Exam 10 (71–76)¶

2007 Nov¶

Notation $ℝ$ is the set of real numbers and $ℝ^n$ is $n$-dimensional Euclidean space. Denote by $m$ Lebesgue measure on $ℝ$ and $m_n$ $n$-dimensional Lebesgue measure. Be sure to give a complete statement of any theorems from analysis that you use in your proofs below.

Problem 71

Let $μ$ be a positive measure on a measure space $X$. Assume that $E_1, E_2, \dots$ are measurable sets in $X$ with the property that for $n ≠ m, μ (E_n ∩ E_m) = 0$. Let $E$ be the union of these sets. Prove that

\[μ E = ∑_{n=1}^∞ μ E_n\]

Problem 72

State a theorem that illustrates Littlewood’s principle for pointwise a.e. convergence of a sequence of functions on $ℝ$.
Suppose that $f_n ∈ L_1(m)$ for $n=1,2,\dots$. Assuming that $\|f_n-f\|_1 → 0$ and $f_n → g$ a.e. as $n → ∞$, what relation exists between $f$ and $g$? Make a conjecture and then prove it using the statement in Part a.

Problem 73

Let $K$ be a compact set in $ℝ^3$ and let $f(x) = \mathrm{dist}(x,K)$.

Prove that $f$ is a continuous function and that $f(x) = 0$ if and only if $x ∈ K$.
Let $g = \max(1-f,0)$ and prove that $\lim_{n → ∞} ∭ g^n$ exists and is equal to $m_3 K$.

Problem 74

Let $E$ be a Borel subset of $ℝ^2$.

Explain what this means.
Suppose that for every real number $t$ the set $E_t = \{(x,y) ∈ E ∣ x = t\}$ is finite. Prove that $E$ is a Lebesgue null set.

Problem 75

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

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

Problem 76

Suppose that $1 < p < ∞$ and that $q = p/(p-1)$.

Let $a_1, a_2, \dots$ be a sequence of real numbers for which the series $∑_n a_n b_n$ converges for all real sequences $\{b_n\}$ satisfying the condition $∑ _n |b_n|^q < ∞$. Prove that $∑_n |a_n|^p < ∞$.
Discuss the cases of $p = 1$ and $p = ∞$. Prove your assertions.

Define $F_1 = E_1, \, F_2 = E_2 \setminus E_1,\, F_3 = E_3 \setminus (E_1 \cup E_2), \dots$, and, in general,

\[F_n = E_n \setminus \bigcup_{i=1}^{n-1} E_i \quad (n=2, 3, \dots).\]

If $\mathfrak{M}$ is the $\sigma$-algebra of $\mu$-measurable subsets of $X,$ then $F_n\in \mathfrak{M}$ for each $n\in \mathbb N,$ since $\mathfrak{M}$ is a $\sigma$-algebra. Also, $F_i\cap F_j = \emptyset$ for $i\neq j$, and $F_1\cup F_2 \cup \cdots \cup F_n = E_1\cup E_2 \cup \cdots \cup E_n$ for all $n\in \mathbb N$. Thus,

\[\bigcup_{n=1}^\infty F_n = \bigcup_{n=1}^\infty E_n \triangleq E,\]

and, by $\sigma$-additivity of $\mu$,

\[\mu(E) = \mu(\bigcup_{n=1}^\infty F_n) = \sum_{n=1}^\infty \mu(F_n).\]

Therefore, if we can show $\mu(E_n) = \mu(F_n)$ holds for all $n\in \mathbb N$, the proof will be complete.

