$ x \in \mathbb{R} $ , $ f: \mathbb{R} \to \mathbb{R} $, $ f $ is measurable ,$\mu$ is Lebesgue outer measure.
$ A_{n, a}=\{x\ |\ f(a)\le f(x)<f (a)+\frac {1}{n}\}$ for any $ a \in \mathbb{R} $
$\lim_{n \to \infty}\mu(A_{n,a})=0$ for all $ a$ implies continuity of $ f(x)$ a.e.?
@GEdgar reduced the problem to if there is a measurable $ f$ such that $$ \lim_{n\to\infty} \mu (A_{n,a}) = \mu\left\{x\ |\ f(a) = f(x)\right\} =0 $$ holds for all $ a $ and discontinuous ae ?
Fix $a$. The set $A_{n,a}$ decreases as $n$ increases, and $$ \bigcap_n A_{n,a} = \left\{x\ |\ f(a) = f(x)\right\} $$ so $$ \lim_n \mu (A_{n,a}) = \mu\left\{x\ |\ f(a) = f(x)\right\} $$ Thus, your condition means: $f$ takes each value on a set of measure zero, possibly empty.
For example, as Netchaief noted, this holds if $f$ is injective ($f$ takes each value on a set with at most one element).
That does not imply $f$ is continuous a.e.
How about this example: $f(x) = x$ for $x$ rational, and $f(x) = -x$ for $x$ irrational. It is injective, but continuous only at the point $0$.
added
For the new definition of "discontinuous a.e." use this example. Let $A \subseteq \mathbb R$ be a Bernstein set. So both $A$ and its complement have inner measure zero. Then our example is $f(x) = x$ for $x \in A$ and $f(x) =-x$ for $x \notin A$.