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Let $(X,\mathcal X,\mu)$ be a fixed measure space, and suppose that $\mu$ is stationary and ergodic with respect to the (left) action of a topological group $G$. Stationarity means that $\mu = g_* \mu := \mu \circ g^{-1}$ for all $g \in G$. Ergodicity means that $$\mbox{if $\mu(A \triangle g^{-1}A) = 0$ for all $g \in G$, then $\mu(A) = 0$ or $\mu(A^c) = 0$,}$$ where $A \triangle g^{-1}A$ denotes the symmetric difference of $A$ and $g^{-1}A$.

Fix some $p \in [0,\infty]$, and let $L^p := L^p(X,\mathbb R; \mu)$ denote the space of $p$-integrable functions. Let $\mathcal B^p := \mathcal B(L^p)$ denote the Borel $\sigma$-algebra of $L^p$ with respect to the natural topology. There is a natural (right) action of $G$ on $L^p$, given by precomposition: $g^* \ell := \ell \circ g$. There is also a natural (left) action, given by conjugation: $c_g \ell := g \circ \ell \circ g^{-1}$.

Does there exist a probability measure on $(L^p,\mathcal B^p)$ which is stationary and ergodic with respect to one (or both) of these actions of $G$?

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I'll do it for the precomposition action. Assume $X = \mathbb{R}$ with Lebesgue measure. Let $F:X \to L^p$ be $F(x) = x 1$, where $1$ is the constant function. Then the pushforward $\mu_* F$ is stationary and ergodic.

Now any nice-enough (nonatomic and somehow regular, see Caratheodory's theorem) measure space has an isomorphism to $(\mathbb{R},\mathcal{B},Leb)$, and everything ought to transfer over. Does this work?

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    $\begingroup$ If I've understood the notation correctly, if $X = \mathbb{R}$ with Lebesgue measure $m$, then $L^p$ is supposed to be $L^p(\mathbb{R}, m)$, which doesn't contain the constants. $\endgroup$ Dec 1, 2013 at 16:12
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    $\begingroup$ Yeah, you're right, sorry. If $\mu$ is finite, replace with $([0,1], Leb)$. $\endgroup$
    – user61891
    Dec 1, 2013 at 16:46
  • $\begingroup$ Let $p \in (0,1)$, and consider $L^p(\mathbb R, \mathbb R, \operatorname{Leb})$. $\endgroup$ Dec 3, 2013 at 18:17

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