# Cardinality argument in proving that not every Lebesgue measurable set is a Borel set

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?

• 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 $c$. – symplectomorphic Jun 13 '15 at 4:11
• 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? – bof Jun 13 '15 at 4:50
• ". . . 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." – bof Jun 13 '15 at 4:58

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.

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.