Exam 8 (57–63)¶

2004 Apr¶

Instructions Use a separate sheet of paper for each new problem. Do as many problems as you can. Complete solutions to five problems will be considered as an excellent performance. Be advised that a few complete and well written solutions will count more than several partial solutions.

Notation: \(f ∈ C(X)\) means that \(f\) is a real-valued, continuous function defined on \(X\).

Problem 57

Let \(S\) be a Lebesgue measurable set in \(ℝ\) and let \(f, g: S → ℝ\) be measurable functions. Prove
1. \(f+g\) is measurable and
2. if \(φ ∈ C(ℝ),\) then \(φ(f)\) is measurable.
Let \(f: [a,b] → [-∞, ∞]\) be a measurable function. Suppose that \(f\) takes the value \(± ∞\) only on a set of Lebesgue measure zero. Prove that for every \(ε > 0\) there is a positive number \(M\) such that \(|f| ≤ M,\) except on a set of measure less than \(ε\).

Problem 58

State Egoroff’s theorem.
State Fatou’s lemma.
Let \(\{f_n\} ⊂ L_p[0,1],\) where \(1 ≤ p < ∞\). Suppose that \(f_n → f\) a.e., where \(f ∈ L_p[0,1]\). Prove that \(\|f_n - f\|_p → 0\) if and only if \(\|f_n\|_p → \|f\|_p\).

Problem 59

Let \(S = [0,1]\) and let \(\{f_n\} ⊂ L_p(S),\) where \(1< p< ∞\). Suppose that \(f_n → f\) a.e. on \(S,\) where \(f ∈ L_p(S)\). If there is a constant \(M\) such that \(\|f_n\|_p ≤ M\) for all \(n,\) prove that for each \(g ∈ L_q(S), \frac{1}{p} + \frac{1}{q} = 1,\) we have

\[\lim_{n\to \infty}\int_S f_n g = \int_S fg.\]
Show by means of an example that this result is false for \(p=1\).

Problem 60

State and prove the closed graph theorem.

Problem 61

Prove or disprove:

For \(1≤ p< ∞\), let \(ℓ_p := \{x = \{x_k\} ∣ \|x\|_p = (∑_{k=1}^∞ |x_k|^p)^{1/p} < ∞\}\). Then for \(p ≠ 2,\) \(ℓ_p\) is a Hilbert space.
Let \(X = (C[0,1], \| \cdot \|_1),\) where the linear space \(C[0,1]\) is endowed with the \(L_1\)-norm: \(\|f\|_1 = ∫_0^1 |f(x)|\,dx\). Then \(X\) is a Banach space.
Every real, separable Hilbert space is isometrically isomorphic to \(ℓ_2\).

Problem 62

Give a precise statement of some version of Fubini’s theorem that is valid for nonnegative functions.
Let \(f,g ∈ L_1(ℝ)\).
1. Prove that the integral
  
  \[h(x) = ∫_ℝ f(x-t) g(t)\, dt\]
  
  exists for almost all \(x ∈ ℝ\) and that \(h ∈ L_1(ℝ)\).
2. Show that \(\|h\|_1 ≤ \|f\|_1 \|g\|_1\).

Problem 63

State the Radon-Nikodym theorem.
Let \((X, ℬ(X), μ)\) be a complete measure space, where \(μ\) is a positive measure defined on the σ-algebra \(ℬ(X)\) of Borel sets of \(X\).

Suppose \(μ X < ∞\). Let \(S\) be a closed set in \(ℝ\) and let \(f ∈ L_1(μ),\) where \(f: X → [-∞, ∞]\) is an extended real-valued function defined on \(X\).

If

\[A_E(f) = \frac{1}{μ E}∫_E f\, dμ ∈ S\]

for every \(E ∈ ℬ(X)\) with \(μ E > 0,\) prove that \(f(x) ∈ S\) for almost all \(x ∈ X\).

Proof 1. Since \(f\) and \(g\) are real measurable functions of \(S\), and since the mapping \(Φ:ℝ × ℝ → ℝ\) defined by \(Φ(x,y) = x+y\) is a continuous function, Theorem Ru.1.8 implies that the function \(f+g = Φ(f,g)\) is measurable.

