Exam 4 (27–35)

1998 Apr

Instructions Do at least four problems in Part A, and at least two problems in Part B.

Part A.

Problem 27

Let \(\{x_n\}_{n=1}^∞\) be a bounded sequence of real numbers, and for each positive \(n\) define

\[\hat{x}_n = \sup_{k\geq n} x_k\]

1. Explain why the limit \(ℓ = \lim_{n→∞}x̂_n\) exists.

2. Prove that, for any \(ε > 0\) and positive integer \(N\), there exists an integer \(k\) such that \(k ≥ N\) and \(|x_k - ℓ| < ε\).

Problem 28

Let \(C\) be a collection of subsets of the real line \(ℝ\), and define

\[A_σ (C) = ⋂ \{A : C ⊂ A = \text{ a } σ\text{-algebra of subsets of } ℝ\}.\]

1. Prove that \(A_σ (C)\) is a \(σ\)-algebra, that \(C ⊂ A_σ (C)\), and that \(A_σ (C) ⊂ A\) for any other \(σ\)-algebra \(A\) containing all the sets of \(C\).

2. Let \(O\) be the collection of all finite open intervals in \(ℝ\), and \(F\) the collection of all finite closed intervals in \(ℝ\). Show that

   \[A_\sigma (O) = A_\sigma (F).\]

Problem 29

Let \((X, 𝔐, μ)\) be a measure space, and suppose \(X = ⋃_n X_n\), where \(\{X_n\}_{n=1}^∞\) is a pairwise disjoint collection of measurable subsets of \(X\). Use the monotone convergence theorem and linearity of the integral to prove that, if \(f\) is a non-negative measurable real-valued function on \(X\),

\[∫_X f\, dμ = ∑_n ∫_{X_n} f\, dμ.\]

Problem 30

Using the Fubini/Tonelli theorems to justify all steps, evaluate the integral

\[∫_0^1 ∫_y^1 x^{-3/2}\cos\left(\frac{π y}{2x}\right) \, dx\, dy.\]

Problem 31

Let \(I\) be the interval \([0,1]\), and let \(C(I)\) (resp., \(C(I × I)\)) denote the space of real valued continuous functions on \(I\) (resp., \(I\times I\)), with the usual supremum norm. Show that the collection of finite sums of the form

\[f(x,y) = ∑_i φ_i(x) ψ_i(y),\]

where \(φ_i, ψ_i ∈ C(I)\) for each \(i\), is dense in \(C(I × I)\).

Problem 32

Let \(m\) be Lebesgue measure on the real line \(ℝ\) and for each Lebesgue measurable subset \(E\) of \(ℝ\) define

\[μ E = ∫_E \frac{1}{1+x^2} \, dm(x).\]

Show that \(m\) is absolutely continuous with respect to \(μ\) and compute the Radon-Nikodym derivative \(dm/dμ\).

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Part B.

Problem 33

Let \(φ(x,y) = x^2 y\) be defined on the square \(S = [0,1] × [0,1]\) in the plane, and let \(m\) be two-dimensional Lebesgue measure on \(S\).

Given a Borel subset \(E\) of the real line \(ℝ\), define

\[μ \,E = m\, φ^{-1}(E).\]

1. Show that \(μ\) is a Borel measure on \(ℝ\).

2. Let \(χ_E\) denote the characteristic function of the set \(E\). Show that

   \[∫_ℝ χ_E \, dμ = \int_S χ_E ∘ φ \, dm.\]

3. Evaluate the integral

   \[∫_{-∞}^∞ t^2 \, dμ(t).\]

Problem 34

Let \(f\) be a real-valued increasing function on the real line \(ℝ\), such that \(f(-∞)=0\) and \(f(∞)=1\).

Prove that \(f\) is absolutely continuous on every closed finite interval if and only if

\[∫_ℝ f' \, dm = 1.\]

Problem 35

Let \(F\) be a continuous linear functional on the space \(L_1[-1,1]\), with the property that \(F(f) = 0\) for all odd functions \(f\) in \(L_1[-1,1]\). Show that there exists an even function \(φ\) such that

\[F(f) = ∫_{-1}^1 f(x) φ(x) \, dx, \quad \text{ for all } f∈ L_1[-1,1].\]

[Hint: One possible approach is to use the fact that every function in \(L_p[-1,1]\) is the sum of an odd function and an even function.]

