Let $\Omega$ be a probability space, and $\{ \tau \}_{y\in \mathbb R^d}$ an ergodic group of measure-preserving transformations, $\tau_y:\Omega \to \Omega$.

The ergodic theorem says that if $f \in L^1(\Omega)$, and we define $\hat f:\mathbb R^d\times\Omega \to \mathbb R$ by $\hat f(y,\omega):= f(\tau_y\omega)$, then \begin{equation*} \lim_{R\to \infty} |B_R|^{-1} \int_{B_R} \hat f(y,\omega) \, dy = \mathbb E[f], \end{equation*} at least almost surely in $\omega$.

In particular, for any $p\geq 1$, the quantity $\| \hat f \|_{L^p(B_R)}$, for large $R$ and up to a scaling factor, should be approximately $\| f \|_{L^p(\Omega)}$.

My question is this: can we say the same thing about Lorentz norms?

Let me define the Lorentz norm $\| \cdot \|_{L^{d,1}}$, which is the particular one I care about. Let $(X,\mu)$ be a measure space, and say $\mu(X) < \infty$. Then for $f:X\to\mathbb R$ we define \begin{equation*} \| f \|_{L^{d,1}(\Omega)} := d \int_0^\infty \mu \left( \{ x\in X: |f(x)| \geq t \} \right)^{1/d} \ dt \end{equation*}

What I think should be true: \begin{equation*} \lim_{R\to \infty} R^{-1} \| \hat f \|_{L^{d,1}(B_R)} = c_d \| f \|_{L^{d,1}(\Omega)}, \end{equation*} where $c_d = |B_1|$, the volume of the unit ball in $\mathbb R^d$.

An attempt at the proof: check that \begin{equation*} R^{-1} \| \hat f \|_{L^{d,1}(B_R)} = |B_1| d \int_0^\infty \left( |B_R|^{-1} \int_{B_R} I_{\{|f|\geq t \}}(\tau_y\omega) \ dy \right)^{1/d} \ dt. \end{equation*} Here $I_E$ is the indicator function of the event $E$.

Okay, now notice that the integrand converges to $\mathbb P[f \geq t]^{1/d}$, again by the ergodic theorem. So if we could pass to limits under the integral, we'd be done! Unfortunately, I don't know how to do this.

Is there a simple trick that I am missing? Or do I have to revisit the proof of the ergodic theorem and work from scratch?