Now, for each $n = 2, 3, \dots$,

(50)¶\[F_n = E_n \cap (\bigcup_{i=1}^{n-1}E_i)^c\]

and

(51)¶\[\mu(E_n) = \mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)^c) + \mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)).\]

Equation (51) holds because $\bigcup_{i=1}^{n-1}E_i$ is a measurable set for each $n = 2, 3, \dots$. Finally, note that

\[E_n \cap (\bigcup_{i=1}^{n-1}E_i) = \bigcup_{i=1}^{n-1}(E_n\cap E_i),\]

which implies

\[\mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)) \leq \sum_{i=1}^{n-1}\mu(E_n\cap E_i),\]

by $\sigma$-subadditivity. By assumption, each term in the last sum is zero, and therefore, by (50) and (51),

\[\mu(E_n) = \mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)^c) = \mu(F_n) \quad \text{holds for each $n=2, 3, \dots$}.\]

For $n=1$, we have $F_1 = E_1$, by definition.

☐

One of Littlewood’s principles involving a.e. convergence of a sequence of functions on $ℝ$ is Egoroff’s theorem.

Another is this: if $f_n → f$ in $L_1(ℝ)$, then there is a subsequence $\{f_{n_j}\}$ converging to $f$ a.e. in $ℝ$.
Claim. $f = g$ a.e.

We give two alternative proofs.

Proof 1. By the second of Littlewood’s principles mentioned in Part a, $f_{n_j} → f$ a.e.

If $f_n → g$ a.e., then every subsequence of $\{f_n\}$ converges to $g$ a.e. In particular, $f_{n_j} → g$ a.e.

Let $A := \{x ∈ ℝ: f_{n_j}(x) ↛ f(x) \}$ and $B := \{x ∈ ℝ: f_{n_j}(x) ↛ g(x) \}$.

Then $m(A ∪ B) ≤ m A + m B = 0$, and off of $A ∪ B$ we have,

\[|f(x) - g(x)| ≤ |f(x) - f_{n_j}(x)| + | f_{n_j}(x)- g(x)| → 0 \; \text{ as } \; j → ∞.\]

Therefore, $f = g$ on $ℝ - (A ∪ B)$ and $m(A ∪ B) = 0$, so $f = g$ a.e.

☐

Proof 2. First, we claim that if $f = g$ a.e. on $[-n,n]$ for every $n ∈ ℕ$, then $f=g$ a.e. in $ℝ$.

To see this, let $B_n = \{x ∈ [-n,n]: f(x) ≠ g(x)\}$. Then $m B_n = 0$ for all $n ∈ ℕ$, so that if $B = \{x ∈ ℝ: f(x) ≠ g(x)\}$, then $B = ⋃_n B_n$ and $m B ≤ ∑ m B_n = 0$, as claimed.

Thus, to prove the conjecture it is enough to show that $f = g$ for almost every $x ∈ [-n, n]$, for an arbitrary fixed $n ∈ ℕ$.

Fix $n ∈ ℕ$, and suppose we know that $f - g ∈ L_1([-n,n], m)$. (This will follow from the fact that $f, g ∈ L_1([-n,n],m)$, which we prove below.)

Then, for all $ε > 0$ there is a $δ > 0$ such that $∫_E |f-g|\, dm < ε$ for all measurable $E ⊆ [-n,n]$ with $m E < δ$.

Now apply Egoroff’s theorem to find a set $A ⊆ [-n,n]$ such that $m([-n,n] - A) < δ$ and $f_n → g$ uniformly on $A$. Then

\[\begin{split}∫_{-n}^n|f-g|\,dm &= ∫_{[-n,n]- A} |f-g|\, dm + ∫_{A} |f-g|\,dm \\ &≤ ε + ∫_{A} |f-f_n|\, dm + ∫_{A} |f_n-g|\,dm \\ &≤ ε + \|f-f_n\|_1 + mA\,\sup_{x∈A}|f_n(x)-g(x)|,\end{split}\]

where $\|f-f_n\|_1 → 0$ (by assumption) and $\sup_{x ∈ A}|f_n(x)-g(x)| → 0$ since $f_n → g$ uniformly on $A$.

Since $ε > 0$ was arbitrary, it follows that $∫_{-n}^n |f - g| \, dm = 0$ and, for functions $f, g ∈ L_1([-n,n],m)$, this implies that $f = g$ a.e. on $[-n,n]$.

