Exam 1 (1–9)¶

1991 Nov¶

Instructions. Work as many problems as you can. Do each problem on a separate sheet. Justify all important steps.

Problem 1

Let $\{f_n\}$ be a sequence of continuous real-valued functions on $[0,1]$ that converges uniformly to $f$ . Prove that $\lim_{n→ ∞}f_n(x_n) = f(1/2)$ for every sequence $\{x_n\}$ that converges to $1/2$ .
Must the conclusion still hold if the convergence is only point-wise? Explain.

Problem 2

Let $f: ℝ → ℝ$ be differentiable and assume there is no $x ∈ ℝ$ such that $f(x) = f'(x) = 0$ . Show that $S := \{x ∣ 0 ≤ x ≤ 1, f(x)=0\}$ is a finite set.

Problem 3

Show that if $(X, 𝔐, μ)$ is a measure space and if $f$ is $μ$ -integrable, then for every $ε > 0$ there is $E ∈ 𝔐$ such that $μ E < ∞$ and

$∫_{X-E} |f| \, dμ < ε.$

Problem 4

Show that if $(X, 𝔐, μ)$ is a measure space, $f$ is a nonnegative measurable function, and $ν E = ∫_E f\, dμ$ , then $ν$ is a measure.

Problem 5

Suppose $f$ is a bounded real-valued function on $[0,1]$ . Show that $f$ is measurable if and only if

$\sup ∫ ψ \, dm = \inf ∫ φ \, dm$

where $m$ is Lebesgue measure on $[0,1]$ , and $ψ$ and $φ$ range over all simple functions, $ψ≤ f≤ φ$ .

Problem 6

Show that if $f$ is Lebesgue integrable on $[0,1]$ and $ε >0$ , then there is $δ > 0$ such that if $E ⊆ [0,1]$ is measurable and $m E < δ$ , then

$\left|∫_E f \, dm \right| < ε.$

Problem 7

Suppose $f$ is a bounded real-valued measurable function on $[0,1]$ such that $∫ x^n f \, dm = 0$ for $n = 0,1,2,\ldots$ , with $m$ Lebesgue measure. Show that $f(x) = 0$ a.e.

Problem 8

Show that if $μ$ and $ν$ are finite measures on the measurable space $(X, 𝔐)$ , then there is a nonnegative measurable function $f$ on $X$ such that for all $E$ in $𝔐$ ,

(1)¶

$∫_E (1-f) \, dμ = ∫_E f \, dν.$

Problem 9

Show that if $f$ and $g$ are integrable functions on $(X, 𝔐, μ)$ and $(Y, 𝔑, ν)$ (resp.) and $F(x,y) = f(x)\, g(y)$ , then $F$ is integrable on $X × Y$ and

$∫ F \, d(μ × ν) = ∫ f \,dμ \, ∫ g \, dν.$

Let $\{x_n\}$ be a sequence in $[0,1]$ with $x_n → 1/2$ as $n → ∞$ .

Fix $ε > 0$ and let $N_0 ∈ ℕ$ be such that $n ≥ N_0$ implies $|f_n(x)-f(x)| < ε/2$ , for all $x ∈ [0,1]$ .

Let $δ > 0$ be such that $|f(x)-f(y)| < ε/2$ , for all $x, y ∈ [0,1]$ with $|x-y|< δ$ .

Finally, let $N_1 ∈ ℕ$ be such that $n ≥ N_1$ implies $|x_n-1/2| < δ$ .

Then $n ≥ \max\{N_0, N_1\}$ implies

$|f_n(x_n)-f(1/2)| ≤ |f_n(x_n)-f(x_n)|+|f(x_n)-f(1/2)| < ε/2+ε/2= ε.$

☐
Suppose the convergence is only point-wise.

Then the conclusion is false, as the following counterexample demonstrates.

Define $f_n(x)$ to be the function

$\begin{split}f(x) = \begin{cases} 0, & 0 ≤ x < \frac{1}{2}-\frac{1}{2n},\\ 2nx-(n-1), & \frac{1}{2}-\frac{1}{2n} ≤ x < \frac{1}{2},\\ 1, & \frac{1}{2} ≤ x ≤ 1. \end{cases}\end{split}$

That is, $f_n(x)$ is constantly zero for $x$ less than $\frac{1}{2}-\frac{1}{2n}$ , then it increases linearly until it reaches one at $x=1/2$ , and then it remains constantly one for $x$ bigger than $1/2$ .

Define the sequence $x_n = \frac{1}{2}-\frac{1}{n}$ .

