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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?

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    $\begingroup$ 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). $\endgroup$
    – user296602
    Nov 27 '18 at 18:18
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    $\begingroup$ @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 $\endgroup$
    – Bananach
    Nov 27 '18 at 19:20
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    $\begingroup$ Chuck Norris knows the last digit of pi. $\endgroup$
    – gerw
    Nov 28 '18 at 7:58
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    $\begingroup$ @gerw Well, Gauß knew it before him! $\endgroup$
    – Dirk
    Nov 29 '18 at 8:33
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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.

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