Do the normal numbers have Lebesgue measure 1?

Question

In this question I found out that the normal numbers in $[0,1]$ form a Borel set. Since every Borel set is Lebesgue measurable, I now wonder what would be the Lebesgue measure of the set of normal numbers?

Intuitively, normal numbers are what you would 'expect' to get when you pick a 'random number' in $[0,1]$. Therefore, I suspect that almost all numbers are normal. Then the Lebesgue measure would be 1.

In this question they prove it by using that the non-normal numbers have measure 0. However, how do you prove that?

If it helps, it should be proven in the book Inevitable Randomness in Discrete Mathematics from Beck. (it's used as the reference for that fact on wikipedia) I don't have the full version of the book, so I can't help you with that. — Vincent Batens, Dec 11 '23 at 00:18
There is a reference to a book by Beck for this result in the Wikipedia page on normal numbers. I don't have the book, so I can't confirm this. — Rob Arthan, Dec 11 '23 at 00:23
Of possibly related interest is the result stated in this MSE answer. Also, the 1994 paper Normal numbers and subsets of $\mathbb N$ with given densities and the 2014 paper Normal numbers and the Borel hierarchy. — Dave L. Renfro, Dec 11 '23 at 00:41

score 2 · Accepted Answer · answered Dec 11 '23 at 00:32

I will prove this for simply normal numbers in base $2$. Let $a$ be a uniformly random number in $[0,1]$. Write $a=\sum_{n=1}^{\infty}a_n2^{-n}$. Then the $a_n$ are i.i.d. random variables with mean $0.5$. By the Law of Large Numbers, it follows that the sample average $\frac{a_1+\dots+a_n}{n}$ converges to the expected value $0.5$ with probability $1$. With other words, the set of real numbers $x=\sum x_n2^{-n}\in[0,1]$ with $\lim_{n\rightarrow\infty}\frac{x_1+\dots+x_n}{n}=\frac12$ has Lebesgue measure $1$.

The case for higher bases can probably be proven in a similar way.

Now we can take the intersection for all bases. This is the intersection of countably many sets with measure $1$. Therefore, the interection also has measure $1$.

Do the normal numbers have Lebesgue measure 1?

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