| "text": "Sigma Algebra\nMeasure Theory, Sigma Algebra\nSearch\nSite map\nContact us\nOperation Clog\nMain Page\nCalculus\nIntegral Calculus\nMeasure Theory\nUse the arrows\nbelow to step through Measure Theory.\nCustom Links:\nSigma Algebra\nBefore I define a sigma algebra,\nI want to emphasise that many of the\nnotions that we will come across in measure theory have analogues in\ntopology.\nFor example, a sigma algebra,\nas we will see shortly,\nis similar to a topology on a set,\ni.e. a collection of subsets that obey certain properties.\nAnd measurable functions are analogous to continuous functions , and so on.\nLet x be a set.\nA sigma algebra on X, sometimes denoted σ, is a collection of subsets of X\nsuch that:\nX is in σ.\nA is in σ iff the complement of A is in σ.\nThe countable union of sets in σ also lies in σ.\n(The term algebra is somewhat unfortunate, since this has nothing to do with rings.)\nWe call the pair (X,σ) a \"measurable space\",\nsimilar to a topological space.\nIf A is in σ, we say A is \"measurable\",\nor a \"measurable set\",\nanalogous to an open set.\nFrom (1) and (2), it follows that the empty set is in σ.\nUsing Demorgan's law,\none can show that the countable intersection of sets in σ also lies in σ.\nNote that σ could be X and the empty set.\nThis is the indiscrete algebra, similar to the indiscrete topology.\nAt the other extreme, σ could contain every subset of X.\nThis is the discrete algebra, similar to the discrete topology.\nOne sigma algebra may be stronger than another, if it includes more subsets of X.\nThus algebras can be partially ordered, just as topologies on X are partially ordered.\nNote that the intersection of algebras on X produces another valid algebra.\nThus we may talk about the weakest algebra on X that contains a given collection C of sets within X.\nIt is the intersection of all algebras\nthat contain C.\nThere is at least one such algebra, namely the discrete algebra on X,\nthus the \"generated algebra\" makes sense,\nand is denoted σ[C].\nWe might think of C as a base for σ.\nIf C, in the above paragraph, is itself a topology on X,\nthen σ[C] is a Borel sigma algebra.\nThis will become clearer later on.\nUnion and Intersection\nThe arbitrary intersection of sigma algebras is another sigma algebra,\nbut not so for unions.\nLet X be two points, and let Y be two more disjoint points.\nLet Z be X and Y taken together, consisting of 4 points.\nLet A be an algebra on Z with any combination of the points of X\nand none or both of the ppoints of Y.\nLet B be an algebra on Z with any combination of the points of Y\nand none or both of the ppoints of X.\nThe union of these algebras A∪B consists of 12 sets out of 16.\nIf this is sigma, then the union of any two of these 12 sets gives another of the 12 sets.\nLet one set have one point from X, and let another set have one point from Y.\nThere union is not in A∪B, hence A∪B is not sigma.\nSubset\nLet Y be a measurable set within X.\nRestrict σ to the sets that are entirely contained in Y.\nThis includes Y and the empty set,\nand the union of any countable collection of sets in Y.\nFinally, if A is measurable in X, and contained in Y,\nthen Y intersect (A complement) is also in σ.\nThis is of course the complement of A relative to Y.\nTherefore σ restricted to Y becomes a sigma algebra on Y.\nSigma Algebra on the Reals\nRecall that the standard topology of the reals is based on the open intervals.\nLet this act as a base for σ.\nFor any point z in R, a countable collection of open intervals closes in on z.\nTherefore every point belongs to σ.\nJoin the open intervals with their endpoints to produce the half open intervals and the closed intervals.\nClosed intervals include points, i.e. the degenerate closed intervals,\nand thus the endpoints of any closed interval lie in σ.\nRemove these to get the open intervals, as described above.\nHalf open intervals pointing up can be used to converge on any given point z.\nThus the endpoints of these half open intervals are in σ, and can be removed to give open intervals.\nA similar result holds for half open intervals pointing down.\nClosed intervals can be combined to produce closed rays pointing up and down,\nand two such rays intersect in a closed interval.\nSimilarly, open intervals combine to produce open rays extending up and down,\nand open rays overlap in open intervals.\nIn summary, the following sets imply the same sigma algebra on the reals.\nOpen intervals, closed intervals,\nhalf open intervals pointing up,\nhalf open intervals pointing down,\nclosed rays, and open rays.\nIn contrast,\nstart with the points of R, and produce a σ algebra.\nNote that σ must contain every countable collection of points, and the complements thereof.\nShow that these sets satisfy the criteria of a sigma algebra.\nNow, the countable union of points cannot create a closed interval,\nhence this is a weaker algebra on the reals.\nMeasurable Function\nA function from X to Y is measurable if the preimage of every measurable set in Y is measurable in X.