real analysis - Show if $f(x+y)=f(x)+f(y)$ for all $x,y$ and $f$ is Lebesgue measurable, then $f$ is continuous.

I'm trying to solve this problem $(5.8)$ from Bass' Real Analysis for Graduate Students:

Show that if $f:\Bbb{R}\to\Bbb{R}$ is Lebesgue measurable and $$f(x+y)=f(x)+f(y)$$ for all $x,y\in\Bbb{R}$, then $f$ is continuous.

The first thing I note is that it suffices to show that $f$ is continuous at $0$, since if this is the case then for any $\epsilon>0$ and $x_0\in\Bbb R$, there exists $\delta>0$ such that $|x|<\delta$ implies $|f(x)|<\epsilon$; in particular, replacing $x$ with $x-x_0$, we see $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|=|f(x-x_0)|<\epsilon$.

Now, what I need to show then is that $f^{-1}(-\epsilon,\epsilon)$ contains some interval at the origin. I know that this set is Lebesgue measurable because $(-\epsilon,\epsilon)$ is Borel measurable, and I know it contains $0$ since $f(0)=0$, but I'm not sure how to show it contains $(-\delta,\delta)$ for some $\delta$. I haven't used the fact that the set is Lebesgue measurable, so obviously theres some way I can use that but I'm not seeing it.

Any suggestions?

edit: because this has been tagged as a possible duplicate, I should point out that I'd prefer to figure out if the solution I am working on will work, rather than just copy a completely different solution outlined by some other user.

Answer

We show that $f$ is right continuous at zero by considering $f$ on the set $[0,1]$. A similar argument applied to $f$ on $[-1,0]$ will show that $f$ is left continuous at the origin. By Lusin's theorem, there exists a compact set $K\subset [0,1]$ with $\mu(K)\geq 2/3$ on which $f$ is uniformly continuous. Hence there exists $0<\delta<1/3$ such that for $x,y\in K$ if $\lvert x-y \rvert <\delta$, then $\lvert f(x)-f(y)\rvert<\epsilon$. Consider the translate $K+h$ where $0 $$\lvert f(h)\rvert=\lvert f(a)-f(a-h)|<\epsilon$$
whenever $0