If \(φ ∈ C(ℝ),\) then \(φ(f)\) is measurable by part b. of Theorem Ru.1.7. ☐

Proof 2. Let \(\{q_i\}_{i=1}^∞\) be an enumeration of the rationals. Then, for every \(α ∈ ℝ\),

\[\{x ∈ S ∣ f(x)+g(x) < α\} = ⋃_{i=1}^∞ \{x ∈ S ∣ f(x) < α - q_i\} ∩ \{x ∈ S ∣ g(x) < q_i\}.\]

Since each set on the right is measurable, and since σ-algebras are closed under countable unions and arbitrary intersections, \(\{x ∈ S ∣ f(x)+g(x) < α\}\) is a measurable set. Since \(α\) was arbitrary, \(f+g\) is a measurable function.

The function \(φ f\) is measurable if and only if for every open subset \(U\) of \(ℝ\) the set \((φ f)^{-1}(U)\) is measurable.

Let \(U\) be open in \(ℝ\). Then \(φ^{-1}(U)\) is open, since \(φ \in C(ℝ)\), and so \((φ f)^{-1}(U) = f^{-1}(φ^{-1}(U))\) is measurable, since \(f\) is measurable. Therefore, \(φ f\) is measurable. ☐
Fix \(ε > 0\). For \(n ∈ ℕ\), define \(A_n = \{x ∈ [a,b] : |f(x)| ≤ n\}\). Then

(39)¶\[[a,b] = ⋃_{n=1}^{∞} A_n ∪ A_{∞},\]

where \(A_∞ := \{x: f(x) = ± ∞\}\). 1

Also, \(A_1⊆ A_2 ⊆ \cdots\), and since \(f\) is measurable each \(A_n\) is measurable.

Therefore, \(μ A_n → μ(⋃_n A_n),\) as \(n → ∞\). Note that all sets are contained in \([a,b]\) and thus have finite measure.

Let \(M ∈ ℕ\) be such that \(μ(⋃_n A_n) - μ A_M < ε\). Then \(|f| ≤ M\) except on \([a,b]- A_M,\) and by (39),

\[\begin{split}μ([a,b]- A_M) &= μ(⋃_n A_n ∪ A_∞ - A_M)\\ &≤ μ(⋃_n A_n - A_M) + μ A_∞ \\ &= μ(⋃_n A_n - A_M) < ε.\end{split}\]

The second equality holds since we assumed \(f(x) = ± ∞\) only on a set of measure zero; i.e.,\(μ A_∞ = 0.\) ☐

See Egoroff’s theorem.
See Fatou’s lemma.
(⇒) By Minkowski’s inequality,

\[\|f_n\|_p = \|f_n-f + f\|_p ≤ \|f_n-f\|_p + \|f\|_p.\]

Similarly, \(\|f\|_p ≤ \|f_n-f\|_p + \|f_n\|_p\).

Together, the two inequalities yield \(\bigl| \|f_n\|_p-\|f\|_p \bigr| ≤ \|f_n-f\|_p\).

Therefore, \(\|f_n-f\|_p → 0\) implies:math:bigl| |f_n|_p-|f|_p bigr| → 0. This proves necessity.

(⇐) There are at least three proofs of sufficiency.

The second is similar to the first, only much shorter as it exploits the full power of the general version of Lebesgue’s dominated convergence theorem, whereas the first proof merely relies on Fatou’s lemma. 2

The third proof uses Fatou’s lemma and Egoroff’s theorem. 3

Note that none of the proofs uses the assumption that the measure space is finite, so we may as well work in the more general space \(L_p(X, 𝔐, μ)\). Both Proofs 1 and 2 make use of the following basic fact:

Lemma. If \(α, β ∈ [0,∞)\) and \(1 ≤ p < ∞\), then \((α + β)^p ≤ 2^{p-1}(α^p + β^p)\).

When \(p ≥ 1\), \(φ x = x^p\) is convex on \([0, ∞)\). Thus, for all \(α, β ∈ [0,∞)\),

\[\left(\frac{α+β}{2}\right)^p = φ\left(\frac{α+β}{2}\right) ≤ \frac{1}{2}[φ(α) + φ(β)] = \frac{1}{2}(α^p + β^p).\]

When \(α, β \in ℝ\), the triangle inequality followed by the lemma yields

(40)¶\[|α - β|^p ≤ \bigl||α| + |\beta|\bigr|^p ≤ 2^{p-1}\bigl(|α|^p + |β|^p\bigr).\]

—

Proof 1. By (40), \(|f_n - f|^p ≤ 2^{p-1}(|f_n|^p + |f|^p)\). In particular, \(f_n - f ∈ L_p\), for each \(n ∈ ℕ\). Moreover, the functions

(41)¶\[g_n = 2^{p-1}(|f_n|^p + |f|^p) - |f_n - f|^p.\]

are nonnegative.

