Suppose we have a dynamical system and a sequence of functions $0=f\_0\leq f\_1\leq\cdots\leq f\_k$. Define $J\_{r,\lambda}$ to be the set of points $x$ such that there are $j\_0<j\_1<\cdots<j\_r$ so that, for each $i<r$, $\sup\_n A\_n(f\_{j\_{i+1}}-f\_{j\_i})(x)>\lambda$. Using the maximal ergodic theorem, we see that $\mu(J\_{1,\lambda})\leq ||f\_k||\_1/\lambda$. Is there a similar bound on $\mu(J\_{r,\lambda})$ (maybe even $||f\_k||\_1/r\lambda$)?


Note that this doesn't seem to trivially reduce to multiple applications of the maximal ergodic theorem, because the sequences of $j$'s might overlap between different points, and one point might see $r$ jumps from $0$ to $j\_r$, while another point doesn't see any jumps until some $j'>j\_r$. 


(We have some ideas that might show this, at least for a slightly more restrictive definition of $J\_{r,\lambda}$---that there are $n\_0<\cdots<n\_{r-1}$ so that, for each $i<r$, $A\_{n\_i}(f\_{j\_{i+1}}-f\_{j\_i})(x)>\lambda$---but they're messy, and it seems like a clever use of the maximal ergodic theorem should give a slick proof, but we don't see how to do it.)



The Maximal Ergodic Theorem more than once
You can't do substantially better than $\|f\_k\|\_\lambda$. Here's a simple example: Consider $X=\{0,1,\ldots,2^{N}-1\}$, equipped with normalized counting measure. The transformation is $T(x)=x+1\bmod 2^{N}$. The functions are 
$$
f\_i(x)=\begin{cases}
2&\text{if $x<i$;}\\
0&\text{otherwise.}
\end{cases}
$$
for $i=0,\ldots,2^{N}$. 


Let $\lambda=1$. Now if $j<2^N-2^{N/2}$ and $1\le n\le N/2$, consider $f\_{j+2^{n}}-f\_{j+2^{n-1}}$. This takes values 2 on the range $j+2^{n-1}$ to $j+2^n-1$
and 0's elsewhere. The maximum of the Birkhoff averages evaluated at $j$ is therefore 1. Hence for each $j\in \{0,\ldots,2^N-2^{N/2}-1\}$, $j\in J\_{N/2,1}$. That is, $\mu(J\_{N/2,1})\ge 1-2^{-N/2}$, but $\|f\_{2^N}\|=2$. You cannot do better than $\mu(J\_{r,\lambda})\le \|f\_k\|/(2\lambda)$. 