It remains to show that $f, g ∈ L_1([-n,n],m)$.

It’s clear that $f ∈ L_1$ since $\|f\|_1 ≤ \|f - f_n\|_1 + \|f_n\|_1 < ∞$.

To prove $g ∈ L_1([-n,n],m)$ note that $f_n \xrightarrow{a.e.} g$ implies $\lim_n |f_n(x)| =|g(x)|$ for almost all $x$, so by Fatou’s lemma,

\[\|g\|_1 = ∫ |g|\, dm = ∫ \lim \inf |f_n| \, dm ≤ \lim \inf ∫ |f_n| \, dm = \lim \|f_n\|_1 = \|f\|_1 < \infty.\]

The last equality holds because $\bigl| \|f_n\|_1 - \|f\|_1 \bigr| ≤ \|f_n - f\|_1 → 0$, by the triangle inequality.

☐

Define $\mathrm{dist}(x,K) = f(x) = \inf_{k∈K} |x-k|$.

Clearly, for all $k ∈ K$, $f(x) ≤ |x-k|$. Therefore, by the triangle inequality, for all $x, y ∈ ℝ^3$,

\[f(x) ≤ |x-y|+|y-k|, \quad ∀ k ∈ K,\]

and so, taking the infimum over $k ∈ K$ on the right,

(52)¶\[f(x) ≤ |x-y|+f(y).\]

Similarly,

(53)¶\[f(y) ≤ |x-y|+f(x).\]

Obviously, for each $x ∈ ℝ^3$, $f(x)$ is finite. Therefore, (52) and (53) together imply

\[|f(x)-f(y)| ≤ |x-y|, \quad ∀ x, y ∈ ℝ^3.\]

Whence $f$ is Lipschitz continuous.

Now, if $x ∈ K$, then it’s clear that $f(x) = 0$.

Suppose $x ∉ K$; that is, $x ∈ K^c$. Since $K$ is closed, $K^c$ is open and we can find an $ε$-neighborhood about $x$ fully contained in $K^c$, in which case $f(x) > ε$.

We have thus proved that $f(x) = 0$ if and only if $x ∈ K$.

☐
First, observe that $f(x)=0$ for all $x ∈ K$, and $f(x)>0$ for all $x ∉ K$.

Define $K_1$ to be a closed and bounded set containing $K$ on which $f(x) ≤ 1$. That is, $K_1$ is the set of points that are a distance of not more than 1 unit from the set $K$. In particular, $K ⊂ K_1$.

Notice that $g = \max(1-f,0) = (1-f)χ_{K_1}$. Also, if $x ∈ K_1 - K$, then $0 ≤ 1-f(x) < 1$, so $g^n → 0$ on the set $K_1 - K$, while on the set $K$, $g^n = 1$ for all $n ∈ ℕ$. Therefore, $g^n → χ_K$.

Finally, note that $g^n ≤ χ_{K_1} ∈ L_1(ℝ^3)$ so the dominated convergence theorem applies and yields $\lim_{n → ∞} ∭ g^n = ∭ χ_K = m_3 K.$

☐

Recall, the Borel σ-algebra of $ℝ^2$, denoted $ℬ(ℝ^2)$, is the smallest σ-algebra that contains the open subsets of $ℝ^2$. The sets belonging to $ℬ(ℝ^2)$ are called Borel sets in $ℝ^2$.

☐
Observe that if $G$ is a finite set in $ℝ$, then $G$ is a Lebesgue null set. That is, $m G = 0$.

In problems involving 2-dimensional Lebesgue measure, distinguishing $x$ and $y$ coordinates sometimes clarifies things. To wit, let $(X, ℬ(X), μ)$ and $(Y, ℬ(Y), ν)$ be two copies of the measure space $(ℝ, ℬ(ℝ), m)$, and represent Lebesgue measure on $ℝ^2$ by $(X × Y, ℬ(X) ⊗ ℬ(Y), μ × ν) =(ℝ^2, ℬ(ℝ^2), m_2)$. 1

Our goal is to prove that $m_2 E = 0$.

