Exam 9 (64–70)¶

2006 Apr¶

Instructions. Do as many problems as you can. Complete solutions (except for minor flaws) to 5 problems would be considered a good performance. Fewer than 5 complete solutions may still be passing, depending on the quality.

Problem 64

Use only “epsilon-delta” techniques to prove the following discrete version of the Lebesgue dominated convergence theorem: Suppose \(f_k: ℕ → ℝ\) is a sequence of functions that converges pointwise to 0, and there is a function \(f: ℕ → [0, ∞)\) with \(∑_{n∈ ℕ} f(n) < ∞\) and \(|f_k(n)| ≤ f(n)\) for all \(n\) and for all \(k\). Then \(\lim\limits_{k→∞}∑_{n∈ ℕ} f_k(n)=0\).

Problem 65

For real numbers \(a\) and \(b\) with \(0<a<b\), evaluate the integral

\[∫_0^∞ \frac{e^{-ax} - e^{-bx}}{x} \, dx := \lim\limits_{c→+∞}∫_{1/c}^c \frac{e^{-ax} - e^{-bx}}{x} \, dx.\]

Justify (in writing) all crucial steps using well-known results. Hint. Consider \(∫_a^b e^{-xy}\, dy\).

Problem 66

Let \(f∈ L_2[0,1]\) and define \(g(x) = ∫_0^x f(t)\, dt\), \(0≤ x ≤ 1\). Show that \(\|g\|_2 ≤ \frac{1}{2}\|f\|_2\).

Problem 67

Let \(\{f_n\}\) be a sequence of measurable functions on the measurable space \((X, 𝔐)\). Prove, from the definition of measurability, that \(\lim\sup_{n→∞} f_n\) is measurable.
Let \(\{f_n\}\) be a sequence of measurable functions on the measurable space \((X, 𝔐)\). Prove that the set of points where \(\lim_{n→∞} f_n\) exists is measurable.

Problem 68

Let \(f: ℝ→[0,∞)\) be uniformly continuous with \(∫_{-∞}^∞ f(x)\, dx < ∞\). Prove that \(f\) is bounded.

Problem 69

A function \(f\) satisfies a Lipschitz condition on an interval if there is a constant \(M\) such that \(|f(x) - f(y)| ≤ M|x-y|\) for all \(x\) in the interval.

Show that a function satisfying a Lipschitz condition is absolutely continuous.
Show that an absolutely continuous function on an interval \(J\) satisfyies a Lipschitz condition if and only if \(f' ∈ L_∞(J)\).

Problem 70

Let \(μ\) and \(ν\) be finite positive measures on the measurable space \((X, 𝔐)\) such that \(ν ≪ μ ≪ ν\), and let \(\frac{dν}{d(μ+ν)}\) represent the Radon-Nikodym derivative of \(ν\) with respect to \(μ+ν\). Show that

\[0< \frac{dν}{d(μ+ν)} < 1 \quad μ\text{-a.e.}\]

It suffices to show that for all \(ε>0\) there exists \(K>0\) such that \(k > K\) implies \(\bigl|∑_{n∈ ℕ}f_k(n)\bigr| < ε\).

Fix \(ε>0\). Since \(∑_n f(n) < ∞\), there exists \(N>0\) such that \(∑_{n≥N} f(n) < ε/2\).

For each \(0≤ j < N\), there exists \(n_j\) such that

\[k > n_j \quad ⟹ \quad |f_k(j)| < \frac{ε}{2N}.\]

Define \(K := \max \{n_j ∣ 0≤ j< N\}\).

To complete the proof, we show that \(k ≥ K\) implies \(\bigl|∑_{n∈ ℕ}f_k(n)\bigr| < ε\).

