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Consider Cauchy's Functional Equation

$$\phi(t+s)=\phi(t)+\phi(s).$$

Can we say that any right continuous with left limits (cadlag) solution is Borel measurable? Obviously continuous solutions are Borel measureable, however, discontinuous solutions are not. I've also read that measurability can be reduced to cadlag, but I don't see how?

Any clarification is greatly appreciated.

In addition, we can observe that solutions to Cauchy's functional equation are deterministic analogues of Levy processes, however in the definition of Levy processes Cauchy's functional equation is satisfied only up to equality in distribution. Does this enable solutions to become Borel or Lebesgue measurable?

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Any measurable solution is continuous. Any nonmeasurable solution is very "wild": in particular its graph is dense in $\mathbb R^2$, so it has no one-sided limits anywhere.

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  • $\begingroup$ Thanks for your answer. Are you suggesting that any cadlag solution is non measurable? $\endgroup$ – John Nov 24 '14 at 2:02
  • $\begingroup$ No, I'm saying that any cadlag solution is continuous. $\endgroup$ – Robert Israel Nov 24 '14 at 2:50
  • $\begingroup$ Ok right. Thanks for the clarification. Did you see my edit of the original question? $\endgroup$ – John Nov 24 '14 at 2:54

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