I want to prove that for AR process:
$$Y_t=aY_{t-1}+\epsilon_t$$
Where $E(\epsilon_t)=0$ and $|a|<1$, that $E(Y_t)=0$
It's not assumed that there is such a thing as $Y(0)$ (in which case the proof would go through by induction).
It's trivial to prove by induction that:
$$E(Y_t) = a^n E(Y_{t-n})$$
And for any real number $b$: $\text{lim}_{n\rightarrow\infty}a^n b=0$, which almost completes the proof.
I am left with proving that $E(Y_{t-n})$ doesn't grow with $n$, which would follow from $E(Y_t)=E(Y_{t-1})$. I am not sure how to prove this, should it be an axiom? I think some assumption is needed like with the finitistic case.
EDIT: It does seem like the above property or property that leads to it is just assumed. I would still very much like to see a formal proof / the axioms involved.
As |a|<1, you can invert the process to give it its (infinite) moving average (MA) representation. Using the lag operator, $L^jx_t=x_{t-j}$, we have $(1-aL)Y_t =\epsilon_t$ and \begin{equation} \begin{split} Y_t &=(1-aL)^{-1}\epsilon_t = (1+aL+a^2L^2 +a^3L^3 + \ldots )\epsilon_t \\ E(Y_t) &= (1+aL+a^2L^2 +a^3L^3 + \ldots )E(\epsilon_t) \\ &=0 \end{split} \end{equation} Think that does it.
If you like you can go, as $|a|<1$, $Y_t$ is stationary. Hence $E(Y_t)=E(Y_{t-1}) =\mu_y$ (say). Hence $\mu_y = E(\epsilon_t)/(1-a) = 0$.