# Cardinality Of Borel Sets

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

• Are you sure the smaller refers to the cardinality? – gaoxinge Mar 1 '14 at 10:22
• @gaoxinge yes i am refering to the cardinality and c is cardinality of $2^{N_0}$ – happymath Mar 1 '14 at 10:24
• See this. – David Mitra Mar 1 '14 at 10:26
• What's $N_0$? Do you mean $\aleph_0$? – Asaf Karagila Mar 1 '14 at 10:28
• @AsafKaragila yes – happymath Mar 1 '14 at 10:29

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 $\aleph_1\cdot\frak c=c$.
• 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 2 '14 at 4:54