Indeed, if \(k≥ K\), then

\[\begin{split}\bigl|∑_{n ∈ ℕ} f_k(n)\bigr| & ≤ ∑_{n ∈ ℕ} |f_k(n)| &\\ & = ∑_{n=0}^{N-1} |f_k(n)| + ∑_{n=N}^∞ |f_k(n)| &\\ & = N \frac{ε}{2N} + ∑_{n=N}^∞ f(n)&(∵ k > K)\\ & = \frac{ε}{2} + \frac{ε}{2} = ε &(\text{by definition of } N).\end{split}\]

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Following the hint, consider

\[∫_a^b e^{-xy}\, dy = \left.\frac{-e^{-xy}}{x}\right|_{y=a}^{y=b} = \frac{-e^{-xb}+e^{-xa}}{x} = \frac{e^{-xa}-e^{-xb}}{x}.\]

Therefore,

\[∫_0^∞ \frac{e^{-ax} - e^{-bx}}{x} \, dx = ∫_0^∞ \bigl(∫_a^b e^{-xy}\, dy\bigr)\, dx = \lim\limits_{c→+∞}∫_{1/c}^c\bigl( ∫_a^b e^{-xy}\, dy \bigr)\, dx.\]

By Tonelli’s theorem we have

(46)¶\[∫_Y \left(∫_X g(x,y)\, dx\right)\, dy = ∬_{X×Y}g(x,y)\, dm(x×y) = ∫_X \left(∫_Y g(x,y)\, dy\right)\, dy,\]

whenever \(g: X × Y → [0,∞]\) is measurable with respect to product measure on \(X × Y\).

In the present case, \(g(x,y) := e^{-xy}\), which is continuous, hence measurable, and \(g≥0\).

Therefore, we can compute the integral as either of the iterated integrals in (46). We choose to integrate with respect to \(x\) first.

\[\begin{split}∫_a^b \bigl(\lim\limits_{c→+∞}∫_{1/c}^c e^{-xy}\, dx \bigr)\, dy &= ∫_a^b \left[\frac{-e^{-xy}}{y}\right]_{x=0}^{x → ∞}\, dy \\ &= ∫_a^b \lim\limits_{c→∞}\left(\frac{-e^{-cy} + e^{-y/c}}{y}\right)\, dy \\ &= ∫_a^b \frac{1}{y}\, dy = \left. \log y \right|_a^b = \log b - \log a = \log (b/a).\end{split}\]

☐

This is false as the following example shows:

Let \(f(x) = 1-x\). Then

\[\begin{split}\|f\|_2^2 &= ∫_0^1 |f(x)|^2\, dx=∫_0^1 (1-x)(1-x)\, dx= ∫_0^1 1-2x +x^2\, dx \\&= \left[x-x^2 +\frac{x^3}{3}\right]_0^1 = 1-1 +\frac{1}{3} = \frac{1}{3}.\end{split}\]

\[\begin{split}\|g\|_2^2 &= ∫_0^1 |g(x)|^2\, dx =∫_0^1 \left|∫_0^x f(t)\, dt\right|^2 \, dx= ∫_0^1 \left|x-\frac{x^2}{2}\right|^2 \, dx\\ &= ∫_0^1 (x^2-x^3 + \frac{x^4}{4}) \, dx= \left[\frac{x^3}{3} - \frac{x^4}{4}+ \frac{x^5}{20}\right]_0^1\\ &=\frac{1}{3} - \frac{1}{4}+ \frac{1}{20} = \frac{1}{12} + \frac{1}{20} = \frac{32}{240} = \frac{4}{30}.\end{split}\]

Then,

\[\|g\|_2 = \sqrt{\frac{4}{30}} = \frac{2}{\sqrt{2}\sqrt{3}\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5} \sqrt{3}} \quad \text{ and } \quad \|f\|_2 = \frac{1}{\sqrt{3}}.\]

Claim. \(\frac{2\sqrt{2}}{\sqrt{5} \sqrt{3}} ≰ \frac{1}{2\sqrt{3}}\).

\[\begin{split}\frac{\sqrt{2}}{\sqrt{5} \sqrt{3}} &≤ \frac{1}{2\sqrt{3}}\\ \frac{2\sqrt{3}\sqrt{2}}{\sqrt{5} \sqrt{3}} &≤1\\ \frac{2\sqrt{2}}{\sqrt{5}} &≤ 1\\ 1.264911064 &≰ 1\\\end{split}\]

☐

Let \((X, 𝔐)\) be the measurable space under consideration, and define

\[g_k(x) := (\sup_{n≥k} f_n)(x).\]

We first show that this supremum of a sequence of measurable functions is measurable. This requires we prove for all \(α∈ ℝ\) and all \(k∈ ℕ\) that \(g_k^{-1}(α, ∞] ∈ 𝔐\).

