(a) Prove the following relative of the Markov-Chebyshev inequality: Let f : R
Lebesgue measurable and let O > 0. Then
(Here f2 is the function f2(x) = f(x)2)
(b) Let fn : R
R, n = 1,2,...,
be a sequence of Lebegue measurable functions.
Assume that mj fn < 80. Using part (a), prove that for every € > 0 we have
(c) Use part (b) and the Borel-Cantelli lemma to conclude that fn 0 pointwise a.e.
[Note: the conclusion that fn 0 pointwise a.e. can also be deduced from the monotone
convergence theorem (MCT); this problem outlines a proof that avoids directly applying
Problem 2. Recall: A subset E of R is called nowhere dense if its closure has empty
interior (i.e., the complement of the closure of E is open and dense in R). A subset E of R
is called meager if it is contained in a countable union of nowhere dense sets.
(a) Prove that for every € > 0 there exists a closed, nowhere dense subset C of R with
m(R - C)
(b) Prove that there exists a meager subset M of R with m(R - M)
(c) Prove or disprove: there exists a nowhere dense subset C of R with m (IR - C) = 0.
Problem 3. Let (fn) and (gn) be sequences in L¹(R) and let g € L' (R). Assume that
(i) Ifnl < In a.e. for all 72
(ii) In converges pointwise a.e. to g
(iii) S 9n
(iv) fn converges pointwise a.e. to a function f.
Prove that f € L' (R) and that j fn
[Note and Hint: The special case where In = g for all n recovers the Dominated Conver-
gence Theorem. To prove this, follow all the steps in the proof the Dominated Convergence
Theorem, making any necessary changes along with way.]
These solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction
of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice.
Unethical use is strictly forbidden.