Use the fact that the Lebesgue measure is complete, i.e., every subset of a measurable set of measure zero is measurable and has measure zero (at worse, every measure has a completion). Now, let $I:=[0,1]$ , and let $S \subset [0,1]$ be the set of measure zero.
Then $I= S \cup (S^c)$. We can use the classical definition that $f$ is measurable iff the inverse image of an open set is measurable, while breaking up $I:=[0,1]$ into the union of $S$ ; the measure-zero subset where $f$ is not continuous, and $I\S$ , where $f$ is continuous, and consider the inverse image intersected with each of the two sets (and then consider the union of the inverse images):
Now, let $U$ be open in $\mathbb R$ . Then , $f^{-1}(U)=[f^{-1}(U) \cap(S)] \cup [f^{-1}(U)\cap(S^c)] $. Now, $f$ is continuous in $S^c$, so that $f^{-1}(U)\cap S^c $ is open in $[0,1]$ , and $f^{-1}(U) \cap S $ is a subset of the measurable set $S$ of measure zero, so that $f^{-1}(U) \cap S:=V$ is measurable, ( with measure zero). Then $f^{-1}U)$ is the union of an open set --which is measurable , and a measurable set $V$ ( with measure zero, but we only care that it is measurable), so the inverse image of the open set $U$ of $\mathbb R$ is the union of two measurable sets, and so it follows, it is measurable, so that $f$ is measurable.
We find that for all $x\in \Bbb R^n$ and $F\subset \Bbb R^n$
$$
m^*F = m^*(F\cap B(x,r))+m^*(F\setminus B(x,r))
$$ since $B(x,r)$ is measurable. Let $F=E\cap B(y,r)$ to obtain
$$
f(y) =m^*(E\cap B(y,r)\cap B(x,r))+m^*\left(\left(E\cap B(y,r)\right)\setminus B(x,r)\right).
$$ By interchanging $x$ and $y$, we also obtain
$$
f(x) =m^*(E\cap B(x,r)\cap B(y,r))+m^*\left(\left(E\cap B(x,r)\right)\setminus B(y,r)\right).
$$ This yields
$$
f(x)-f(y) =m^*\left(\left(E\cap B(x,r)\right)\setminus B(y,r)\right)-m^*\left(\left(E\cap B(y,r)\right)\setminus B(x,r)\right),
$$ hence
$$\begin{eqnarray}
f(x)-f(y)&\le& m^*\left(\left(E\cap B(x,r)\right)\setminus B(y,r)\right)\\
&\le&m^*\left( B(x,r)\setminus B(y,r)\right)\\
&\le& m^*\left( B(x,r+|x-y|)\setminus B(y,r)\right)\\
&=&m^*B(x,r+|x-y|)-m^*B(y,r)\\&=&\omega_n \left[(r+|x-y|)^n-r^n\right]
\end{eqnarray}$$ where $\omega_n=m^*B(0,1)$ is the volume of the unit ball. By interchanging $x$ and $y$ we also find that
$$
f(x)-f(y)\ge -\omega_n \left[(r+|x-y|)^n-r^n\right].
$$Therefore
$$
|f(x)-f(y)|\le \omega_n \left[(r+|x-y|)^n-r^n\right] \stackrel{|x-y|\to 0}\longrightarrow 0,
$$ which proves (uniform) continuity of $f$ as desired.
Note: Since
$$
|f(x)-f(y)|\le \omega_n |x-y|\left[(r+|x-y|)^{n-1} +(r+|x-y|)^{n-2}r +\cdots +r^{n-1}\right],
$$ your conjecture that $f$ is locally Lipschitz is correct.
Best Answer
Using Lusin's theorem, there exists a compact $K \subset [0,1]$ such that $\mu(K)>2/3$ and $f$ continuous on $K$. Let $\epsilon>0$. In fact, $f$ is uniformly continuous on $K$, so there exists $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$; without loss of generality, suppose $\delta<1/3$.
Let $h<\delta$. Notice that the intersection between $K$ and $K-h$ is nonempty; otherwise, $1+h= \mu([-h,1]) \geq \mu(K \cup K-h)= \mu(K)+\mu(K-h)=2\mu(K)>4/3$, so $h>1/3$ whereas $h<\delta<1/3$ by assumption.
Let $x_0 \in K \cap (K-h)$. We have $|f(x_0+h)-f(x_0)|<\epsilon$, hence $|f(h)|<\epsilon$ because $f$ is additive. You deduce that $f$ is continuous at $0$.
Finally, it is straightforward to conclude that $f$ is continuous on $\mathbb{R}$.
Edit: For a simple proof of Lusin's theorem, see for example: Marcus B. Feldman, A proof of Lusin's theorem, The American Mathematical Monthly Vol. 88, No. 3 (Mar., 1981) (pp. 191-192).