\nThis assumes X and Y are measurable spaces, with sigma algebras.\nThis definition should remind you of a continuous function from one topological space into another,\nwhere the preimage of every open set is open.\nThe composition of measurable functions,\nfrom X to Y to Z, is measurable from X to Z.\nThe identity map on X is measurable, for every σ on X.\nAny function from a discrete space, or into an indiscrete space, is measurable.\nUsing the Base\nAssume the measurable sets of Y are generated by a base C.\nA function f from X into Y is measurable iff the preimage of every base set in C is measurable.\nThe reasoning is the same as that used in topology,\nwhere it is sufficient to show the preimage of every base open set is open.\nArbitrary unions in the range pull back to arbitrary unions in the domain, and that proves continuity.\nLet's do the same thing here.\nAssume each set in C has a measurable preimage under f.\nEach measurable set in Y is constructed by taking the union or complement of preexisting measurable sets.\nProceed by induction on the number of steps in the construction of our measurable set.\nWhen n = 1, we have the assertion \"belongs to C\", and the preimage is measurable by assumption.\nIf T is measurable in Y, with measurable preimage S in X,\ntake complements, and the preimage of the complement of T is the complement of S.\nBoth these sets are measurable, by the properties of a sigma algebra.\nSimilarly, the countable union of measurable preimages produces a measurable set in X\n- and that completes the proof.\nLinear Combination of Measurable Functions is Measurable\nIf σ on the reals is generated by closed intervals, and f is measurable,\nand b is a real constant, then b×f is measurable.\n(This clearly holds if b = 0.)\nBy the above, we only need look at base sets.\nGiven a closed interval, divide it by b, and the preimage of the scaled closed interval, under f,\nbelongs to σ.\nThis is the preimage of the original open interval under b×f.\nTherefore b×f is measurable.\nAs you move to the sum of two measurable functions f+g,\nbeware of the topology trap.\nIn topology, we considered an arbitrary point x in the preimage of a given base open set,\nthen surrounded x with an open neighborhood that maps into our open set.\nThe preimage is the union of these open neighborhoods, and is open.\nBut that assumes the arbitrary union of open sets is open.\nIn a sigma algebra, unions must be countable.\nWe have a little more work to do.\nStart with a closed interval [a,b], and graph, in the xy plane, a ≤ x+y ≤ b.\nThis is a closed strip slanting down and to the right.\nCover the strip with closed squares.\nThis is not a partition; the squares overlap at their boundaries.\nThe covering is fractal in nature - I'll leave the details to you.\nSince each square has a distinct rational point inside, the covering is countable.\nNow, each square implies a range of x and a range of y, i.e. two closed intervals.\nThese pull back, through f and g respectively, to preimages that belong to #s'.\nTheir intersection also belongs to #s'.\nThese points in the domain put f in the proper range, and g in the proper range,\nas described by our square, and the square lies in our strip, so that f+g lies in [a,b].\nThe countable union of these preimages also lies in #s', and that covers the strip.\nThe preimage of [a,b] is a measurable set, and f+g is measurable.\nThe proof for f×g works the same way.\nGraph a ≤ xy ≤ b, and cover the region between these two hyperbolas with squares.\nSimilarly, a ≤ 1/x ≤ b is used to prove 1/g is measurable,\nprovided g is nonzero.\nCombine this with multiplication, and f/g is measurable.\nConcatenation\nIf A is any measurable set in X, and f is a measurable function on X,\nthen f is measurable on A.\nThe preimage of a measurable set is measurable in X, so intersect with A, and find another measurable set.\nConversely, let X be the countable union of disjoint measurable regions, and assume f is measurable on each region.\nThe various preimages under f are measurable inside each region,\nand measurable in X.\nTake their union to find a measurable set, which is the preimage of f throughout X.\nThis is analogous to topology, where a function is continuous on a space iff it is continuous on the individual components.\nMin Max\nIf the topology of R includes rays, then the absolute value of a measurable function is measurable.\nPartition X into two regions, that map to values ≥ 0, and values < 0.\nBy the above, f is measurable on both regions.\nNegate f on the second region, then put the two pieces back together to find the measurable function |f|.\nMax(f,g) is derived in the same way.\nSince f-g is measurable, we can separate X into regions where f ≥ g, and f < g.\nApply f and g to these regions respectively, and max(f,g) is measurable.\nSimilarly, min(f,g) is measurable.\nThis generalizes to any finite set of measurable functions.\nCustom Links:\n© 2002-2023 Karl Dahlke",
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