Problems and Solutions in - Mathematics - University of Hawaii

1.5 2001 November 26 1 REAL ANALYSIS
1.5 2001 November 26
Instructions Masters students do any 4 problems Ph.D. students do any 5 problems. Use a separate sheet of paper for
each new problem.
1. Let {fn} be a sequence of Lebesgue measurable functions on a set E ⊂ R, where E is of finite Lebesgue measure.
Suppose that there is M > 0 such that |fn(x)| ≤ M for n ≥ 1 and for all x ∈ E, and suppose that
limn→∞ fn(x) = f(x) for each x ∈ E. Use Egoroff’s theorem to prove that
f(x) dx = lim fn(x) dx.
n→∞
E
Solution: First note that |f(x)| ≤ M for all x ∈ E. To see this, suppose it’s false for some x0 ∈ E, so that
|f(x0)| > M. Then there is some ɛ > 0 such that |f(x0)| = M + ɛ. By the triangle inequality, then, for all n ∈ N,
|f(x0) − fn(x0)| ≥ ||f(x0)| − |fn(x0)|| = |M + ɛ − |fn(x0)|| ≥ ɛ,
which contradicts fn(x0) → f(x0). Thus, |f(x)| ≤ M for all x ∈ E.
Next, fix ɛ > 0. By Egoroff’s theorem (A.8), there is a G ⊂ E such that µ(E \ G) < ɛ and fn → f uniformly
on G. Furthermore, since |fn| ≤ M and |f| ≤ M and µ(E) < ∞, it’s clear that {fn} ⊂ L 1 and f ∈ L 1 , so the
following inequalities make sense (here we’re using the notation fG = sup{|f(x)| : x ∈ G}):
f dµ − fn dµ
E
E
≤
|f − fn| dµ = |f − fn| dµ + |f − fn| dµ
E
E\G
G
≤ |f| dµ + |fn| dµ + f(x) − fn(x)Gµ(G)
E\G
E\G
≤ 2Mµ(E \ G) + f(x) − fn(x)Gµ(G)
< ɛ + f(x) − fn(x)Gµ(G).
Finally, µ(G) ≤ µ(E) < ∞ and f(x) − fn(x)G → 0, which proves that
E fn dµ →
2. Let f(x) be a real-valued Lebesgue integrable function on [0, 1].
(a) Prove that if f > 0 on a set F ⊂ [0, 1] of positive measure, then
f(x) dx > 0.
(b) Prove that if x
f(x) dx = 0, for each x ∈ [0, 1],
then f(x) = 0 for almost all x ∈ [0, 1].
Solution: (a) Define Fn = {x ∈ F : f(x) > 1/n}. Then
and m(F ) > 0 implies
0
F
F1 ⊆ F2 ⊆ · · · ↑
Fn = F,
0 < m(F ) = m(∪nFn) ≤
m(Fn).
20
n
E
n
E
f dµ. ⊓⊔