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  "body": "\n## Statement\n\nLet $X$ be a [[Polish space|584GXXY73M]] and let $A \\subseteq X$.\nThen $A$ belongs to the [[Borel sigma-algebra|0WQXY32Q8D]] of $X$ if and only if both $A$ and its [[complement|0RKF1ZT5JJ]] $X \\setminus A$ are [[analytic sets|3RWWZYPAFV]].\n\n## Remarks\n\nOne direction is easy: every Borel set is analytic, and the complement of a Borel set is Borel.\nFor the other direction, the Lusin separation theorem says that two disjoint analytic sets $A$ and $B$ can be separated by a Borel set $C$, that is, $A \\subseteq C$ and $C \\cap B = \\emptyset$.\nApplied to $A$ and $X \\setminus A$, it gives a Borel set $C$ with $A \\subseteq C \\subseteq A$, so $C = A$.\n\nThe hypothesis on the complement cannot be dropped.\nEvery uncountable Polish space contains an analytic set that is not Borel, and the complement of such a set is not analytic.\n\nIn the notation of the projective hierarchy, the theorem says $\\boldsymbol{\\Delta}^1_1 = \\boldsymbol{\\Sigma}^1_1 \\cap \\boldsymbol{\\Pi}^1_1$ is exactly the class of Borel sets.\n\n## Explanations\n\n### L1 human\n\nAnalytic sets are the images of Borel sets under continuous maps, and they can be more complicated than Borel sets.\nThe theorem says when an analytic set is Borel after all: exactly when its complement is analytic too.\nSo \"Borel\" means \"simple to describe from both sides\".\n\n### L2 human\n\nThe theorem is the standard way to prove that a set is Borel without writing it explicitly from open sets by countable unions and complements.\nOne shows that the set and its complement are both projections of Borel sets, and the theorem does the rest.\nA common use is the result that a bijective Borel map between Polish spaces has a Borel inverse.\n",
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