real analysis - Solutions to matrix-valued multiplicative Cauchy equation under local boundedness

Let $f : (0, \infty) \to \mathbb{R}^{n \times n}$, $n \in \mathbb{N}$ satisfy the functional equation $f(x + y) = f(x) f(y)$. In general, $f$ need not be measurable (by the usual constructions of non-measurable solutions based on a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$).

Is it true that if $f$ is uniformly bounded on $(0, \infty)$ (or only on some subinterval of $(0, \infty)$) then $f$ is measurable. (It then can also be shown that $f$ is continuous and of the form $f(x) = \exp(A x)$ for some $A \in \mathbb{R}^{n \times n}$.)

Remarks:

1. For $n=1$ this follows from here. For $n \geq 2$ the above functional equation translates into a system of one-dimensional functional equations $f_{ij}(x+y) = \sum_{k=1}^n f_{ik}(x) f_{kj}(y)$. Are similar arguments applicable?


2. For "$n=\infty$" the above claim is not true, see e.g. Doob "Topics in the theory of Markoff chains" (1942), p. 2.

Edit: For the case of stochastic solutions ($f(x)$ is a stochastic matrix for each $x > 0$) the above question was solved in the affirmative by W. Doeblin in the 30s.)