A subadditive maximal ergodic theorem Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $\tau:\Omega\to\Omega$ be a measurable map on $(\Omega,\mathcal A)$ with $\operatorname P\circ\:\tau^{-1}=\operatorname P$, $Y_n:\Omega\to[-\infty,\infty)$ be $\mathcal E$-measurable for $n\in\mathbb N$ with $\operatorname E\left[Y_1^+\right]<\infty$ and $$Y_{m+n}\le Y_m+Y_n\circ\tau^m\;\;\;\text{for all }m,n\in\mathbb N\tag1$$ and $$M_n:=\max(Y_1,\ldots,Y_n)\;\;\;\text{for }n\in\mathbb N.$$

It's easy to show the following extension of the maximal ergodic theorem: $$\operatorname E[Y_1;M_n\ge0]\ge0\;\;\;\text{for all }n\in\mathbb N.\tag2$$

The ordinary maximal ergodic theorem is given by the special case, where $$Y_n=\sum_{i=0}^{n-1}X\circ\tau^i\;\;\;\text{for all }n\in\mathbb N$$ for some integrable real-valued random variable on $(\Omega,\mathcal A,\operatorname P)$. In that special case, it can be deduced from $(2)$ that $$\operatorname P\left[\sup_{n\in\mathbb N}\left|\frac{Y_n}n\right|\ge c\right]\le\frac1c\operatorname E[|Y_1|]\;\;\;\text{for all }c>0\tag3.$$

Can we extend this result to the general case?

 A: The answer is no in general, but yes if the sequence $Y_n$ is non-negative.
First, let us focus on the case where $Y_n$ is non-positive. Then, $\sup \frac{1}{n}|Y_n|=-\inf \frac{1}{n}Y_n$. If you assume moreover that all the $Y_n$ are $L^1$, then by Kingman's subadditive theorem, $\frac{1}{n}Y_n$ converges to $Y=\inf \frac{1}{n}Y_n$ almost surely. Notice then that $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=\mathbb{P}\left (Y\leq -c\right )$. Whenever $Y$ has positive probability of taking the value $-\infty$, you cannot bound $\mathbb{P}\left (Y\leq -c\right )$ by something converging to 0 as $c$ goes to infinity.
Here is a concrete counter-example. Let $Y_n$ be the constant function $Y_n=-n^2$. Then $Y_n$ is subadditive and satisifes all your assumptions. You have $\sup \frac{1}{n}|Y_n|=+\infty$ and so for any $c$, $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=1$, so you don't have $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|)=\frac{1}{c}$.
However, small remark : the answer is yes for non-positive $Y_n$ if you have the property that $\mathbb{E}(\frac{1}{n}|Y_n|)\leq \mathbb{E}(|Y_1|)$. Indeed, using Markov inequality, you get $\mathbb{P}\left (\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(\frac{1}{n}|Y_n|)\leq \frac{1}{c}\mathbb{E}(|Y_1|)$ and this is true for all $n$, so this is true for the almost sure limit, using dominated convergence.

About the non-negative case now. Inequality (3) is usually stated without the absolute values in literature :
$$\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|).$$
This statement is true in general and so in particular, the answer to your question is yes whenever $Y_n$ is non-negative.
Indeed, consider a subbaditive sequence $Y_n$, that is satisfying your condition $Y_{n+m}\leq Y_m+Y_n\circ \tau^m$. Let $Z_n=\sup_{k=1,...,n}\frac{1}{k}Y_k$. Also let $\widetilde{Y}_n=\sum_{j=0}^{n-1}Y_1\circ \tau^j$ and finally, let $\widetilde{Z}_n=\sup_{k=1,...,n}\frac{1}{k}\widetilde{Y}_k$. As you claim, the result is true for the sequence $\widetilde{Y}_n$.
Note that since $Z_n$ in non-decreasing, you have $\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )=\lim_n\mathbb{P}(Z_n\geq c)$ so we just need to prove that $\mathbb{P}(Z_n\geq c)\leq \frac{1}{c}\mathbb{E}(Y_1)$.
Now for fixed $n$, for every $x$, there exists $1\leq k(x)\leq n$ such that $Z_n=\frac{1}{k(x)}Y_{k(x)}$. Because of subadditivity, you have $Y_{k(x)}\leq \sum_{j=0}^{k(x)-1}Y_1\circ \tau^j(x)=\widetilde{Y}_{k(x)}(x)$.
So $\frac{1}{k(x)}Y_{k(x)}\leq \frac{1}{k(x)}\widetilde{Y}_{k(x)}(x)\leq \widetilde{Z}_n(x)$. This proves that for any $x$, $Z_n(x)\leq \widetilde{Z}_n(x)$ so $\mathbb{P}(Z_n\geq c)\leq \mathbb{P}(\widetilde{Z}_n\geq c)$. Using that $\widetilde{Z}_n$ also is non-decreasing, you get $\mathbb{P}(\widetilde{Z}_n\geq c)\leq \mathbb{P}\left (\sup\frac{1}{n} \widetilde{Y}_n\geq c\right )$ and so you can use the result for $\widetilde{Y}_n$.