Fix \(α∈ ℝ\) and \(k∈ ℕ\). Then,

(47)¶\[\begin{split}g_k^{-1}(α, ∞] &= \{x∈ X ∣ g_k(x) > α\}\\ & = ⋃_{n≥ k} \{x ∈ X ∣ f_n(x) > α\}\\ & = ⋃_{n≥ k} f_n^{-1}(α, ∞].\end{split}\]

Now, since \(f_n^{-1}(α, ∞] ∈ 𝔐\) by measurability of \(f_n\), and since \(𝔐\) is a σ-algebra, we have \(g_k^{-1}(α, ∞] ∈ 𝔐\) by (47).

Next observe

\[\lim\limits_{k→∞} g_k = \limsup\limits_{k→∞} f_n = \inf\limits_k \sup\limits_{n≥k} f_n = \inf\limits_k g_k.\]

To complete the proof then, we must show that \(\inf\limits_k g_k\) is measurable whenever \(\{g_k\}\) is a sequence of measurable functions.

Observe that \(\inf\limits_{k≥ 0} g_k = -\sup\limits_{k≥ 0} (-g_k)\), and \(-g_k\) is measurable since each \(g_k\) is measurable.

We have already proved that the supremum of a sequence of measurable functions is measurable, so the proof for Part a is complete.

☐
We must show that the set \(E:= \{x∈ X ∣ \lim\limits_{n→∞} f_n(x) \text{ exists }\}\) is measurable.

Fix \(x∈ X\) and recall that \(\lim\limits_{n→∞} f_n(x)\) exists if and only if \(\liminf\limits_{n→∞} f_n(x) = \limsup\limits_{n→∞} f_n(x)\). Thus,

\[E= \{x∈ X ∣ \liminf\limits_{n→∞} f_n(x) = \limsup\limits_{n→∞} f_n(x)\}.\]

It follows from Part a above that if \(\{f_n\}\) is a sequence of measurable functions on \((X, 𝔐)\), then \(\liminf\limits_{n→∞} f_n(x)\) is measurable. Indeed (dropping \((x)\) for brevity),

\[\liminf\limits_{n→∞} f_n = \sup\limits_{k≥ 0}\inf\limits_{n≥k}f_n = \sup\limits_{k≥ 0}(-\sup\limits_{n≥k}(-f_n)) = -\inf\limits_{k≥ 0}\sup\limits_{n≥k}(-f_n).\]

The following lemma completes the proof.

Lemma. If \(g\) and \(h\) are measurable functions, then the set \(\{x∈ X ∣ g(x) = h(x)\}\) is measurable.

Proof. Let \(\{r_n ∣ n ∈ ℕ\}\) be an enumeration of the rationals. Then,

\[\begin{split}E_1 &:= \{x ∈ X ∣ g(x) < h(x)\}\\ &= ⋃_{n=0}^∞\left(\{x∈ X ∣ g(x) < r_n\} ∩ \{x∈ X ∣ r_n < h(x)\}\right),\end{split}\]

is measurable since each set involved in the union is measurable. For the same reason, the set \(E_2 := \{x ∈ X ∣ g(x) > h(x)\}\) is also measurable.

Therefore, \(\{x∈ X ∣ g(x) = h(x)\} = X - E_1 - E_2\) is measurable.

☐

We must show there exists \(M>0\) such that \(∀ x∈ ℝ\), \(f(x) < M\).

Define \(I := ∫_{-∞}^∞ f(x) \, dx\) and keep in mind that \(I<∞\).

Suppose on the contrary that \(f\) is unbounded.

By uniform continuity, \(∀ ε>0\), \(∃ δ>0\), \(∀ x, y ∈ ℝ\),

(48)¶\[|x-y| < δ \; ⟹ \; |f(x) - f(y)|<ε.\]

Consider \(M := I/δ\). Suppose there exists \(x∈ ℝ\) such that \(f(x) > M + ε\).

By (48), for all \(y∈ B(x;δ)\) we have \(|f(x) - f(y)|<ε\), so \(f(y) > M\) for all \(y∈ B(x;δ)\). Therefore,

\[I := ∫_{-∞}^∞ f(x) \, dx ≥ ∫_{B(x;δ)} f(x) \, dx > M ⋅ m(B(x;δ)) = M ⋅ 2δ = 2I,\]

a contradiction.

