[Also as postscript and pdf]
Maximum marks possible: 50
You are permitted to bring the book `Adams and Guillemin: Measure Theory And Probability' with you - just the book, nothing else is permitted (and no notes in your book!) You may use theorems, lemmas and propositions from the book.
Solution:- The are continuous, hence measurable (or check this directly). Moverover increases with so the monotone convergence theorem applies and shows that the limit is measurable and has integral
Note: The limit is not bounded, hence not Riemann integrable in the strict sense (it has an improper Riemann integral).
Solution:- The collection of all subsets, if is a -ring containing Consider the non-empty intersection
Constructing directly is more painful.
Solution:- Given a finite, but countably additive, measure, on we first define the outer measure for any set as the infimum of where the cover (or as if no such cover exists). Thus if are two such measures on then for all subsets Then we define as the collection of subsets which are -approximable by elements of that is if and only if there exists a sequence in with If is approximable it is certainly approximable, since Thus The measurable sets, forming are by definition just the countable unions of elements of so again On the extended measure is just so necessarily on where they are both defined.
Solution:- For a non-negative measurable function the integral is
Solution:- Given divide the interval into equal subintervals, each of which has measure If is bounded for each then so is their sum, which is Thus for at least one interval the sequence is unbounded. Since the sequence of functions is increasing, so is this sequence, hence it tends to
Richard B. Melrose 2004-05-24