How many sigma algebras exist on $\mathbb{R}$? On the one hand, there are at least $2^\mathfrak{c}$ sigma algebras on $\mathbb{R}$: one can take any subset $A$ of $\mathbb{R}$ and consider a sigma algebra $\{\emptyset, A, \bar A, \mathbb{R}\}$
On the other hand, number of sigma algebras can be bounded as $2^{2^\mathfrak{c}}$ since a sigma algebra is a family of subsets of $\mathbb{R}$.
Let's assume generalized continuum hypothesis to simplify. In this case there are only two possible answers. Unfortunately I can't either find $2^{2^\mathfrak{c}}$ different sigma algebras, or prove their number is exactly $2^\mathfrak{c}$.
 A: The other answer gives many $\sigma$-ideals, but I don't think it settles the question about $\sigma$-algebras.
Assume GCH. Find a family $\mathcal{F}$ of $\aleph_2$ many subsets of $\mathbb{R}$, any two of which have countable intersection. There are $\aleph_3 = 2^{2^c}$ distinct subsets of $\mathcal{F}$ which contain more than one element.
Say that $A \subseteq \mathbb{R}$ essentially contains $B$ if $B \setminus A$ is countable. Then for each $\mathcal{F}_0 \subseteq \mathcal{F}$ with more than one element the family of subsets of $\mathbb{R}$ which either essentially contain every set in $\mathcal{F}_0$ or no set in $\mathcal{F}_0$, is a $\sigma$-algebra which contains no set in $\mathcal{F}_0$ and every set in $\mathcal{F}\setminus\mathcal{F}_0$. So there are as many distinct $\sigma$-algebras as there are subsets of $\mathcal{F}$ with more than one element.
(To construct $\mathcal{F}$, identify $\mathbb{R}$ with $\{f: \alpha \to \{0,1\}\, |\, \alpha$ is a countable ordinal $\}$. For each function from $\aleph_1$ to $\{0,1\}$, its restrictions to all countable ordinals is a subset of this set, and any two of these have countable intersection.)
A: This is a continuation of my previous answer.
Suppose $F$ is a family of $2^{2^{|\mathbb{R}|}}$ sigma-ideals on $\mathbb{R}$. By throwing away the principal ideals, we can assume that they are all non-principal. Let $I, J \in F$ be distinct and towards a contradiction suppose they generate the same sigma-algebra. Note that the sigma-algebra generated by a sigma-ideal is just the ideal together with the dual filter. Say $X \in I \setminus J$. Then $ \mathbb{R} \setminus X \in J$ and $J$ restricted to $X$ is a non-principal prime sigma-ideal on $X$. But this is impossible as there is no measurable cardinal below $|Y|$. Am I missing something obvious?
A: Since $|\mathbb{R}|^{\aleph_0}= |\mathbb{R}|$, there is a sigma-independent (all countable boolean cominations are non-empty) family $F \subseteq \mathcal{P}(\mathbb{R})$ of size $2^{|\mathbb{R}|}$. Now for each subfamily $G$ of $F$ consider the sigma-ideal that contains each member of $G$ and the complement of each member of $F\setminus G$. So there are $2^{2^{2^{\aleph_0}}}$ different sigma-ideals on $\mathbb{R}$.
