# Are almost everywhere equal functions asymptotic?

I'm quite new at measure theory and all I know I read by myself, so bear with me if this question is too trivial.

Given the definition of asymptotic, can we say that for $x \rightarrow \infty$ two almost everywhere equal functions are asymptotic? I really think that this is not the case, but I haven't been able to find a counterexample. Maybe, if in general this does not hold, would it be right for real functions?

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The short answer is no; I assume that you are working with the real numbers and the usual measure with the usual definition of asymptotic. The counterexample goes like this: any countable set has measure 0, and so choose some sequence xi such that xi goes to infinity (say, the positive integers). Then take any function f and define a new function g by g(x)=f(x) if x is not one of the xi and g(xi)=2f(xi).

The two functions are equal almost everywhere since they differ only on the sequence xi, which as a sequence is countable and thus have measure 0.

Then f(x)/g(x) equals 1 when x is not one of the xi and equals 2 on the xi, which means that the value as you go to infinity doesn't converge, i.e. it always oscillates between 1 and 2 as you go farther and farther out.

Even more awesomely, you can do the same trick not with the sequence xi, but say, with the rational numbers, which are also countable and also have a measure of 0. Then the function given by h(x)=f(x)/g(x) is going to be continuous nowhere in the extended real line (i.e. no limit of h(x) as x goes to any number, or to positive or negative infinity will exist).

The long answer, which I lack time to think through and write up right now, would discuss why countable sets have measures 0 and the interplay between Borel measures, topology, and the equivalence classes of functions given by a.e. equality.

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Right, thanks for the clarification; I think that I might have forgotten the part where it said that all countable sets have measure zero. Certainly it was my mistake not putting enough effort in it. Still, thanks for the answer. –  Andy Aug 22 '10 at 20:41