Let $f$ be a entire function (stable on $\mathbb{R}$), and $E_{\mathbb{R}}$ its real escaping set : $$E_{\mathbb{R}} = \{ x \in \mathbb{R} : f^{(k)}(x) \rightarrow_{k \to \infty} \infty \} $$
We put the Lebesgue measure on $\mathbb{R}$.
Question : If $E_{\mathbb{R}}$ is a (measurable) null set, is it also empty ?
For every continuous function $g$ on the real line and for every positive continuous function $\epsilon$ on the real line, there exists an entire function $f$ such that $|f(x)-g(x)|<\epsilon(x)$ for all real $x$ (This is due to Carleman). So if you can construct a real continuous function with your property then you can also construct an entire one. But construction of a real continuous function with this property does not seem to be difficult. Here is a sketch.
First, let us make the negative semi-axis invariant: $x\leq 0$ implies $f(x)\leq 0$. On the positivde ray, let $f(x)\leq 0$ except some small disjoint intervals $I_k$ tending to $+\infty$. Let $E=\{ x:f(x)\leq 0\}$ be the complement of these intervals. On the intervals we arrange like this: Let $J_k\subset I_k$ be a smaller interval near the middle of $I_k$, where our function is large and has a local max, but this $J_k$ is mapped to $E$, and on the two subintervals $I_k\backslash J_k$ has very large derivative (by absolute value) and the image of these two subintervals contains only one interval of the $I_j$, namely $I_{k+1}$.
Then the escaping set consists of points whose orbits $x_k$ are in the intervals for $k$ large enough. It is clear that this is not empty, and on each interval the set of escaping points is a Cantor set, which can be easily made of measure $0$ by making the derivative very large on the "side" subintervals.