Then $f_n(x_n)=0$ for all $n ∈ ℕ$ and $x_n → 1/2$ , while the sequence $f_n$ approaches the characteristic function $f := χ_{[\frac{1}{2},1]}$ which is one on $[\frac{1}{2},1]$ and zero elsewhere. Therefore, $f(1/2) = 1 ≠ 0 = \lim_n f_n(x_n)$ .

☐

Consider $f^{-1}(\{0\})$ . Since $\{0\}$ is closed and $f$ continuous, $f^{-1}(\{0\})$ is closed. Therefore $S = [0,1] ∩ f^{-1}(\{0\})$ is a closed and bounded subset of $ℝ$ . Hence, $S$ is compact.

Assume by way of contradiction that $S$ is infinite.

Then by the compactness theorem there is a limit point $x ∈ S$ ; i.e., there is a sequence $\{x_n\}$ of distinct points in $S$ that converges to $x$ .

Also, as all points are in $S$ , $f(x_n) = f(x) = 0$ for all $n ∈ ℕ$ .

We now show that $f'(x) = 0$ which will give us our desired contradiction.

Since $|x_n-x| → 0$ , we can write the derivative of $f$ as follows:

$f'(x) = \lim_{n→ ∞}\frac{f(x+(x_n-x)) - f(x)}{x_n-x} = \lim_{n→ ∞}\frac{f(x_n) - f(x)}{x_n-x} = 0.$

The last equality holds since $f(x) = f(x_n) = 0$ holds for all $n ∈ ℕ$ .

For $n = 1, 2, \ldots$ , define $A_n = \{x ∈ X : 1/n ≤ |f(x)| < n\}.$ Clearly,

$A_1 ⊆ A_2 ⊆ \cdots → A := ⋃_{n=1}^∞ A_n$

and each $A_n$ is measurable (why?). 1

$A_0 = \{x ∈ X : f(x)=0\} \quad \text{ and } \quad A_∞ = \{x ∈ X : |f(x)|=∞\}.$

Then $X = A_0 ∪ A ∪ A_∞$ is a disjoint union, and

(2)¶

$∫_X |f| = ∫_{A_0} |f| + ∫_A |f| + ∫_{A_∞} |f| = ∫_A |f|.$

The first term in the middle expression is zero since $f$ is zero on $A_0$ , and the third term is zero since $f∈ L_1(μ)$ implies $μ A_∞ = 0$ .

To prove the result then, we must find a measurable set $E$ such that $∫_{A-E} |f| < ε$ , and $μ E< ∞$ .

Define $f_n = |f|χ_{A_n}$ . Then $\{f_n\}$ is a sequence of nonnegative measurable functions and $\lim_{n→∞}f_n(x) = |f(x)|χ_A(x)$ for each $x ∈ X$ .

Since $A_n ⊆ A_{n+1}$ , we have $0≤ f_1(x) ≤ f_2(x)≤ \cdots$ , so the monotone convergence theorem implies $∫_X f_n → ∫_A |f|$ . 2

$\lim_{n→∞}∫_{A_n} |f| \, dμ = ∫_A |f| \, dμ = ∫_X |f| \, dμ.$

Therefore, there is some $N>0$ for which

$∫_{X-A_N} |f| \, dμ < ε.$

Finally, note that $1/N ≤ |f| < N$ on $A_N$ , so

$μ A_N ≤ N ∫_{A_N} |f| \, dμ ≤ N ∫_X |f| \, dμ < ∞.$

Therefore, the set $E = A_N$ meets the given criteria.

Clearly $μ E = 0 → ν E = 0$ . Therefore, $ν ∅ = μ ∅ = 0$ .

In particular $ν$ is not identically $∞$ , so we need only check countable additivity.

Let $\{E_1, E_2, \ldots\}$ be a countable collection of disjoint measurable sets. Then,

$\begin{split}ν (⋃_n E_n) &= ∫_{⋃_n E_n} f \, dμ = ∫ f χ_{⋃_n E_n}\, dμ &\\ &= ∫ ∑_{n=1}^∞ f χ_{E_n}\, dμ & (∵ E_i ∩ E_j = ∅,\, i≠j)\\ &= ∑_{n=1}^∞ ∫ f χ_{E_n}\, dμ & (∵ f χ_{E_n} ≥ 0,\, n=1,2, \ldots )\\ &= ∑_{n=1}^∞ ν E_n. &\end{split}$

The penultimate equality follows from the monotone convergence theorem applied to the sequence of nonnegative measurable functions $g_m = ∑_{n=1}^m f χ_{E_n}\; (m=1,2,\ldots)$ .

We prove a more general result that does not restrict $f$ to $[0,1]$ .