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Part A Solutions

Solution to Problem 27

1. By assumption

   \[x̂_n = \sup_{k≥n} x_k = \sup \{x_n, x_{n+1}, \dots \} ≥ \sup \{x_{n+1}, x_{n+2}, \dots \} = x̂_{n+1},\]

   and there exists \(M>0\) such that \(\{x_n\} ⊆ [-M, M]\) for all \(n\).

   Therefore, \(x̂_n ∈ [-M,M]\) for all \(n\), so \(\{x̂_n\}\) is a bounded monotone decreasing sequence of real numbers.

   By the Bolzano-Weierstrass theorem, \(x̂_n\) converges to some \(ℓ := \lim_{n→ ∞} x̂_n ∈ [-M,M]\).

   ☐

2. Fix \(ε > 0\) and \(N>0\). By definition of \(ℓ\), there exists \(N_0 ≥ N\) such that \(|x̂_k - ℓ| < ε/2\) for all \(k ≥ N_0\).

   By definition of supremum, there exists \(j>0\) such that

   \[|x_{N_0+j} - \hat{x}_{N_0}| = |x_{N_0+j} - \sup \{x_{N_0}, x_{N_0+1}, \dots\}| < ε/2.\]

   Therefore,

   \[|x_{N_0+j} - ℓ| ≤ |x_{N_0+j} - x̂_{N_0}| + |x̂_{N_0} - ℓ| ≤ ε/2 + ε/2 = ε.\]

   Thus, for arbitrary \(ε > 0\) and \(N>0\), we have found an integer \(k = N_0 + j ≥ N\) such that \(|x_k - ℓ| < ε\), as desired.

   ☐

Solution to Problem 28

(coming soon)

Solution to Problem 29

Define \(f_n = ∑_{k=1}^n f χ_{X_k} = f χ_{⋃_1^n X_k}\).

Then it is clear that the hypotheses of the monotone convergence theorem are satisfied. That is, for all \(x ∈ X\),

1. \(0≤ f_1(x) ≤ f_2(x) ≤ \cdots ≤ f(x)\), and

2. \(\lim_{n → ∞} f_n(x) = f(x)χ_X(x) = f(x)\).

Therefore,

\[\begin{split}∑_{k=1}^∞ ∫_{X_k} f\, dμ &= \lim_{n → ∞} ∑_{k=1}^n ∫_X f χ_{X_k}\, dμ &\\ &= \lim_{n→∞} ∫_X ∑_{k=1}^n f χ_{X_k}\, dμ & \text{ (by linearity of the integral)}\\ &= \lim_{n→∞} ∫_X f_n \, dμ & (\text{by definition of} f_n)\\ &= ∫_X \lim_{n → ∞} f_n \, dμ & (\text{by the monotone convergence theorem})\\ &= ∫_X f\, dμ. &\end{split}\]

☐

Solution to Problem 30

By Tonelli’s theorem, if \(f(x,y)≥ 0\) is measurable and one of the iterated integrals \(∬ f(x,y)\,dx\,dy\) or \(∬ f(x,y)\,dy\,dx\) exists, then they both exist and are equal.

Moreover, if one of the iterated integrals is finite, then \(f(x,y) ∈ L_1(dx,dy)\).

Fubini’s theorem states that \(f(x,y)∈ L_1(dx,dy)\) implies the iterated integrals exist and are equal.

Now let \(g(x,y) = x^{-3/2}\cos(π y/2x)\), and apply Tonelli’s theorem to the (nonnegative measurable) function \(|g(x,y)|\) as follows:

\[\begin{split}∫_0^1 ∫_0^x |g(x,y)|\,dy\,dx &= ∫_0^1 ∫_0^x |x|^{-3/2} \left|\cos\left(\frac{π y}{2x}\right)\right|\,dy\,dx\\ & ≤ ∫_0^1 ∫_0^x x^{-3/2} ⋅ 1 \,dy\,dx\\ &= ∫_0^1 x^{-1/2} \,dx = 2.\end{split}\]

Thus one of the iterated integrals is finite which, by Tonelli’s theorem, implies \(g(x,y)∈ L_1(dx,dy)\).

