Generalizing the ptwise or $L^1$ ergodic theorem

Question

I was able to write a proof based on the hardy-littlewood maximal function of the following statement:

Let $(X, \Sigma, \mu)$ be a $\sigma$-finite measure space, and $d=n \geq 1$ be fixed. Let $\{\theta_t\}_{t \in \mathbb{Z}^d}$ be a group of measure preserving bijections. Then if $f: X \rightarrow \mathbb{R}$ is measurable integrable ($\mu$) then we have

$(2k+1)^{-n}\sum_{t \in [-k, k]^n}f \circ \theta_t$ converges a.s. as $k \rightarrow \infty$ and in $L^1$ if $\mu$ is finite and $f$ is bounded.

Notice that I can only approximate over n-dimensional discrete cubes centered at the origin. But that's all I could possibly hope for directly from Hardy-Littlewood arguments, because that maximal function is at first defined on Euclidean balls. You require a weak-1,1 type estimate, and you can only get that if the shapes you choose are comparable in some nice sense, where for instance, cubes, rectangles with a fixed aspect ratio, or discrete cubes, discrete rectangles all satisfy this. ("Discrete" means finite sets, and all the lebesgue measures appearing in Hardy-Littlewood turn into cardinalities of finite subsets of the integer lattice.)

But I write this question in hopes that someone will tell me how to append additional arguments to go from here to two generalizations.

1. What if I want to be able to take the limit in the strongest sense? I.e. I don't restrict to cubes, or rectangles with a fixed aspect ratio. Instead I want convergence as arbitrary rectangles approach the point at infinity in $\mathbb{Z}^d$ We already know that the ergodic theorem stated above would also hold if I replaced the limit over cubes with an iterated limit, one variable at a time. (By the Birkhoff ptwise ergodic theorem) But even these two combined is not obviously sufficient for this strong limit.
2. What if I want to take averages over $[0, k]^n$ instead of symmetric about the origin? This is clearly stronger than what I know. Is it equivalent?

Dunford and Schwartz make the strongest claim in the sense of both 1 and 2 at the same time. But I found at least a gap in their proof, if not a mistake. However, this link seems to say that the reason I found a gap is because the claim would be false, and even false if I wanted to require merely 1. Am I understanding correctly?