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Given that $f,g:\mathbb{C}\rightarrow \mathbb{C}$ are holomorphic, $A=\{x\in\mathbb{R}:f(x)=g(x)\}$. The minimum requirement for $f=g$ is

  1. $A$ is uncountable

  2. $A$ has positive lebesgue measure

  3. $A$ contains a nontrivial interval

  4. $A=\mathbb{R}$

By identity theorem to be $f=g$ we just need a limit point inside $A$, so If $A$ has positive lebesgue measure then it will contain a interval so which will have one limit point. so $2$ is correct?

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    I think the answer is 1.2012-07-15
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    As has been pointed out, $A$ having positive measure does NOT imply $A$ contains an interval; nevertheless, by the [Steinhaus theorem](http://en.wikipedia.org/wiki/Steinhaus_theorem) the difference set DOES indeed contain an interval.2012-07-15

2 Answers 2

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The minimum is the list is the requirement 1. The discrete subsets of $\Bbb C$ are necessarily countable, since $\Bbb C$ is separable (*). So if $A$ is uncountable, it's not a discrete set and it has a limit point, which proves that $f\equiv g$.

(*) Let $D\subset\Bbb C$ be discrete. For $z\in D$, consider $r_z$ such that $B(z,r_z)\cap (D\setminus \{z\})=\emptyset$. Then by separability, extract from the open cover $(B(z,r_z),z\in D)$ a countable subcover of $D$.

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    could you tell me why seperability of $\mathbb{C}$ confirm a discrete subset of $\mathbb{C}$ is countable? suppose $A$ be a discrete subset of $\mathbb{C}$ then if $A$ is uncountable then $A$ has a limit point hence contradiction? where seperability is used?2013-06-02
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    @TaxiDriver I've added details.2013-06-02
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The irrationals have positive measure and do not contain any interval so your argument is not correct. On the other hand, being uncountable is the weakest condition and it already imply the existence of a limit point on $A$. So the answer is 1.