Yes. Let $\nu$ be a probability measure on $[0,1]$. Let $T$ be the left shift on a sequence space $X:=[0,1]^{\mathbb N}$ equipped with the product $\sigma$-field and the product measure $\mu:=\nu^{\mathbb N}$. Then for **every** strictly increasing sequence
$\{n_k\}$ and $f \in L^1 (X)$we have 
$$(*) \quad  \lim_{K \to \infty} \frac{1}{K} \sum_{k = 0}^{K-1} f(T^{n_k} (x)) = \int f d\mu \text{ for a.e } x \in X \, ,$$ 
so $E(T)=0$. To verify $(*)$, it suffices to check it for a dense collection of functions in $L^1$. If $f$ depends only on the first $q$ coordinates, then $(*)$ follows from the law of large numbers if you separate $n_k$ into at $q$ subsequences where each subsequence has gaps at least $q$. Since such functions $f$ are dense, this completes the proof.