We have thus proved that \(ε + I/δ\) is an upper bound on \(f\). By assumption, \(f\) is bounded below by 0, so the proof is complete.

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Assume \(f\) is Lipschitz continuous in \([a,b]\). We must prove \(f ∈ AC[a,b]\).

That is, we must show for all \(ε>0\) there exists \(δ>0\) such that for every finite collection \(\{(a_i,b_i) ∣ i=1,2,\dots, n\}\) of open intervals in \((a,b)\), we have

\[∑_{i=1}^n (b_i - a_i) < δ \; ⟹ \; ∑_{i=1}^n |f(b_i)- f(a_i)| < ε.\]

Fix \(ε>0\). Let \(M\) be the Lipschitz constant and define \(δ := ε/M\).

If \(∑_{i=1}^n (b_i - a_i) < δ\), then

\[∑_{i=1}^n |f(b_i)- f(a_i)| ≤ ∑_{i=1}^n M ⋅ (b_i-a_i) =M ∑_{i=1}^n (b_i-a_i) < M ⋅ δ = ε.\]

☐
Assume \(f∈ AC(J)\).

(⇒) Suppose \(f\) is Lipschitz continuous with Lipschitz constant \(M\).

For \(x\) in the interior of \(J\),

\[f'(x) = \lim\limits_{h→ 0} \frac{f(x+h) - f(x)}{h}.\]

For \(h ≠ 0\) and \(x + h ∈ J\),

\[\left|\frac{f(x+h) - f(x)}{h}\right| ≤ \frac{Mh}{h} = M.\]

Thus if \(h ≠ 0\) and \(x + h ∈ J\), then

\[-M ≤ \frac{f(x+h) - f(x)}{h} ≤ M,\]

so \(-M ≤ f'(x) ≤ M\) since limits preserve finite bounds.

Therefore, \(f'(x)\) is bounded for almost every \(x∈ J\).

Measurability of \(f'\) follows from Problem 4. Whence \(f'∈ L_∞(J)\).

(⇐) Suppose \(f' ∈ L_∞(J)\), so \(|f'| ≤ M\) a.e. for some \(M>0\).

Suppose \(x ≤ y\) in \(J\). Then,

\[\begin{split}|f(x) - f(y)| &= \left|∫_x^y f'(t) \, dt\right| \qquad (∵ f ∈ AC(J))\\ &≤ ∫_x^y |f'(t)| \, dt ≤ M(y-x) = M |x - y|.\end{split}\]

Therefore, \(f\) satisfies a Lipschitz condition on \(J\).

☐

Clearly \(μ ≪ μ+ν\) and \(ν ≪ μ+ν\), so the Radon-Nikodym theorem gives unique nonnegative functions \(f,g ∈ L_1(μ+ν)\) such that

(49)¶\[μ E = ∫_E f \, d(μ+ν) \quad \text{ and } \quad ν E = ∫_E g \, d(μ+ν),\]

for every \(E ∈ 𝔐\). Here \(g = dν/d(μ+ν)\).

We must show \(0<g(x)<1\) holds for \(\mu\)-almost every \(x ∈ X\).

Since \(μ+ν\) and \(ν\) are both positive, it follows from (49) that \(g(x) ≥ 0\) for all \(x∈ X\).

Suppose there exists \(E ∈ 𝔐\) such that \(g(x) = 0\) for all \(x ∈ E\).

We will show that \(μ E = 0\), as this will allow us to conclude \(g>0\) \(μ\)-a.e.

If \(g(x) = 0\) for all \(x∈ E\), then

\[ν E = ∫_E g \, d(μ+ν) = 0 \quad \text{ and } \quad μ ≪ ν \; ⟹ \; μ E = 0.\]

So, to finish the proof, we must show \(g(x) < 1\) for \(μ\)-a.e. \(x∈X\).

Suppose \(g(x) ≥ 1\) for \(x ∈ E\). Then,

\[ν E = ∫_E g\, d(μ+ν) ≥ ∫_E 1 \, d(μ+ν) = μ E + ν E.\]

Now, since \(ν E\) is finite, we can subtract it from both sides of the last expression yielding \(0≥ μ E\). On the other hand, \(μ ≥ 0\), so \(μ E = 0\).

We have thus proved that \(g(x) < 1\) for \(μ\)-almost every \(x ∈ X\).

☐