Observe that

\[m_2 E = (μ × ν)(E) = ∫_{X × Y} χ_E \, d(μ × ν).\]

The integrand $χ_E$ is non-negative and measurable (since $E$ is Borel). Therefore, by Tonelli’s theorem),

(54)¶\[m_2 E = ∫_Y ∫_X χ_E(x,y) \, dμ(x)\, dν(y) = ∫_X ∫_Y χ_E(x,y) \, dν(y) \, dμ(x).\]

Now, let

\[G_t = \{y ∈ ℝ: (x,y) ∈ E \text{ and } x=t\}.\]

This is the $x$-section of $E$ at the point $x=t$, which is a subset of $ℝ$, though we can view it as a subset of $ℝ^2$ by simply identifying each point $y ∈ G_t$ with the point $(t,y) ∈ E_t = \{(x,y) ∈ E: x = t\}$.

Then, for each $t ∈ ℝ$, $G_t$ is a finite subset of $ℝ$, so $m G_t = 0$. 2

Finally, by (54),

\[m_2 E = ∫_X ∫_Y χ_{G_t}(y) \, dν(y) \, dμ(t) = ∫_X ν (G_t) \, dμ(t) = 0.\]

since $ν G_t := m G_t = 0$.

☐

First note that $ν ≪ μ$ implies $ν ≪ μ + ν$, so, by the Radon-Nikodym theorem, there is a unique $f ∈ L_1(μ + ν)$ such that

\[ν E = ∫_E f \, d(μ+ν) \quad ∀ E ∈ 𝒜.\]

Indeed, $f$ is the Radon-Nikodym derivative; i.e., $f = \frac{dν}{d(μ + ν)}$. We want to show $0 < f(x) < 1$ holds for $\mu$-almost every $x \in X$. Since we’re dealing with positive measures, we can assume $f(x) ≥ 0$ for all $x ∈ X$.

If $B_0 = \{x ∈ X: f(x) = 0\}$, then

\[ν B_0 = ∫_{B_0} f \, d(μ + ν) = 0.\]

Therefore, $μ B_0 = 0$, since $μ ≪ ν$, which proves that $f(x) > 0$, $μ$-a.e.

If $B_1 = \{x ∈ X: f(x) ≥ 1\}$, then

\[ν B_1 = ∫_{B_1} f \, d(μ + ν) ≥ (μ + ν)(B_1) = μ B_1 + ν B_1.\]

Since $ν$ is finite by assumption, we can subtract $ν B_0$ from both sides to obtain $μ B_1 = 0$. This proves $f(x) < 1$, $μ$-a.e.

☐

For each $k ∈ ℕ$, define $T_k: ℓ_q(ℕ) → ℝ$ by $T_k b = ∑_{n=1}^k a_n b_n$, for $b ∈ ℓ_q(ℕ)$.

Then $\{T_k\}$ is a family of pointwise bounded linear functionals. That is, each $T_k$ is a linear functional, and for each $b ∈ ℓ_q(ℕ)$ there is an $M_b ≥ 0$ such that $|T_k b| ≤ M_b$ holds for all $k ∈ ℕ$.

To see this, simply note that a convergent sequence of real numbers is bounded, and, in the present case, we have

\[S_k := ∑_{n=1}^k a_n b_n → ∑_{n=1}^∞ a_n b_n = x ∈ ℝ.\]

Thus, $\{T_k b\} = \{S_k\}$ is a convergent sequence of real numbers, so, if $N ∈ ℕ$ is such that $k ≥ N$ implies $|S_k - x| < 1$, and if $M_b'$ is defined to be $\max\{|S_k|: 1 ≤ k ≤ N\}$, then for each $k ∈ ℕ$,

\[|T_k b| ≤ M_b := \max\{ M_b', x+1\}.\]

Next note that $ℓ_q$ is a Banach space, so the (Banach-Steinhauss) principle of uniform boundedness implies that there is a single $M > 0$ such that $\|T_k\| ≤ M$ for all $k ∈ ℕ$.

