{\aleph_0}$).

I see no reason to expect there is any nice "natural" or "minimal" set of functions you could use. (Indeed, with the usual meaning of "minimal", I would expect that there does not exist any minimal such set at all, though that seems rather difficult to prove.)

Note that this question is basically the same as asking for a set of generators for $\mathcal{P}(\mathbb{R})$ as a $\sigma$-algebra, by picking a bijection between $\mathbb{R}$ and each uncountable Borel null set.

  • $\begingroup$ Wow, it’s surprising that not even continuum many functions suffice. $\endgroup$ Commented Oct 14, 2019 at 1:53

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