Exam 3 (17–26)¶

1995 Nov¶

Instructions. Each of the following 10 problems will be scored from 0 to 10. Please begin each problem on a separate sheet and write on only one side of each sheet.

The following notations are used in the problem statements:

\(ℝ\) is the set of real numbers;
\(ℕ\) is the set \(\{0,1,2,\dots\}\);
\(m^∗\) is Lebesgue outer measure, which is defined on the set of all subsets of \(ℝ\).
\(m\) is Lebesgue measure, which is the restriction of \(m^∗\) to the set of Lebesgue measurable susbsets of \(ℝ\).

Problem 17

Define the following terms:

Lebesgue outer measure (which is denoted by \(m^∗\)).
Lebesgue measurable subset of \(ℝ\) (give Caratheodory’s definition).
Borel subsets of \(ℝ\).
Lebesgue measurable function (defined on a subset of \(ℝ\) and taking values in \(ℝ\)).
Borel function (defined on a subset of \(ℝ\) and taking values in \(ℝ\)).

Problem 18

Suppose that \(A ⊆ ℝ\) and \(m^∗ A = 0\). Prove that \(A\) is a Lebesgue measurable subset of \(ℝ\).

Problem 19

Suppose that \(f: [0,∞) → ℝ\), \(f\) is uniformly continuous, and \(∫_0^∞ |f(x)|\, dm(x) < ∞\). Prove that \(\lim\limits_{x→ ∞} f(x) = 0\).

Problem 20

Suppose that \(f, g: ℝ → ℝ\), \(f\) and \(g\) are continuous, \(∫_ℝ |f(x)|\, dm(x) < ∞\) and \(∫_ℝ |g(x)|\, dm(x) < ∞\). For each \(x∈ ℝ\), define \(f ⋆ g (x) := ∫_ℝ f(x-t)g(t)\, dm(t)\). Prove that \(∫_ℝ |f ⋆ g (x)|\, dm(x) < ∞\).

Problem 21

State a version of Fatou’s lemma (which concerns a relationship between integrals of limits and limits of integrals). Give an example to show that strict inequality can obtain in your version of Fatou’s lemma.

Problem 22

Suppose that \(f: [0,1]→ ℝ\), \(f\) is a Lebesgue measurable function, and, whenever \(φ: [0,1]→ℝ\) and \(φ\) is continuous, we have \(∫_0^1 f(x) φ(x)\, dm(x) = 0\). Prove that \(f(x) = 0\) for \(m\)-almost every \(x∈ [0,1]\). (Hint: Prove that \(∫_0^xf(t) \, dm(t) = 0\) for all \(x∈ [0,1]\) and finish by citing a major theorem.)

Problem 23

Suppose that \((X, 𝔐, μ)\) is a measure space, \((A_n)_{n=1}^∞\) is a sequence from \(𝔐\), and \(A_n ⊆ A_{n+1}\) whenever \(n∈ ℕ\). Prove that

\[\lim\limits_{n↑∞} μ A_n = μ \left(⋃_{n=0}^∞ A_n\right).\]

Problem 24

Suppose that \(μ\) is a measure on the Borel subsets of \([0,1]\), \(μ[0,1]<∞\), \(f:[0,1] → ℝ\), \(f\) is a Borel function, and \(f\) is integrable with respect to \(μ\) on \([0,1]\). Define \(F: [0,1] → ℝ\) by the rule \(F(x) = ∫_{[0,x]} f(t) \, dμ(t)\). Prove that \(F\) is of bounded variation on \([0,1]\). State conditions on \(f\) and \(μ\) which imply that \(F\) is absolutely continuous on \([0,1]\).

Problem 25

Let \((Ω_j, τ_j)\) be a topological space for each \(j∈ J\). Define the corresponding product topology, \(τ\), on the Cartesian product \(∏_{j∈ J} Ω_j\). State Tychonoff’s theorem.

Problem 26

State the Baire category theorem for a general complete metric space and cite an application of the theorem; no details are required in the citation.

See Lebesgue measure.
See Lebesgue measurable set and measurable set.
See Borel set.
See Lebesgue measurable function.
See Borel measurable function.

Let \(A ⊆ ℝ\) and assume \(m^∗ A = 0\). We must show that \(A\) is Lebesgue measurable subset of \(ℝ\).

Equivalently, we show that \(m^∗(B) = m^∗(B ∩ A) + m^∗(B ∩ A^c)\) for every \(B ⊆ ℝ\).

Indeed, by subadditivity of \(m^∗\) we have \(m^∗(B) ≤ m^∗(B ∩ A) + m^∗(B ∩ A^c)\), so we need only prove \(≥\).

Recall, a measure is called complete if every subset of a negligible set is measurable and negligible.

