Does "All additive functions are linear" imply every set of reals is measurable?

Suppose we are working in ZF set theory without choice. An additive function is a function defined over the real line such that $f(x+y)=f(x)+f(y)$. It is known that if every set of reals is measurable, then every additive function is linear. Is the converse true in ZF?

No.

We know that "Every set of reals has the Baire property" will also have the same consequence, and that ZF+"Every set of reals has the Baire property" is actually a weaker(!) theory than ZF+"Every set of reals is Lebesgue measurable", as shown by Shelah.

Example of two dependent random variables that satisfy $E[f(X)f(Y)]=Ef(X)Ef(Y)$ for every $f$

If $f$ is uniformly differentiable then $f '$ is uniformly continuous?

Nature of the series $\sum\limits_{n}(g_n/p_n)^\alpha$ with $(p_n)$ primes and $(g_n)$ prime gaps

Jobs in industry for pure mathematicians [closed]

distance from the centre of a $n$-cube as $n \rightarrow \infty$

Given an $n\times n\times n$ cube, what is the largest number of $1\times 1\times 1$ blocks that a plane can cut through?

What is the relation between connections on principal bundles and connections on vector bundles?

How do you pronounce (partial) derivatives?

Continuous $f\colon [0,1]\to \mathbb{R}$ all of whose nonempty fibers are countably infinite?

If $\lim_n f_n(x_n)=f(x)$ for every $x_n \to x$ then $f_n \to f$ uniformly on $[0,1]$?

How the dual LP solves the primal LP

How to prove there exists $c$ such $f(c)f'(c)+f''(c)=0$