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 $\mathbb{R}$. 



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> **Question** : If $E\_{\mathbb{R}}$ is a (measurable) null set, is it also empty ?
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Entire functions with a null real escaping set
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.


