What's wrong with the following proof that almost all real numbers are normal?

Question

Say $U \sim \text{U}[0,1]$. Then $\mathbb{P}[\,U\text{ is normal}\,] = 1$. This is evident from the fact that a Bernoulli process can be used to sample the decimal digits of $U$, leading to a normal number with probability 1. But the probability measure for a uniform random variable is just the Lebesgue measure. Hence almost all numbers in [0,1] are normal. And then all of $\mathbb{R}$ can then be covered by such intervals.

Answer 1

There's nothing wrong with this proof; it is correct (modulo the proof that a Bernoulli process actually produces a normal number, which is an easy consequence of the strong law of large numbers). More complicated proofs you may have seen are proofs that prove it from first principles without using any machinery from probability theory such as the strong law of large numbers.