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Hello I need help understanding a proof. The proof is from Rudin's Real and Complex Analysis. The proof is supposed to address the question: Is every Lebesgue measurable set a Borel set?

Let $c$ be the cardinality of the continuum (the real line or, equivalently, the collection of all sets of integers). We know that $\mathbb R^k$ has a countable base (open balls with rational radii and with centers in some countable dense subset of $\mathbb R^k$) and that $\mathcal B_k$ (the collection of all Borel sets of $\mathbb R^k$) is the $\sigma$-algebra generated by this base. It follows from this that $\mathcal B_k$ has cardinality $c$. On the other hand, there exist Cantor sets $E\subset\mathbb R^1$ with $m(E)=0$ (where $m$ is a measure). The completeness of $m$ implies that each of the $2^c$ subsets of $E$ is Lebesgue measurable. Since $2^c>c$, most subsets of $E$ are not Borel sets

I do not understand why the $2^c$ is needed. Neither do I understand how it is so suddenly introduced. Where does the $2^c$ come from? Also, how does the fact that $2^c>c$ show that most subsets of $E$ are not Borel sets? What is the relation between cardinality and Borel sets?

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  • $\begingroup$ A cantor set has cardinality $c$. So its power set, the set of all its subsets, has cardinality $2^c$. But Rudin just showed the set of Borel sets has cardinality $c$. $\endgroup$ – symplectomorphic Jun 13 '15 at 4:11
  • $\begingroup$ I need help understanding your question. You said: "I do not understand why the $2^c$ is needed." I guess that means you think it can be proved without $2^c$? So why don't you show us how you would prove without $2^c$ that there are measurable suts which are not Borel sets? $\endgroup$ – bof Jun 13 '15 at 4:50
  • $\begingroup$ ". . . it is so suddenly introduced. Where does the $2^c$ come from?" Yes, Rudin's explanation, "The completeness of $m$ implies that each of the $2^c$ subsets of $E$ is Lebesgue measurable," is a bit terse. A verbose paraphrase: "The completeness of $m$ implies that each subset of $E$ is Lebesgue measurable. How many sets is that? Well, the set $E$ has $c$ elements, so it has $2^c$ subsets." $\endgroup$ – bof Jun 13 '15 at 4:58
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On the one hand he’s shown that $\Bbb R^k$ has $\mathfrak{c}$ Borel sets. On the other hand, there is a Cantor set $E\subseteq\Bbb R^1$ such that $m(E)=0$. Since $m$ is complete, this implies that every subset of $E$ is Lebesgue measureable with measure $0$. Since $|E|=\mathfrak{c}$, $E$ has $2^{\mathfrak{c}}$ subsets, so we know that $\Bbb R^k$ has $2^\mathfrak{c}$ Lebesgue measurable subsets. However, it has only $\mathfrak{c}$ Borel subsets, and $\mathfrak{c}<2^{\mathfrak{c}}$.

Let $\mathscr{S}_k$ be the family of subsets of $\Bbb R^k$ that are Lebesgue measurable but not Borel. If $|\mathscr{S}_k|$ were less than $2^{\mathfrak{c}}$, the family of Lebesgue measurable subsets of $\Bbb R^k$, being the union of $\mathscr{B}_k$ and $\mathscr{S}_k$, would have cardinality

$$|\mathscr{S}_k|+|\mathscr{B}_k|=|\mathscr{S}_k|+\mathfrak{c}=\max\{|\mathscr{S}_k|,\mathfrak{c}\}<2^{\mathfrak{c}}\;,$$

which is false. Thus, we must have $|\mathscr{S}_k|=2^{\mathfrak{c}}$: $\Bbb R^k$ has as many Lebesgue measurable subsets that are not Borel as it has subsets altogether. Since it has only $\mathfrak{c}<2^{\mathfrak{c}}$ Borel subsets, it’s fair to say that most of its Lebesgue measurable subsets are not Borel.

More generally, if $\kappa$ is an infinite cardinal, $A$ is a set of cardinality $\kappa$, and $B\subseteq A$ is of cardinality $\lambda<\kappa$, then $|A\setminus B|=\kappa$: throwing away a smaller subset of an infinite set does not reduce the cardinality.

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The question is what does "most subsets" mean. Especially in the context of measure, it should be clear that "most" could mean different things (e.g. most numbers in $[0,1]$ are irrational; but also most of them lie outside the Cantor set; and even if you have a fat Cantor set of measure $0.9$ in another sense [it is nowhere dense], still most numbers lie outside of it).

But one thing seems to be very clear, if $A\subseteq X$ and there is a significant cardinality difference between $A$ and $X\setminus A$, then we can say that most elements lie in the larger set.

So if there are only $\frak c$ Borel sets, but $2^\frak c$ Lebesgue measurable sets, then almost all the Lebesgue measurable sets are not Borel sets, since $2^\frak c$ is significantly larger than $\frak c$. Similar reasoning shows that most subsets of $\Bbb R$ are not Borel either.

You can ask whether or not "most" subsets of $\Bbb R$ are Lebesgue measurable, since both collections (measurable and non-measurable) are of size $2^\frak c$, this requires finer tools than cardinality, and this is a whole different question.

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