# First test, solutions

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Maximum marks possible: 50

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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.

1. Consider the sequence of functions on for Explain why these functions are measurable with respect to Lebesgue measure. Assuming the equality of the Lebesgue and Riemann integrals for continuous functions what can you deduce, and how, about the Lebesgue integrability of on

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

By the equality of Riemann and Lebesgue integrals for continuous functions we can compute the integrals of the

Thus is Lebesgue integrable with integral

Note: The limit is not bounded, hence not Riemann integrable in the strict sense (it has an improper Riemann integral).

2. Recall that a collection of subsets of a given set is a -ring if the difference of any two elements is an element and it contains the union of any countable collection of its elements. Given any collection of subsets of a set show that there is a smallest -ring of sets containing

Solution:- The collection of all subsets, if is a -ring containing Consider the non-empty intersection

Then us closed under differences and countable unions, since each of the collections is so closed and it contains Moreover, by definition, any -ring containing must be one of the so must contain which is therefore the smallest.

Constructing directly is more painful.

3. Briefly recall the manner in which a countably additive measure on a ring of subsets of a set is completed to a countably additive measure on a -ring of subsets of Show that if for are two such countably additive finite measures on a ring of subsets of and for all then and that the extended measures satisfy for all

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.

4. Suppose that is a bounded measurable function on explain what its Lebesgue integral is and why it is finite.

Solution:- For a non-negative measurable function the integral is

 with simple measurable

For simple functions implies and since we know that implies so and the supremum on the right is finite, hence so is the integral.

5. Let be an increasing sequence of functions on ( for all and all which are Lebesgue measurable with finite for each but unbounded as Show that, given there is a measurable subset with such that

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