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I was trying to show that Borel $\sigma$ algebra is smaller than lebesgue measurable sets. I could come up with a proof for the cardinality of lebesgue measurable sets being $2^c$. Cardinality of Borel sets is at least as big as reals. So I am left with showing that its cardinality is less than $2^c$. But I have no idea how to do this.This also seems to be contradictory because continuum hypothesis states that there is no cardinality in between $C$ and $2^c$. If anyone can help it would be great. Thanks

happymath
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1 Answers1

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The Borel sets themselves is a set of size $\frak c$, which is indeed smaller than $2^\frak c$, the cardinality of the Lebesgue measurable sets. So there's no contradiction with any form of continuum hypothesis.

The simplest proof I know involves transfinite induction, and the internal hierarchy of Borel sets. So you don't just view Borel sets as "the smallest $\sigma$-algebra including the open sets", but rather construct it in $\omega_1$ steps from the open sets. Then one can prove by induction that every step along the way has cardinality $\frak c$, so the entire algebra has size $\aleph_1\cdot\frak c=c$.

Asaf Karagila
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    isn't there any proof which avoids transfinite induction or its equivalent formulations? – happymath Mar 01 '14 at 10:35
  • Not that I know of. And I don't know how to circumvent it either. In any case, as far as using ordinals and transfinite recursion goes, this case is quite simple. – Asaf Karagila Mar 01 '14 at 10:38
  • So is there something like it is not provable without transfinite induction. But transfinite induction is not justified all the times. So how can we be sure that the proof is right? – happymath Mar 01 '14 at 10:42
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    I don't understand your last comment. – Asaf Karagila Mar 01 '14 at 10:52
  • We can use transfinite induction only on well ordered sets. But we do not know whether $c$ is well ordered. So can we assume transfinite induction to hold even here. – happymath Mar 02 '14 at 04:54
  • We don't do the transfinite induction on the cardinality of the continuum. We do it on $\omega_1$. But since you mention the axiom of choice, you do need to assume some small part of it, otherwise it is possible that every set of reals is Borel. – Asaf Karagila Mar 02 '14 at 05:35
  • @happymath, every set can be well ordered. This fact is equivalent with the axiom of choice. What the answer was saying was that if one considers a well ordering of $\omega_1$ and for every ordinal less than $\omega_1$ (which is a countable ordinal) one does countable unions of all the sets so far obtained, one gets all borel sets. So for each countable ordinal, we got continuum many sets and we repeated this $\omega_1$ times, so we have $\omega_1$ times continuum which is still only continuum. – Máté Wierdl Oct 31 '19 at 00:13