In other terms,

\[(∃ M>0) \, (∀ b ∈ ℓ_q) \, (∀ k ∈ ℕ) \; |T_k b| ≤ M \|b\|.\]

Define

\[T b := ∑_{n=1}^∞ a_n b_n = \lim_{k → ∞} T_k b,\]

which exists by assumption. Since $| \; |$ is continuous, we conclude that $\lim_{k → ∞}|T_k b| = |T b|$.

Finally, since $|T_k b| ≤ M \|b\|$ for all $k ∈ ℕ$, we have $|T b| ≤ M \|b\|$. That is $T$ is a bounded linear functional on $ℓ_q(ℕ)$.

Now, by the Riesz representation theorem, if $1 ≤ q < ∞$, then every bounded linear functional $T ∈ ℓ_q^∗$ is uniquely representable by some $α = (α_1, α_2, \dots) ∈ ℓ_p$ as

(55)¶\[T b = ∑_{n=1}^∞ α_n b_n.\]

On the other hand, by definition $T b = ∑_{n=1}^∞ a_n b_n$ for all $b ∈ ℓ_q$.

Since the representation in (55) is unique, $a = α ∈ ℓ_p$. That is, $∑_n |a_n|^p < ∞$.

☐

Consider the case $p = 1$ and $q = ∞$.

Recall that the Riesz representation theorem says that every $T ∈ ℓ_q^∗ \, (1 ≤ q < ∞)$ is uniquely representable by some $α ∈ ℓ_p$ (where $p = q/(q-1)$, so $1 < p ≤ ∞$). That is, $ℓ_p$ is the dual of $ℓ_q$, when $1 ≤ q < ∞$ and $p = q/(q-1)$.

In the present case we have $q = ∞$ and $p = 1$ and $ℓ_1$ is not the dual of $ℓ_∞$. 3 So we can’t use the same method of proof for this case.

Nonetheless, the result still holds by the following simple argument: Define $b = (b_1, b_2, \dots)$ by

\[\begin{split}b_n = \mathrm{sgn}(a_n) = \begin{cases} ā_n/|a_n|, & a_n ≠ 0,\\ 0, & a_n= 0, \end{cases} \quad (n ∈ ℕ).\end{split}\]

Then $∑_n |a_n| = ∑_n a_n b_n$ converges by the hypothesis, since $|b_n| ∈ \{0, 1\}$ implies $b ∈ ℓ_∞$. Therefore, $a ∈ ℓ_1$.

Finally, in case $p = ∞$ and $q=1$, the Riesz representation theorem can be applied as in Part a.

☐

1: The expression $ℬ(X) ⊗ ℬ(Y)$ denotes the σ-algebra generated by all sets $A × B ⊆ X × Y$ with $A ∈ ℬ(X)$ and $B ∈ ℬ(Y)$. (See Fubini’s theorem. In this case, $ℬ(X) ⊗ ℬ(Y)$ is the same as $ℬ(ℝ^2)$.
2: In fact, it is easy to prove that if $G$ is any countable subset of $ℝ$, then $m G = 0$. Just fix $ε > 0$ and cover each point $x_n ∈ G$ with a set $E_n$ of measure less than $ε 2^{-n}$. Then $m G ≤ ∑_n m E_n < ε$.
3: One way to see that $ℓ_1$ is not the dual of $ℓ_∞$ this is to note that $ℓ_1$ is separable but $ℓ_∞$ is not. For the collection of $a\in \ell_\infty$ such that $a_n\in \{0, 1\}, \,n\in \mathbb N$, is uncountable and, for any two distinct such sequences $a, b\in \{0, 1\}^{\mathbb N}$, we have $\|a - b\|_\infty = 1$, so there cannot be a countable base, so $\ell_\infty$ is not second countable, and a metric space is separable iff it is second countable.)