Therefore, Fubini’s theorem applies to \(g(x,y)\) and gives the first of the following equalities:

\[\begin{split}∫_0^1 ∫_y^1 x^{-3/2}\cos\left(\frac{π y}{2x}\right)\,dx\,dy &= ∫_0^1 ∫_0^x x^{-3/2}\cos\left(\frac{π y}{2x}\right)\,dy\,dx \\ &= ∫_0^1 x^{-3/2} \cdot \frac{2x}{π} \left[\sin\left(\frac{π y}{2x}\right)\right]_{y=0}^{y=x} \, dx \\ &= ∫_0^1 \frac{2}{π}x^{-1/2}\, dx = \frac{2}{π} \left[2 x^{1/2}\right]_{x=0}^{x=1} = \frac{4}{π}.\end{split}\]

☐

Solution to Problem 31

(coming soon)

Solution to Problem 32

Obviously both measures are nonnegative.

We must first prove \(m ≪ μ\). To this end, let \(𝔐\) denote the σ-algebra of (Lebesgue) measurable sets and suppose \(E ∈ 𝔐\) and \(m E >0\).

Then, if we can show \(μ E > 0\), this will establish that the following implication holds for all \(E ∈ 𝔐\): \(μ E = 0 \, ⇒ \, m E=0\); that is, \(m ≪ μ\).

For \(n = 1, 2, \ldots\), define

\[A_n = \left\{x∈ ℝ: \frac{1}{n+1} < \frac{1}{1+x^2} ≤ \frac{1}{n}\right\}.\]

Then \(A_i ∩ A_j = ∅\) for all \(i≠ j\) in \(ℕ\), and, for all \(n = 1, 2, \ldots\),

\[\mu(A_n) \geq \frac{1}{n+1} m(A_n).\]

Also, \(ℝ = ⋃_n A_n\), since \(0< \frac{1}{1+x^2} \leq 1\) holds for all \(x ∈ ℝ\). Therefore,

\[\mu(E) = \mu(E\cap (\cup_n A_n)) = \mu(\cup_n (A_n \cap E)) = \sum_n \mu(A_n \cap E).\]

(The last equality might need justification. Since \(f(x) = \frac{1}{1+x^2}\) is continuous, hence measurable, the sets \(\{A_n\}\) are measurable, so the last equality holds by countable additivity of disjoint measurable sets.)

Now note that \(m E = ∑_n m(A_n ∩ E) >0\) implies the existence of an \(n ∈ ℕ\) such that \(m (A_n ∩ E) >0\). Therefore,

\[μ E ≥ μ(A_n ∩ E) ≥ \frac{1}{n+1} m(A_n ∩ E) > 0,\]

which proves that \(m ≪ μ\).

By the Radon-Nikodym theorem, there is a unique \(h ∈ L_1(μ)\) such that

\[m(E) = ∫ h \, dμ, \quad \text{ and } \quad ∫ f\, \, dm = ∫ f h \, dμ \quad ∀ f ∈ L_1(m).\]

In particular, if \(E ∈ 𝔐\) and \(f(x) = \frac{1}{1+x^2} χ_E\), then

\[μ E = ∫_E \frac{1}{1+x^2} \, dm(x) = ∫_E \frac{h(x)}{1+x^2}\, dμ(x).\]

That is, \(∫_E \, dμ = ∫_E \frac{h(x)}{1+x^2}\, dμ(x)\) holds for all measurable sets \(E\). It follows by integral extensionality that \(\frac{h(x)}{1+x^2} = 1\) for \(μ\)-almost every \(x ∈ ℝ\). Therefore,

\[\frac{\, dm}{dμ}(x) = h(x) = 1+x^2.\]

One final note: \(h\) is uniquely defined only up to an equivalence class of functions that equal \(1+x^2\) for \(μ\)-almost every \(x ∈ ℝ\).

☐

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Part B Solutions

Solution to Problem 23

(coming soon)

Solution to Problem 34

First note that \(f\) is increasing, so by differentiability of increasing functions, \(f'\) exists for a.e. \(x ∈ ℝ\), and \(f'(x) ≥ 0\) wherever \(f'\) exists.

Also, \(f'\) is measurable. To see this, define

\[g(x) = \limsup_{n → ∞} \, [f(x+1/n)-f(x)] \, n.\]

As a limsup of a sequence of measurable functions, \(g\) is measurable ([Rud87], Theorem 1.14).

Let \(E\) be the set on which \(f'\) exists. Then \(m(ℝ - E)=0\), and \(f'=g\) on \(E\) (by definition of derivative), so \(f'\) is measurable.

