The following result, which I know under the name Fekete's lemma is quite often useful. It was, for example, used in this answer: Existence of a limit associated to an almost subadditive sequence.
If $(a_n)_{n=0}^\infty$ is a subadditive sequence of real numbers, i.e., $$(\forall m,n) a_{m+n} \le a_m + a_n,$$ then $$\lim\limits_{n\to\infty} \frac{a_n}n = \inf_n \frac{a_n}n.$$
Some references are given in Wikipedia article, the original Fekete's paper is available here. Basically the exponential version (for submultiplicative sequences) can be shown in a similar way as Satz II in this paper.
I was wondering, whether some analogous claim is true for functions. I.e. something like: Whenever $f:{(0,\infty)}\to{\mathbb R}$ fulfills $$(\forall x,y)f(x+y) \le f(x)+f(y)$$ (i.e., it is subadditive), then $$\lim\limits_{x\to\infty} \frac{f(x)}x = \inf_x \frac{f(x)}x.$$ (In particular, the above limit exists -- if we include the possibility $-\infty$.)
Clearly, this is not true without any additional assumptions on $f$. (E.g. if $f$ is any non-linear solution of Cauchy's equation, then $\liminf \frac{f(x)}x < \limsup \frac{f(x)}x$ and $f$ is both subadditive and superadditive. Probably even much simpler examples can be given.)
On the other hand, if $f$ is well-behaved, the above claim is true. If I assume that $f$ is bounded on intervals of the form $(0,x]$, then I can basically repeat the proof which is given for sequences here.
So my question is:
Under what assumptions on $f$ the above claim is true.
Can you give some references for this claim?
EDIT: I found a result which shows that measurability of $f$ is sufficient and added this result as an answer. I think this is sufficient for most applications and my guess is that there is not much space to improve this result. However, I will wait a little bit before accepting my own answer - just in case someone would like to add some interesting information or further useful references. I have accepted my own answer, but if you have some interesting information which you can add, I'll be very glad to learn about it.