Claim. Let $(X, 𝔐, μ)$ be a measure space and $f ∈ L_1(μ)$ an integrable function. For all $ε >0$ there exists $δ > 0$ such that if $E ∈ 𝔐$ and $μ E < δ$ , then

$\left|∫_E f \, dμ \right| < ε.$

For example, suppose $f$ is nonnegative and bounded, say, $0≤ f(x) ≤ M$ , for all $x∈ X$ .

Then the result is trivial since, as long as $μ E < \frac{ε}{M}$ , we have

$\left|∫_E f \, dμ \right| = ∫_E f \, dμ ≤ M μ E < ε.$

Now let’s relax the boundedness assumption and only assume $f ∈ L_1(μ)$ and $f ≥ 0$ .

Define a sequence of bounded functions $f_n = f ∧ n$ for all $n∈ ℕ$ , so that $f_n ≤ n$ , $f_0 ≤ f_1 ≤ f_2 ≤ \cdots ≤ f$ , and $f_n ↗ f$ ; in particular, $\lim_n f_n(x)= f(x)$ for all $x∈ X$ .

Thus, $\{f_n\}$ is a monotonically increasing sequence of bounded measurable functions that approaches $f$ as $n→ ∞$ .

By the monotone convergence theorem we have $∫_X f_n\, dμ ↗ ∫_X f \, dμ$ . Therefore, for all $ε > 0$ there exists $N > 0$ such that

$n ≥ N \quad ⟹ \quad ∫_X (f - f_n) \, dμ < \frac{ε}{2}.$

Define $δ = \frac{ε}{2N}$ . If $E ∈ 𝔐$ has measure $μ E < δ$ , then

$\begin{split}0 ≤ ∫_E f\, dμ &= ∫_E (f - f_N) \, dμ + ∫_E f_N \, dμ\\[4pt] &≤ ∫_X (f - f_N) \, dμ + ∫_E f_N \, dμ\\[4pt] &< \frac{ε}{2} + N μ E < \frac{ε}{2} + \frac{ε}{2} = ε.\end{split}$

If $f$ is not nonnegative, then we work instead with the function $|f|$ .

Recall, $f ∈ L_1(μ)$ iff $∫_X |f| \, dμ < ∞$ . Of course, $|f|$ is a composition of a nonnegative continuous function ( $x↦ |x|$ ) and a measurable function ( $f$ ), so $|f|$ is a nonnegative measurable function.

For $n∈ ℕ$ , define $f_n = |f| ∧ n$ so that $f_n ≤ n$ and $f_0 ≤ f_1 ≤ f_2 ≤ \cdots ≤ f$ and $f_n ↗ |f|$ .

As above, there exists $δ > 0$ such that if $E∈ 𝔐$ has measure $μ E < δ$ , then $∫_E |f| \, dμ < ε$ .

Since $0 ≤ \left|∫_E f \, dμ\right| ≤ ∫_E |f| \, dμ < ε$ , this completes the proof.

☐

The first relies on the frequently useful technique employed in Problem 3 in which the domain is written as a union of the nested sets, $A_n = \{x ∈ X : 1/n ≤ |f(x)| < n\}$ .

Proof 1. Let $A_n, \, n=1,2,\ldots$ be the sequence of measurable sets defined in Problem 3.

That is,

$A_n = \{x ∈ X : 1/n ≤ |f(x)| < n\}, \quad \text{where} X = [0,1].$

As we saw in Problem 3,

$\lim_{n→ ∞}∫_{A_n} |f| \, dm = ∫_A |f| \, dm = ∫_X |f| \, dm.$

Let $n>0$ be such that

$∫_{X-A_n} |f| \, dm < ε/2.$

Define $\delta = (2n)^{-1}ε$ , and suppose $E⊆ [0,1]$ is a measurable set with $m E < δ$ .

We must show $|∫_E f \, dm| < ε$ .

$\begin{split}\left|∫_E f \, dm \right|&≤ ∫_E |f| \, dm\\ &= ∫_{(X-A_n)∩ E} |f| \, dm + ∫_{A_n ∩ E} |f| \, dm \\ &≤ ∫_{X-A_n} |f| \, dm + ∫_{A_n ∩ E} |f| \, dm \\ &< ε/2 + n\, m(A_n ∩ E) \\ &< ε/2 + n \, \frac{ε}{2n} = ε.\end{split}$

The penultimate inequality holds because $|f|<n$ on $A_n$ .

☐

Proof 2. The signed measure defined by $ν E = ∫_E f \, dμ$ is finite iff $f ∈ L_1(μ)$ . 4

It is also clearly absolutely continuous with respect to $μ$ .

