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Department of Mathematics, University of Michigan Analysis Qualifying Exam, January 8, 2022 Solutions
Problem 1: Let E be a measurable subset of [0, 1]. Suppose there exists α ∈ (0, 1) such that
m(E ∩ J) ≥ α · m(J) for all subintervals J of [0, 1].
Prove that m(E) = 1.
Solution 1: Let F = [0, 1] \ E. Then
m(F ∩ J) ≤ (1 − α) · m(J) for all subintervals J of [0, 1].
Assume that m(F ) > 0 and choose a cover of F by intervals Jn, n ∈ N such that
∑^ ∞
n=
m(Jn) ≤ (1 + α)m(F ).
Then
m(F ) = m
F ∩
n=
Jn
∑^ ∞
n=
m (F ∩ Jn) ≤ (1−α)
∑^ ∞
n=
m (Jn) ≤ (1−α^2 )m(F ).
This contradiction implies that m(F ) = 0, and thus m(E) = 1.
Solution 2: By the Lebesgue differentiation theorem,
m(E ∩ (a − ε, a + ε) 2 ε
→ (^1) E (a) for almost all a ∈ (0, 1)
as ε → 0. Since the assumption implies that
lim inf ε→ 0
m(E ∩ (a − ε, a + ε) 2 ε
≥ α for all a ∈ (0, 1),
(^1) E = 1 a.e. on [0, 1] which means that m(E) = 1.
Problem 2: Let f, g ∈ L^1 (0, 1). Assume for all functions φ ∈ C∞([0, 1]) with φ(0) = φ(1) that ∫ (^1)
0
f (t)φ′(t) dt = −
0
g(t)φ(t) dt.
Show that f (·) is absolutely continuous and f ′^ = g.
Solution: Let a, b ∈ (0, 1), a < b. Let ε > 0 be such that
ε < min
a 2
b − a 2
1 − b 2
Define the function φε : (0, 1) → R by
φε(x) =
x−a+ε 2 ε ,^ if^ x^ ∈^ [a^ −^ ε, a^ +^ ε] 1 , if x ∈ (a + ε, b − ε) 1 − x− 2 bε+ ε, if x ∈ [b − ε, b + ε] 0 , otherwise.
We can find a sequence of C∞([0, 1]) functions ψn with ψn(0) = ψn(1) = 0 con- verging to φε in the L∞^ norm. In addition to it, the functions ψn can be chosen so that
ψ′ n →
2 ε
(^1) [a−ε,a+ε] − (^1) [b−ε,b+ε]
a.e. and ‖ψ′ n‖∞ ≤
ε
Applying the Lebesgue dominated convergence theorem, we obtain
1 2 ε
(∫ (^) a+ε
a−ε
f (t) dt −
∫ (^) b+ε
b−ε
f (t) dt
0
g(t)φε(t) dt.
Letting ε → 0 and using the Lebesgue differentiation theorem for the left hand side and the Lebesgue dominated convergence theorem for the right hand side, we conclude that
f (a) − f (b) = −
∫ (^) b
a
g(t) dt
for almost all a, b ∈ (0, 1), a < b. Since g ∈ L^1 (0, 1) the result follows.
Problem 3: Let {gn} be a sequence of measurable functions on [0, 1] such that
(a) |gn(x)| ≤ C for a.e. x ∈ [0, 1], (b) limn→∞
∫ (^) a 0 gn(x)^ dx^ =^ 0 for all^ a^ ∈^ (0,^ 1). Prove that if f ∈ L^1 (0, 1) then
lim n→∞
0
f (x)gn(x) dx = 0.
Solution: Let
V = {f ∈ L^1 (0, 1) : lim n→∞
0
f (x)gn(x) dx = 0 }.
Then V is a closed linear subspace of L^1 (0, 1). Indeed, the linearity is obvious. To show that V is closed, take f ∈ L^1 (0, 1), and let h ∈ V. Then ∣ ∣ ∣ ∣lim sup n→∞
0
f (x)gn(x) dx
∣ ≤^ lim sup n→∞
0
|f (x) − h(x)| · |gn(x)| dx ≤ C‖f − h‖ 1.
If f ∈ cl(V ), then the right hand side can be made arbitrarily small which implies that f ∈ V.
