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  "body": "\n## Statement\n\nLet $(X, \\Sigma)$ be a [[measurable space|N4KDT83V3S]] and let $\\nu$ be a [[signed measure|VG2P7R91Z7]] on it.\nA set $A \\in \\Sigma$ is *positive* for $\\nu$ if $\\nu(E) \\ge 0$ for every $E \\in \\Sigma$ with $E \\subseteq A$, and *negative* if $\\nu(E) \\le 0$ for all such $E$.\nThen there are disjoint sets $P, N \\in \\Sigma$ with $P \\cup N = X$ such that $P$ is positive and $N$ is negative for $\\nu$.\nIf $P', N'$ is another such pair, then $P \\triangle P' = N \\triangle N'$ is null for $\\nu$: every measurable subset of it has $\\nu$-measure $0$.\n\n## Remarks\n\nAssume $\\nu$ does not take the value $+\\infty$; otherwise apply the argument to $-\\nu$.\nLet $s$ be the supremum of $\\nu(A)$ over positive sets $A$.\nA maximising sequence of positive sets has a positive union $P$ with $\\nu(P) = s$.\nSuppose $N = X \\setminus P$ contained a set of strictly positive measure.\nThen a lemma gives a positive subset of $N$ of strictly positive measure, and adding it to $P$ contradicts maximality.\n\nExample: let $\\nu(E) = \\int_E f \\, d\\mu$ with $f \\in L^1(\\mu)$ real.\nThen $P = \\set{f \\ge 0}$ and $N = \\set{f \u003c 0}$ is a Hahn decomposition.\nThe set $\\set{f = 0}$ can be put into either part, which shows that the decomposition is unique only up to null sets.\n\n## Explanations\n\n### L1 human\n\nA signed measure has regions where it is positive and regions where it is negative.\nThe theorem says that one can split the whole space into two measurable pieces, one on which every part counts positively and one on which every part counts negatively.\nThe cut is not unique, but the possible cuts differ only by sets that do not matter for $\\nu$.\n\n### L2 human\n\nThe Hahn decomposition is the first step towards the [[Jordan decomposition|WAGJVK2NWE]] $\\nu = \\nu^+ - \\nu^-$, which reduces questions about signed measures to questions about measures.\nIt is also used in the proof of the [[Radon–Nikodym theorem|1ZSZS5G1Q4]], applied to signed measures of the form $\\nu - c \\mu$.\n",
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