Therefore, absolute continuity of measures, applied to the real and imaginary parts of any complex-valued $f∈ L_1(μ)$ , implies that for every $ε >0$ there exists $δ >0$ such that

$|ν E| = \left|∫_E f \, dμ\right| < ε, \quad \text{ whenever } μ E < δ.$

☐

Fix an arbitrary continuous function on $[0,1]$ , say, $φ ∈ C[0,1]$ .

By the Stone-Weierstrass theorem, there is a sequence $\{p_n\}$ of polynomials such that $\|φ - p_n\|_∞ → 0$ as $n→ ∞$ .

(3)¶

$\begin{split}\left| ∫ f φ \right| &= \left| ∫ f (φ - p_n +p_n) \right|\\ &\leq \int |f| |φ - p_n| + \left|∫ f p_n \right| \\ &\leq \|f\|_1 \|φ - p_n\|_∞ + \left|∫ f p_n \right|\\ &= \|f\|_1 \|φ - p_n\|_∞.\end{split}$

The last equality holds since $∫ x^n f = 0$ for all $n=0,1,2,\ldots$ , which implies that $∫ f p_n = 0$ for all polynomials $p_n$ .

Finally, note that $\|f\|_1 < ∞$ , since $f$ is bounded and measurable on the bounded interval $[0,1]$ . Therefore, the right-hand side of (3) tends to zero as $n$ tends to infinity.

Since the left-hand side of (3) is independent of $n$ , we have thus shown that $∫ f φ = 0$ for every $φ ∈ C[0,1]$ .

Now, since $C[0,1]$ is dense in $L_1[0,1]$ , let $\{φ_n\} ⊆ C[0,1]$ satisfy $\|φ_n - f\|_1 → 0$ as $n → ∞$ . Then

$0 ≤ ∫ f^2 = \left| ∫ f (f-φ_n + φ_n) \right| ≤ ∫ |f||f-φ_n| + \left| ∫ f φ_n\right|.$

Therefore, if $M$ is the bound on $|f|$ , we have $0≤ ∫ f^2 ≤ M \|f - φ_n\|_1 → 0$ .

As $∫ f^2$ is independent of $n$ , we have $∫ f^2 = 0$ . Since $f^2≥ 0$ , this implies $f^2=0$ a.e., hence $f=0$ a.e.

Alternative **Solution** Quinn Culver suggests shortening the proof by using the fact that polynomials are dense in $L_1[0,1]$ .

Simply start from the line, “Now, since $C[0,1]$ is dense in $L_1[0,1]$ , let $\{φ_n\} ⊆ C[0,1]$ satisfy…” but instead write,

“Since $\mathrm{Pol}[0,1]$ is dense in $L_1[0,1]$ , let $\{φ_n\} ⊆ \mathrm{Pol}[0,1]$ satisfy…”

This is a nice observation and dispenses with the problem more quickly and efficiently. However, we leave the original proof intact because it shows how one applies the Stone-Weierstrass theorem, which is a theorem students ought to learn how to use.

We will assume that $μ$ and $ν$ are finite positive measures on the measurable space $(X, 𝔐)$ . 5

By the linearity of the integral and since $μ E = ∫_E \, dμ$ , we have

(4)¶

$∫_E (1-f) \, dμ = ∫_E \, dμ - ∫_E f \, dμ = μ E - ∫_E f \, dμ.$

$∫_E (1-f) \, dμ = ∫_E f \, dν.$

$μ E - ∫_E f \, dμ = ∫_E f \, dν.$

Define $f = \frac{dμ}{d(μ+ν)}$ . By the Radon-Nikodym theorem,

(5)¶

$μ E = ∫_E \, dμ = ∫_E f_λ \, dλ,$

where $λ$ is a measure with respect to which $μ$ is absolutely continuous ( $μ ≪ λ$ ) and $f_λ := dμ/dλ$ is the (unique) Radon-Nikodym derivative.

Taking $λ = μ + ν$ we have $μ ≪ μ + ν$ when $μ$ and $ν$ are positive, and setting $f := \frac{dμ}{d(μ+ν)}$ we have, by (5),

(6)¶

$μ E = ∫_E f\, d(μ + ν) = ∫_Ef\, dμ + ∫_E f\, dν.$

Also, $0 ≤ f = \frac{dμ}{d(μ+ν)} ≤ 1$ when $μ$ and $ν$ are positive. From this and the fact that $μ$ is finite, we obtain

$\left|∫_E f\, dμ \right| < ∞.$

Therefore, $∫_Ef\, dμ$ is finite so we can subtract it from both sides of (6) obtaining

$μ E - ∫_Ef\, dμ = ∫_E f\, dν,$

which is equivalent to the goal: $∫_E (1-f)\, dμ = ∫_E f\, dν$ .

