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In Folland's Real Analysis, p.87,

The decomposition $X = P \cup N$ if $X$ as the disjoint union of a positive set and a negative set is called a Hahn decomposition for $\nu$. It is usually not unique ($\nu$-null sets can be transferred from $P$ to $N$ or from $N$ to $P$), but it leads to a canonical representation of $\nu$ as the difference of two positive measures.

It says this decomposition is not unique because $\nu$-null sets can be transferred from $P$ to $N$ and from $N$ to $P$. What does this mean? Could anyone provide me with a simple example?

According to the Hahn decomposition theorem, If $P'$ and $N'$ is another such pair, then $P \triangle P'=Q \triangle Q'$ is null for $\nu$. Does not this mean that the decomposition is unique?

Mike Pierce
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sleeve chen
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    Basically, if $X = P\cup N$ be a decomposition into positive and negative sets. Let $A$ be a null subset of $P$, i.e., $\nu (A) = 0$. Call $P' = P - A$ and $N' = N \cup A$, then $P'$ and $N'$ is also a Hahn decomposition---the decomposition need not be unique. – hrkrshnn Nov 26 '15 at 05:16
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    Uniqueness can be thought of in more than one sense. The decomposition is not strictly unique in the sense that we can find other decompositions that are not exactly the same. But what we can say is that the decomposition is unique up to null sets. This language just means that any other decomposition just differs from the original decomposition by a null set, i.e., the symmetric differences you described are null sets. – layman Nov 26 '15 at 05:19
  • @hrkrshnn Is there always a no empty null subset of $P$? – Tom Ryddle May 14 '20 at 05:09
  • @TomRyddle Not sure if I understand your question. For general measures, there can be non-empty null subsets of $P$, However this is not guaranteed. You could just take a discrete set and create a measure on it to create an example. – hrkrshnn May 14 '20 at 07:32
  • @hrkrshnn Aja, in that case, the Hahn decomposition is unique! – Tom Ryddle May 17 '20 at 08:04

1 Answers1

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A Hahn decomposition is unique in the following sense: If $P \cup N$ is a Hahn decomposition of X, then there is another Hahn decomposition $P' \cup N'$ such that $P \triangle P'$ and $N \triangle N'$ are $\nu$-null sets.

Therefore, all Hahn decompositions form a unique equivalence class, similar to $L^p$ spaces being equivalence classes of functions that are almost-everywhere equal.

Olorun
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