The last digit of pi (in terms of Banach limits)

Let $$\phi : l^\infty \to \mathbb C$$ be a Banach limit, and define the sequence $$\{x_k\}_{k\geq 0}$$ to be the digits in the 10-base decimal expansion of $$\pi$$. Note that $$\{x_k\}_{k\geq 0} \in l^\infty$$ and so we can talk about $$\phi(\{x_k\}_{k\geq 0})$$.

What it is? Note that Banach limits don't have to be unique.

Now consider the real number $$\sqrt 2$$. What is its last number? Finally, consider any element $$x\in \mathbb R$$. Can we say about the last digit of $$x$$, in the sense of Banach limits as considered above?

• Perhaps this is naïve, but the shift invariance property might tell you that $\phi$ is just a long-term average of the digits (and so it's $4.5$ for almost any real number).
– user296602
Nov 27 '18 at 18:18
• @T.Bongers looking at en.m.wikipedia.org/wiki/Almost_convergent_sequence I'd say the opposite is true: for almost all real nmbers there is no unique Banach limit Nov 27 '18 at 19:20
• Chuck Norris knows the last digit of pi.
– gerw
Nov 28 '18 at 7:58
• @gerw Well, Gauß knew it before him!
– Dirk
Nov 29 '18 at 8:33

Concerning last question about general $$x\in\mathbb{R}$$ and concerning the twice upvoted comment (under the original question) that claims the limit is 4.5 for almost all $$x$$:
I think the opposite is true (see proof below): almost all (in the sense of the Lebesgue measure) real numbers do not have a unique Banach limit. Unless $$\pi$$ is an exception to this (and I don't know whether it is), your first question does not have a well-defined answer: The 'last digit' of $$\pi$$ is different for different choices of the Banach limit $$\phi$$.
Proof of claim: According to a theorem of Lorentz, a sequence $$(x_k)_{k=0}^{\infty}$$ has a unique Banach limit $$L\in\mathbb{R}$$ if and only if its averages $$\overline{x}_{n,p}:=\frac{x_n+\dots+x_{n+p-1}}{p}$$ converge to $$L$$ as $$p\to\infty$$, uniformly in $$n$$. In formulas $$\forall \epsilon>0: \exists p_0\in\mathbb{N}: \forall p\geq p_0: \forall n\in\mathbb{N}:|\overline{x}_{n,p}-L|<\epsilon$$. In particular, just by rewriting this definition in set form and choosing $$\epsilon:=1$$, this implies that $$x\in \bigcap_{m\in\mathbb{N}}A_{p_0,m}$$ for some $$p_0\in\mathbb{N}$$, where $$A_{p_0,m}:=\{x:|\overline{x}_{mp_0,p_0}-L|<1\}$$. If we assume for simplicity that $$(x_k)_{k=0}^{\infty}$$ are the digits of $$x\in[0,1]$$, the events $$(A_{p_0,m})_{m\in\mathbb{N}}$$ are independent under the Lebesgue probability measure on $$[0,1]$$ (since the digits themselves are independent random variables) and have probability less than $$1$$. This shows $$P(\bigcap_{m\in\mathbb{N}}A_{p_0,m})=0$$ for all $$p_0\in\mathbb{N}$$, or, if we denote by $$B_L$$ the reals with unique Banach limit $$L$$, that $$P(B_L)=0$$ for all $$L\in\mathbb{R}$$. Since $$\mathbb{R}$$ is uncountable, it could happen that $$P(\bigcup_{L} B_L)>0$$. However, this is not the case since $$P(\bigcup_{L\not= 4.5}B_L)=0$$ by the law of large numbers.