Since \(B∩ A ⊆ A\) and since Lebesgue outer measure is complete, we have

(13)¶\[m^∗(B ∩ A) = 0.\]

By monotonicity of \(m^∗\) under the partial ordering \(⊆\), we have

(14)¶\[m^∗(B ∩ A^c) ≤ m^∗(B) \quad (∵ B∩ A^c ⊆ B).\]

By (13) and (14), \(m^∗(B) ≥ m^∗(B ∩ A) + m^∗(B ∩ A^c)\).

☐

Suppose that \(f: [0,∞) → ℝ\), \(f\) is uniformly continuous, and \(∫_0^∞ |f(x)|\, dm(x) < ∞\). Prove that \(\lim\limits_{x→ ∞} f(x) = 0\).

Suppose on the contrary that \(\lim\limits_{x→ ∞} f(x) ≠ 0\). Then there exists \(ε>0\) such that for all \(M>0\) \(∃ x > M\)

\[(∃ε>0)\, (∀ M>0) \, (∃ x > M), \, |f(x)| > ε.\]

Fix such an \(ε>0\). Since \(f\) is uniformly continuous, \(∃ δ>0\) such that

\[|f(x) - f(y)| < ε/2 \quad \text{ for all }\; |x - y| < δ.\]

Taking \(δ' = \min \{δ, 1/2\}\) if necessary, we can assume without loss of generality that our original \(δ\) is \(< 1\).

Now

\[∑_{n=0}^∞ ∫_{[n,n+1]} |f(x)| \, dm(x) ≤ ∫_0^∞ |f(x)|\, dm(x) < ∞,\]

so there exists \(N>0\) such that

(15)¶\[∑_{n=N+1}^∞ ∫_{[n,n+1]} |f(x)| \, dm(x) < ε ⋅ δ.\]

Next, find \(x > N+2\) such that \(|f(x)| >ε\) and

\(|f(x) - f(y)| < ε/2\) for all \(|x - y| < δ\).

By the triangle inequality, \(|f(x)| ≤ |f(x) - f(y)| + |f(y)|\), so

\[|f(y)| ≥ |f(x)|- |f(x) - f(y)| > ε/2 \quad \text{ for all } \; |x-y|< δ.\]

Therefore,

(16)¶\[∫_{[x-δ , x+δ]} |f(x)| \, dm(x) ≥ \frac{ε}{2} m[x-δ, x+δ] = ε ⋅ δ.\]

On the other hand, since \(x>N+2\), we have \([x-δ , x+δ] ⊆ [N,∞)\).

Therefore, by (15) and (16),

\[ε ⋅ δ ≤ ∫_{[x-δ , x+δ]} |f(x)| \, dm(x) ≤ ∫_{[N+1, ∞)} |f(x)| \, dm(x) = ∑_{n=N+1}^∞ ∫_{[n,n+1]} |f(x)| \, dm(x) < ε ⋅ δ,\]

which is a contradiction. Therefore, \(f(x) → 0\) as \(x → ∞\).

☐

Since the function \(ψ(x,y) := x-y\) is a continuous mapping from \(ℝ^2\) to \(ℝ\), the function

\[F(x,y) := (f ∘ ψ)(x,y) = f(x-y)`\]

is also continuous and therefore measurable.

Since the second projection function \(π_2: ℝ^2 → ℝ\)—defined for all \((x,y) ∈ ℝ^2\) by \(π_2(x,y) := y\)—is continuous, so is the function \(G(x,y) := (g ∘ π_2)(x,y) = g(y)\); the latter is thus measurable.

Therefore, the function

\[H(x,y) = F(x,y) G(x,y) = f(x-y) g(y)\]

is a measurable function on \(ℝ^2\).

Next note the analogy with the functions defined in Fubini’s theorem:

(17)¶\[φ(x) = ∫_Y H(x,y)\, dν(y) = ∫_ℝ f(x-y)g(y) \, dm(y) = (f ∗ g)(x).\]

In the present context, \((X, 𝔐, μ) = (Y, 𝔑, ν) = (ℝ, ℬ(ℝ), m)\).

Now applying Part 2 of Fubini’s theorem, we consider the integral with respect to \(x\) first:

\[\begin{split}∬_{ℝ^2} |f(x-y)||g(y)| \, dm(x) \, dm(y) &=∫_ℝ |g(y)| ∫_ℝ |f(x-y)| \, dm(x) \, dm(y)\\ &=∫_ℝ |g(y)| \, dm(y) ∫_ℝ |f(x)| \, dm(x) < ∞.\end{split}\]

The second equality holds by translation invariance of Lebesgue measure.

Therefore, Fubini’s theorem implies \(H ∈ L_1(m × m)\).

