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I'm currently doing a course on measure theoretic probability. In the course I learned that the cardinality of Borel $ \sigma $-algebra $ \mathscr{B} $ is the same as that of continuum $ \mathbb{R} $. But what is the cardinality of the set of Non-Borel Sets, let's call it $ \mathcal{S} $.

My inital guess is along these lines. That since $ \mathcal{S} = \mathscr{B^{c}} $. We can write that, $$ \mathbb{2^R} = \mathcal{S} ~ \cup \mathscr{B} $$ Assume that $ \mathcal{S} $ has the same cardinality as $ \mathbb{R} $. Now we know that $ \mathscr{B} $ also has the same cardinality as that of $ \mathbb{R} $. But their union $ \mathbb{2^R} $ has a stricly greater cardinality than $ \mathbb{R} $. Again assuming that union of uncountable sets of same cardinality has the same cardinality (I'm aware that this is true for countable sets, but not sure about uncountables), this results in a contradiction.

Hence the set $ \mathcal{S} $ must have a strictly bigger cardinality than $ \mathbb{R} $. Are the approach and the conclusion correct?

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    $\begingroup$ Your argument needs some tweaking. Here's the simplest way to do it, in my opinion. Let $f:\mathbb R\to\mathbb R\setminus{(0,1]}$ be defined as $$f(x)=\begin{cases}x&x\leq0\\x+1&x>0\end{cases}$$I'll leave it to you to show that this map is an injection. Let $X$ be some non-measurable set in $[0.25,0.75]$. Let $g:\mathcal P(\mathbb R)\to\mathcal P(\mathbb R)$ be defined by $g(S)=f(S)\cup X$. I'll leave it to you to show that the image of $g$ consists of only immeasurable sets, and that $g$ is an injection. By CSB the cardinality of $\mathcal S=\left\vert\mathcal P(\mathbb R)\right\vert$. $\endgroup$ – Don Thousand May 23 at 18:40
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    $\begingroup$ You argument is correct. If $A$ and $B$ are infinite sets of the same cardinality, the $A\cup B$ has also the same cardinality. It true no matter $A$ and $B$ is countable or uncountable. $\endgroup$ – Ramiro May 23 at 21:21
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The sum of two infinite cardinals is the larger of the two cardinals.

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  • $\begingroup$ As an $\epsilon$-generalization, this is still true under the assumption that at least one of the cardinals is infinite. $\endgroup$ – Dave L. Renfro May 23 at 21:09
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    $\begingroup$ And it follows that $S$ has the same cardinality as $2^{\mathbb R}$ (which can be stronger than just saying it's strictly bigger than $\mathbb R$). $\endgroup$ – Andreas Blass May 24 at 1:48

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