(⇒) Assume \(f∈ AC[a,b]\) for all \(-∞<a<b<∞\). We must show \(∫_ℝ f' \, dm =1\), which is equivalent to showing

\[\lim_{x\rightarrow \infty}\int_{-x}^x f'(t) \, dm(t) = \lim_{x\rightarrow \infty}[f(x)-f(-x)],\]

since \(f(\infty)-f(-\infty) = 1\), by assumption.

Let \(x∈ ℝ\), \(x>0\), and \(f\in AC[-x,x]\). Then we claim \(f(x)-f(-x) = \int_{-x}^x f' \, dm\). Assuming the claim is true (see [Roy88], page 110 for the proof), we have

\[1 = \lim_{x\rightarrow \infty}[f(x)-f(-x)] = \lim_{x\rightarrow \infty}\int_{-x}^x f'(t) \, dm(t) = \int_\BbbRf' \, dm.\]

(⇐) Assume \(∫_{ℝ} f' \, dm =1\). We must show \(f∈ AC[a,b]\) for all \(-∞ < a < b < ∞\).

Observe that \(f' ∈ L_1(ℝ)\), since

\[1=∫_ℝ f' \, dm = ∫_{ℝ-E} f' \, dm + ∫_{E} f' \, dm = ∫_{E} f' \, dm,\]

and, since \(f\) is increasing, \(f'≥ 0\) on \(E\), so

\[∫_ℝ|f'| \, dm = ∫_{ℝ- E} |f'| \, dm + ∫_{E} |f'| \, dm = ∫_{E} |f'| \, dm = ∫_{E} f' \, dm = 1.\]

Thus, \(f' ∈ L_1(ℝ)\) as claimed.

Since AC is equivalent to being an indefinite integral, it suffices to show that \(∫_ℝ f' \, dm =1\) implies \(\int_a^x f'(t)\, dt = f(x) - f(a)\) holds for all \(-∞ < a < x < ∞\).

By the differentiability of increasing functions, we have \(∫_a^b f'(t) dt ≤ f(b) - f(a)\), so we need only show that strict inequality cannot hold.

Suppose, by way of contradiction, that \(∫_a^b f'(t) dt < f(b) - f(a)\) holds for some \(-∞ < a < b < ∞\). Then,

\[\begin{split}1 &= ∫_ℝ f' \, dm = ∫_{-∞}^a f' \, dm + ∫_a^b f' \, dm + ∫_b^∞ f' \, dm\\ &< [f(a)-f(-∞)]+[f(b)-f(a)]+[f(∞)-f(b)]\\ &= f(∞)-f(-∞) = 1.\end{split}\]

This contradiction proves that \(∫_ℝ f' \, dm =1\) implies \(∫_a^b f'(t) dt = f(b) - f(a)\) holds for all \(-∞ < a < b < ∞\), as desired. ☐

Solution to Problem 35

Since \(F ∈ L_1^∗[-1,1]\), then by the Riesz representation theorem there is a unique \(h∈ L_∞[-1,1]\) such that

\[F(f) = ∫^1_{-1} f(x) h(x) \, dx \qquad (∀ f ∈ L_1[-1,1])\]

Now (using the hint) write \(h = φ + ψ\), where \(φ\) and \(ψ\) are the even and odd functions,

\[φ(x) = \frac{h(x) + h(-x)}{2} \quad \text{ and } \quad ψ(x) = \frac{h(x) - h(-x)}{2}.\]

Similarly, let \(f = f_e + f_o\) be the decomposition of \(f\) into a sum of even and odd functions.

Then, by linearity of \(F\) and since \(F(f_o) = 0\) by hypothesis,

\[F(f) = F(f_e) + F(f_o) = F(f_e) = ∫_{-1}^1 f_e h = ∫_{-1}^1 f_e φ + ∫_{-1}^1 f_e ψ.\]

As the product of even and odd, \(f_e ψ\) is odd; so \(∫_{-1}^1 f_e ψ = 0\) (since \([-1,1]\) is symmetric).

Similarly, \(∫_{-1}^1 f_o φ = 0\). Therefore,

\[F(f) = F(f_e) = ∫_{-1}^1 f_e φ = ∫_{-1}^1 f_e φ+ ∫_{-1}^1 f_o φ = ∫_{-1}^1 (f_e+f_o) φ = ∫_{-1}^1 f φ.\]

(See also Problem 52.)

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Complex Analysis Exams

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