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

Define

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

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

By (2) then,

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

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 \(g_m = ∑_{n=1}^m f χ_{E_n}\; (m=1,2,\ldots)\).

☐

(See also: Problem 29.)

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

\[\left|∫_E f \, dμ \right| < ε.\]

Proof. First consider what additional hypotheses would make the proof easier.

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.

☐

(See also Prop 4.3 of Royden [Roy88].)

We provide two proofs.

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, \(A_n = \{x ∈ X : 1/n ≤ |f(x)| < n\}\).

The second is a shorter proof that relies on a result about absolute continuity of measures that is dangerously close to the statement of the problem.

Students should learn the first proof. The second is included as it helps illuminate the connection with the analogous result about absolutely continuous measures.

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 \(n→ ∞\).

Therefore, since all functions involved are integrable,

(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

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

(The second term on the right is zero by what we proved above.)

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.

An assumption is missing here and one can prove the result is false as stated.

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)¶\[∫_E (1-f) \, dμ = ∫_E \, dμ - ∫_E f \, dμ = μ E - ∫_E f \, dμ.\]

Recall, the equation we wish to prove is

\[∫_E (1-f) \, dμ = ∫_E f \, dν.\]

By (4) then, we must prove

\[μ E - ∫_E f \, dμ = ∫_E f \, dν.\]

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: \(∫_E (1-f)\, dμ = ∫_E f\, dν\).

☐

(See also Problem 75.)

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,

\[F(x,y) = f(x) g(y) = (Φ ∘ Ψ)(x,y).\]

A basic fact about continuous and measurable functions states that a continuous function of a measurable function is measurable. 6

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,

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

Therefore,

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

☐

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.