Finally, Fubini’s theorem Part 3 (iii) says that \(H ∈ L_1(m × m)\) implies \(φ ∈ L_1(m)\).

By (17), this is equivalent to

\[∫ |φ| \, dm = ∫_ℝ | (f ∗ g)(x)| \, dm(x) < ∞.\]

☐

Fatou’s lemma (relating integrals of limits and limits of integrals) asserts the following: if \(\{f_n\}\) is a sequence of nonnegative, extended, real-valued, measurable functions then \(∫\varliminf f_n ≤ \varliminf ∫ f_n\).

We now describe a witness to strict inequality. Define \(f_n = \frac{1}{n}χ_{[0,n]}\) for \(n =1,2,\dots\). Then \(\varliminf f_n(x) = 0\) for all \(x\) and \(∫ f_n = 1\) for all \(n\), so

\[∫\varliminf f_n = 0 < 1 = \varliminf ∫ f_n.\]

☐

Let \(f: [0,1]→ ℝ\) be a Lebesgue measurable function, and assume that for all \(φ ∈ C([0,1],ℝ)\) we have \(∫_0^1 f(x) φ(x)\, dm(x) = 0\).

We wish to prove that \(f(x) = 0\) for almost every \(x∈ [0,1]\).

First observe that \(f∈ L_1[0,1]\). Indeed, take \(φ\) to be the constant function \(φ ≡ 1\), so that

\[∫_0^1f(t) \, dm(t) = ∫_0^1f(t) φ(x)\, dm(t) = 0,\]

so the integral \(∫_0^1f(t) \, dm(t)\) exists, which by definition means that \(∫_0^1 f^+\) and \(∫_0^1 f^-\) both exist, which is true if and only if \(∫_0^1 |f|\) exists. (See the theorem on integrability of a measurable function.)

Next, we prove the assertion given in the hint. That is, \(∫_0^xf(t) \, dm(t) = 0\) for all \(x∈ [0,1]\).

Indeed, if \(x=1\), take \(φ ≡ 1\) so that \(∫_0^xf(t) \, dm(t) = ∫_0^1f(t) φ(x)\, dm(t) = 0\).

If \(0≤ x<1\), then for each \(n =1,2,\dots\) define

\[\begin{split}φ_n(t) = \begin{cases} 1, & 0≤ t ≤ x-1/n\\ -n t,& x-1/n < t≤ x\\0, & x<t≤ 1.\end{cases}\end{split}\]

Then \(φ_n ∈ C([0,1],ℝ)\) and \(\lim_{n→∞}f(x)φ_n(x) = f(x)χ_{[0,x]}(x)\). Also,

\[-|f| ≤ -|f| φ_n ≤ f φ_n ≤ |f| φ_n ≤ |f|,\]

holds for all \(n\), so the bounded convergence theorem gives the penultimate eequality in the following calculation.

\[\begin{split}∫_0^x f(t)\, dm(t) &= ∫_0^1f(x)χ_{[0,x]}(x)\, dm(t)\\ &= ∫_0^1 \lim_{n→∞}f(x)φ_n(x)\, dm(x)\\ &= \lim_{n→∞}∫_0^1f(x)φ_n(x) = 0.\end{split}\]

It follows for all \(0≤ a ≤ b ≤ 1\) that

\[∫_0^a f(t)\, dm(t) = ∫_0^bf(x)\, dm(t) = 0,\]

and therefore,

(18)¶\[∫_a^b f(t)\, dm(t) = ∫_0^bf(x)\, dm(t) - ∫_0^a f(t)\, dm(t) = 0.\]

Now let \(B := \{x ∈ [0,1] : f(x) >0\} = f^{-1}((0,∞])\), which is measurable since \(f\) is measurable. We will prove \(m B = 0\).

Recall the (Borel) measurable sets in \([0,1]\) are generated by the half-open intervals of the form \((a,b]\).

Since \(B\) is measurable, either we can find an interval \((a,b] ⊆ B\), or else \(B\) is a set of measure 0.

If \(m B > 0\), then there exists \(0≤ a < b ≤ 1\) such that \((a,b] ⊆ B\). Since \(f>0\) on \((a,b]\), we then have \(∫_a^b f(x) \, dm(x) >0\), which contradicts (18). Therefore, we must have \(m B = 0\).

Similarly, the set \(C := \{x∈ [0,1] : f(x) <0\}\) is negligible, so

\[B ∪ C = \{x ∈ [0,1] : f(x) ≠ 0\}\]

is negligible. This proves that \(f(x) = 0\) for almost every \(x∈ [0,1]\).

