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I have a problem which asks to show that a function $f$ of bounded variation can be expressed uniquely except for addition of constants as the sum of an absolutely continuous function and a singular function.

I have been able to express $f$ as $f = g + h$ where $f$ is absolutely continuous and $h$ is singular. My problem lies in showing uniqueness and so I need assistance.

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    Assume that $f=g+h = k+\ell$, where $g$ and $k$ are absolutely continuous and $h$ and $\ell$ are singular. This implies that $g-k = \ell-h$. Show that this implies that $g-k$ and $\ell-h$ are constant, so that $g=k+C$ and $h=\ell-C$ for some constant $C$.2012-04-10
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    @ArturoMagidin: Thanks. If I can show that $g-k =0$, then I'll be done, but I'm struggling a bit to show that. any further hints.2012-04-10
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    Note that you want "uniqueness up to addition of constants", so you do not need to show $g-k=0$, you only need to show that $g-k$ is *constant*. Note that you have an equality between $g-k$ and $\ell-h$. Maybe you can leverage the fact that $g-k$ is absolutely continuous (being a difference of two absolutely continuous functions) and that $\ell$ and $h$ are singular to get that...2012-04-10
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    @ArturoMagidin: Okay. this is what I have. Since $l$ and $h$ are singular, $l'=0=h'$ and since $l-h = g-k$, $(g-k)'=0$ and since $g-k$ is absolutely continous, $g-k$ is a constant...right?2012-04-10
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    Well, $\ell$ and $h$ are singular, so the derivative is zero *outside a set of measure $0$*; but you can then deduce that $\ell-h$ is has zero derivative outside of a set of measure $0$. So $g-k$ is an absolutely continuous function whose derivative is $0$ outside of a set of measure $0$, and from this you should be able to deduce that $g-k$ is constant, yes.2012-04-10
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    @ArturoMagidin thanks very much.2012-04-10
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    Might I suggest you post your solution (with details) as an answer? Then people may point any weaknesses (or lack thereof) in your argument, and you will eventually be allowed to accept it! That way the question will not appear as "unanswered". Thank you.2012-04-10

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