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Let $f(x) = x^{-1/2}$ for $ x \in (0,1)$ and zero otherwise.

Let $F(x) = \sum 2^{-n} f(x - r_{n})$ where $ {r_{n}} $ is an enumeration of rationals.

1) Prove that $F(x)$ is integrable and thus its series is convergent almost everywhere.

(I did this part)

2) Prove that any function $G$, where $G$ agrees with $F$ almost everywhere, is unbounded in any interval.

I'm stuck on part 2.

Thank you for any help

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Does this mean (1) pick any interval and any function $G$ that agrees with $F$ almost everywhere, and prove that $G$ is unbounded in that interval, or (2) prove that no matter which interval it is (it could be any interval) and no matter which function $G$ you pick, that agrees with $F$ almost everywhere (it could be any of them), that function is unbounded on that interval? The word "any" is in some contexts ambiguous, and usually the ambiguity can be avoided by saying "every" instead of "any". I think it's really only mathematicians who habitually use "any" as a synonym of "every". – Michael Hardy Nov 28 '11 at 17:03
up vote 1 down vote accepted

Well, fix any interval $I$, pick any $r_n$ in the interior of $I$ and use the fact that $F\geqslant2^{-n}f_n$ everywhere, with $f_n:x\mapsto f(x-r_n)$. If one shows that $f_n$ is unbounded on $I$, the proof is over. Can you?

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