I want to show that if an additive (Cauchy functional) is Lebesgue measurable, and Lebesgue integrable on a neighbourhood of a point on the real line, then it must be continuous, from there it's easy to show that $f(x) = bx$ for some constant $b$.
I'm having trouble with this, my question is, how do I use the above information to show that $f$ must be continuous on a neighbourhood of zero. From there it is simple to show it is continuous everywhere
You need the fact that if $E$ has positive Lebesgue measure then $E-E$ contains an interval around 0. Write $\mathbb R$ as union of the sets $\{x:|f(x)|<n\}$. At least one of these sets must have positive measure. Let $E=\{x:|f(x)|<n\}$ have positive measure. There exists $t>0$ such that $(-t,t) \subset \{x:|f(x)|<n\}-\{x:|f(x)|<n\}$. If $|x|<t$ the we can write $x=a-b$ with $|f(a)|<n$ and $|f(b)|<n$. Additivity gives $|f(x)|<2n$. Given $\epsilon >0$ let $\delta=\frac t k$ where $k$ is such that $\frac {2n} k <\epsilon$. Then $|x|<\delta$ implies $| x k| <t$ , so $k|f(x)|<2n<k\epsilon$ or $|f(x)| <\epsilon$. We have proved that $f$ is continuous at 0. Continuity ate any other point $x$ follows from the fact that $f(y)-f(x)=f(y-x)$.