# Show that the set of numbers normal to base d has Lebesgue Measure 1 (d=2 and d=3).

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.