As mentioned in this link, it shows that For any f on the real line R1, f(x+y)=f(x)+f(y) implies f continuous ⇔ f measurable.
But
how to show there exists such an non-measurable function satisfying f(x+y)=f(x)+f(y)?
I guess we may use the uniform bounded principal and the fact that f is continuous iff it is continuous at zero under the above assumption.
Thanks in advance!
Answer
Considering R as infinite-dimensional Q vector space, any linear map will do. For example, one can extend the function
f(x)=42a+666b if x=a+b√2 with a,b∈Q
defined on Q[√2] to all of R, if one extends the Q-linearly independent set {1,√2} to a basis of R. (This requires the Axiom of Choice, of course)
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