I need to prove this for $d = 2, d= 3$. I'm working on $d =2$. The idea is to show that my $x_n$'s are IID so that I can apply the strong law of large numbers.

Let $N_2 = \{x \in [0,1] \mid x\text{ is normal to base 2}\}$.

Define $x_n(x) = 1$ if the $n$th digit of $x$ is $1$, $0$ if the $n$th digit of $x$ is $0$.

i.e., $x = .0110110 \implies x_1(x) = 0, x_2(x) = 1, x_3(x) = 1, x_4(x) = 0...$

Want to show that these are IID.

To first we show that they each have the same mean. By inspection for example we can see that P$(x_1 = 1) = \lambda\{x=.x_1.x_2... \text{such that the first digit in binary is 1}\}=\frac{1}{2}$ where $\lambda(A)$ denotes the Lebesgue Measure of $A$. This also means that P$(x_1 = 0) = \frac{1}{2}$, and we can check that the other $x_i$'s also have mean $\frac{1}{2}$. My problem is showing that this is the case for all of the $x_i$.

In addition, I need to show independence for all $x_i$, and I am stuck here as well. Once I know that they are IID, I know how to complete the proof.

I believe for $d=3$ I will need to define the functions differently, but after that, it will probably be similar to $d=2$ proof.

Any help is appreciated.


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