To show $F(x,y) = f(x) g(y)$ is integrable, an important but often overlooked first step is to prove that $F(x,y) = f(x) g(y)$ is $(𝔐 ⊗ 𝔑)$ -measurable.

Define $Ψ: X × Y → ℝ × ℝ$ by $Ψ(x,y) = (f(x),g(y))$ , and let $Φ: ℝ × ℝ → ℝ$ be the continuous function $Φ(s,t) = s t$ . Then,

$F(x,y) = f(x) g(y) = (Φ ∘ Ψ)(x,y).$

Therefore, if we can show that $Ψ(x,y)$ is an $(𝔐 ⊗ 𝔑)$ -measurable function from $X × Y$ into $ℝ × ℝ$ , then it will follow that $F(x,y)$ is $(𝔐 ⊗ 𝔑)$ -measurable.

To show $Ψ$ is measurable, let $R$ be an open rectangle in $ℝ × ℝ$ . Then $R = A × B$ for some open sets $A$ and $B$ in $ℝ$ , and

$\begin{split}Ψ^{-1}(R) &= Ψ^{-1}(A × B)\\ &= \{(x,y) : f(x) ∈ A, \, g(y) ∈ B\}\\ &= \{(x,y): f(x) ∈ A\} ∩ \{(x,y): g(y) ∈ B\}\\ &= (f^{-1}(A) × Y) ∩ (X × g^{-1}(B))\\ &= f^{-1}(A) × g^{-1}(B).\end{split}$

Now, $f^{-1}(A)∈ 𝔐$ and $g^{-1}(B)∈ 𝔑$ , since $f$ and $g$ are $𝔐$ - and $𝔑$ -measurable, resp. Therefore, $\Psi^{-1}(R)\in 𝔐 ⊗ 𝔑$ , which proves the claim.

Now that we know $F(x,y) = f(x) g(y)$ is $(𝔐 ⊗ 𝔑)$ -measurable, we can apply part b. of Fubini’s theorem to prove that $F(x,y) = f(x) g(y)$ is integrable if one of the iterated integrals of $|F(x,y)|$ is finite. Indeed,

$\begin{split}∫_X ∫_Y |f(x) g(y)|\, dν(y)\, dμ(x) &=∫_X ∫_Y |f(x)| |g(y)|\, dν(y)\, dμ(x) \\ &=∫_X |f(x)| \left(∫_Y |g(y)| \, dν(y)\right) \, dμ(x)\\ &=∫_X |f(x)|\, dμ(x) ∫_Y |g(y)|\, dν(y) < ∞,\end{split}$

which holds since $f ∈ L_1(μ)$ and $g ∈ L_1(ν)$ .

Therefore, Fubini’s theorem implies that $F(x,y) ∈ L_1(μ × ν)$ .

Finally, we must prove that $∫ F\,d(μ × ν) = ∫ f \, dμ ∫ g \, dν$ . Since $F(x,y) ∈ L_1(μ × ν)$ , Part 3 of Fubini’s theorem asserts that $φ(x) = ∫_Y F(x,y)\, dν(y)$ is defined almost everywhere, belongs to $L_1(μ)$ ; moreover,

$∫_{X × Y} F\, d(μ × ν) = ∫_X ∫_Y F(x,y)\, dν(y) \, dμ(x).$

$\begin{split}∫_{X × Y} F\, d(μ × ν) &= ∫_X ∫_Y f(x) g(y)\, dν(y) \, dμ(x)\\ &= ∫_X f(x) ∫_Y g(y)\, dν(y) \, dμ(x)\\ &= ∫_X f(x) \, dμ(x) ∫_Y g(y)\, dν(y).\end{split}$

Footnotes

1: Answer: $f$ is measurable and $x→ |x|$ is continuous, so $g = |f|$ is measurable. Therefore, $A_n = g^{-1}([1/n,n))$ is measurable (see Theorem Ru.1.7).
2: Alternatively, we could have cited the dominated convergence theorem here since $f_n(x) \leq |f(x)| \; (x\in X; n=1,2,\ldots)$ .
4: For what follows we only need that $ν$ is finite if $f$ is integrable, but the converse is also true.
5: Note that by a “positive” measure we really mean a nonnegative measure; that is, $μ$ is a “positive” iff $μ$ is a measure satsifying $μ E ≥ 0$ for all $E ∈ 𝔐$ .
6: See also Rudin [Rud87] Thm 1.7.