☐

Let \((X, 𝔐, μ)\) be a measure space and \(\{A_n\}_{n=1}^∞ ⊆ 𝔐\) a sequence of measurable sets such that \(A_n ⊆ A_{n+1}\) for all \(n∈ ℕ\). We must show,

(19)¶\[\lim\limits_{n → ∞} μ A_n = μ \left(⋃_{n=1}^∞ A_n\right).\]

Define \(A:= ⋃_{n=1}^∞ A_n\) and consider the sequence \(\{χ_{A_n}\}\) of characteristic functions of the \(A_n\).

Clearly \(A_n ⊆ A_{n+1}\) (\(∀ n\)) implies \(0≤ χ_{A_1} ≤ \cdots ≤ χ_{A_n} ≤ χ_{A_{n+1}} ≤ \cdots ≤ χ_A ≤ χ_X = 1\), and \(\lim_{n→ ∞} χ_{A_n} = χ_A\), so the monotone convergence theorem yields the second equality in the following calculation:

\[\lim_{n→ ∞} μ A_n = \lim_{n→ ∞} ∫χ_{A_n}\, dμ = ∫\lim_{n→ ∞} χ_{A_n}\, dμ = ∫χ_A\, dμ = μA = μ\left(⋃_{n=1}^∞A_n\right).\]

Thus (19) is proved.

☐

Let \(μ\) be a finite measure on the Borel subsets of \([0,1]\). Let \(f:[0,1] → ℝ\) be a Borel function, and assume \(f ∈ L_1([0,1],μ)\). Define \(F: [0,1] → ℝ\) by the rule

(20)¶\[F(x) = ∫_{[0,x]} f(t) \, dμ(t).\]

Claim. \(F\) is of bounded variation on \([0,1]\).

Proof. By (20) we have \(F(x_{i})-F(x_{i-1}) = ∫_{x_{i-1}}^{x_{i}} f(t) \, dμ(t)\),

so

\[|F(x_{i})-F(x_{i-1})| ≤ ∫_{x_{i-1}}^{x_{i}} |f(t)| \, dμ(t).\]

Since \(f∈ L_1\) we have \(∫_0^1 |f(t)| \, dμ(t) < ∞\), and for every partition \(0=x_0 < \cdots < x_n = 1\),

\[\begin{split}∑_{i=1}^n|F(x_{i})-F(x_{i-1})| &≤ ∑_{i=1}^n\left|∫_{x_{i-1}}^{x_{i}} f(t) \, dμ(t)\right|\\ &≤ ∑_{i=1}^n∫_{x_{i-1}}^{x_{i}} |f(t)| \, dμ(t)\\ &= ∫_0^1 |f(t)| \, dμ(t) < ∞,\end{split}\]

which proves that \(F ∈ BV[0,1]\).

☐

Todo

To complete the solution we must give conditions on \(f\) and \(μ\) under which \(F\) is absolutely continuous on \([0,1]\).

For the first part, see the definition of product topology in the appendix of definitions.

For the second part, see the statement of Tychonoff’s theorem in the appendix of theorems.

☐

Here’s a version of the Baire category theorem:

Let \(X\) be a complete metric space and let \(G⊆ X\) be an open set. If \(G = ⋃_{n=1}^∞ G_n\) then \((Ḡ_n)^° ≠ ∅\) for at least one \(n∈ ℕ\).

Here’s another version:

If \(\{A_n: n ∈ ℕ\}\) is a (countable) collection of dense subsets of \(X\), then \(⋂_{n=0}^∞ A_n\) is dense.

Here’s a corollary:

A complete metric space is not a countable union of nowhere dense sets.

Here’s an application:

Often we want to prove that a particular set has a nonempty interior. For example, the Baire category theorem can be used to prove the following:

Proposition. If \(T: X → Y\) is a linear map from one Banach space to another, then \(T\) is bounded iff the set \(T^{-1}\{y∈ Y ∣ \|y\|_Y ≤ 1\}\) has a nonempty interior.

A related application due to Stefan Banach and Hugo Steinhaus is the uniform boundedness principle which asserts the following:

Let \(ℱ\) be a family of mappings from one Banach space \(X\) to another \(Y\). Suppose that for each \(x ∈ X\) the set \(\{f(x) ∣ f ∈ ℱ\}\) is bounded; i.e., \(∃ M_x\) such that \(\|f(x)\|_Y ≤ M_x\) for all \(f ∈ ℱ\). Then \(ℱ\) is uniformly bounded; i.e., \(∃ M\) such that that \(\|f(x)\|_Y ≤ M \|x\|_X\) for all \(x∈ X\) and all \(f ∈ ℱ\). Equivalently, \(\|f\| = \sup\{\|f(x)\|_Y : \|x\|_X = 1\} ≤ M\).