Analytical Methods in Probability Theory. Proc. conf. by Daniel Dugue, E. Lukacs, V. K. Rohatgi

By Daniel Dugue, E. Lukacs, V. K. Rohatgi

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The indicator function of a set A is defined by 1 A (x) = 1 for x ∈ A and 1 A (x) = 0 for x ∈ X \A. If X is uncountable, show that there is a net of indicator functions of finite sets converging to the constant function 1, but that the net cannot be replaced by a sequence. Problems 33 8. (a) Let Q be the set of rational numbers. Show that the Riemann integral of 1Q from 0 to 1 is undefined (the net in its definition does not converge). ) (b) Show that for a sequence 1 F(n) of indicator functions of finite sets F(n) converging pointwise to 1Q , the Riemann integral of 1 F(n) is 0 for each n.

So (II) does imply (III). Now assume (III). If {xn } is a sequence with infinite range, let x be a limit point of the range. Then there are n(1) < n(2) < n(3) < · · · such that d(xn(k) , x) < 1/k for all k, so xn(k) converges to x as k → ∞. If {xn } has finite range, then there is some x such that xn = x for infinitely many values of n. Thus there is a subsequence xn(k) with xn(k) = x for all k, so xn(k) → x. Thus (III) implies (IV). Last, let’s prove that (IV) implies (I). Let U be an open cover of S.

A filter F in a set X is called an ultrafilter iff for all Y ⊂ X , either Y ∈ F or X \Y ∈ F . 36 General Topology The simplest ultrafilters are of the form {A ⊂ X : x ∈ A} for x ∈ X . These are called point ultrafilters. The existence of non-point ultrafilters depends on the axiom of choice. Some filters converging to a point x are included in the point ultrafilter of all sets containing x; but (0, 1/n), n = 1, 2, . . , for example, is a base of a filter converging to 0 in R, where no set in the base contains 0.

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