Now notice that \(\lim \inf g_n = 2^p |f|^p\). Applying Fatou’s lemma to (41), then,

\[∫ 2^p|f|^p = ∫ \lim \inf g_n ≤ \lim \inf ∫ g_n = \lim \inf ∫ \bigl\{2^{p-1}(|f_n|^p + |f|^p) - |f_n - f|^p\bigr\}.\]

Since \(\|f_n\|_p → \|f\|_p\), this implies

\[2^p ∫ |f|^p ≤ 2^p ∫ |f|^p - \lim \sup ∫ |f_n - f|^p.\]

Equivalently \(0 ≤ - \lim \sup ∫ |f_n - f|^p\). This proves \(\|f_n - f\|^p → 0\). ☐

—

Proof 2. By (40),

(42)¶\[|f_n - f|^p ≤ 2^{p-1}(|f_n|^p + |f|^p).\]

In particular, \(f_n - f ∈ L_p\), for each \(n ∈ ℕ\).

Define the functions

\[g_n = 2^{p-1}(|f_n|^p + |f|^p) \quad \text{ and } \quad g = 2^p|f|^p.\]

Then \(g_n → g\) a.e., and \(\|f_n\|_p → \|f\|_p\) implies \(∫ g_n \to ∫ g\).

Also, \(g_n ≥ |f_n - f|^p → 0\) a.e., by (42). Therefore, the dominated convergence theorem implies \(∫ |f_n - f|^p → 0\). ☐

—

Proof 3. Since \(f ∈ L_p\), for all \(ε > 0\), there is a number \(δ >0\) and a set \(B ∈ 𝔐\) of finite measure such that \(f\) is bounded on \(B\), \(∫_{X - B}|f|^p \, dμ < ε/2\), and \(∫_{E}|f|^p \, d\mu < ε/2\), for all \(E ∈ 𝔐\) with \(μ E < δ\).

By Egoroff’s theorem, there is a set \(A ⊆ B\) such that \(μ (B - A) < δ\) and \(f_n → f\) uniformly on \(A\). Therefore,

(43)¶\[\begin{split}\int_X |f|^p &= \int_{X\setminus B} |f|^p + \int_{B\setminus A} |f|^p + \int_{A} |f|^p \\ &< \epsilon/2 + \epsilon/2 + \int_{A} |f|^p\\ &\leq \epsilon + \varliminf \int_{A} |f_n|^p,\end{split}\]

since, by Fatou’s lemma, \(\int_{A} |f|^p = \int_{A} \varliminf |f_n|^p \leq \varliminf \int_{A} |f_n|^p\). By hypothesis, \(\|f_n\|_p \rightarrow \|f\|_p\). Therefore,

\[\varliminf \int_{A} |f_n|^p = \varliminf \left\{\int_{X} |f_n|^p - \int_{X\setminus A} |f_n|^p\right\} =\int_{X} |f|^p - \varlimsup \int_{X\setminus A} |f_n|^p.\]

By (43), then,

\[\int_{X} |f|^p < \epsilon + \int_{X} |f|^p - \varlimsup \int_{X\setminus A} |f_n|^p.\]

Therefore, \(\varlimsup \int_{X\setminus A} |f_n|^p < \epsilon\) (since \(f\in L_p\)). Finally, note that

\[\begin{split}\|f_n-f\|^p &= \|(f_n - f) χ_A + (f_n - f) χ_{X-A}\|_p &\\ &≤ \|(f_n - f) χ_A \|_p + \|(f_n - f) χ_{X- A}\|_p & (∵ \text{Minkowsky})\\ &≤ \|(f_n - f) χ_A \|_p + \|f_nχ_{X- A}\|_p + \|f χ_{X-A}\|_p. &\end{split}\]

Therefore, \(\varlimsup \|f_n-f\|^p\) is bounded above by

\[\varlimsup\{|f_n(x) - f(x)|: x∈ A\} μ(A)^{1/p} + \varlimsup \left(∫_{X-A} |f_n|^p\right)^{1/p} + \left(∫_{X-A} |f|^p\right)^{1/p}.\]

The first term on the right goes to zero since \(f_n → f\) uniformly on \(A\). The other terms are bounded by \(2 ε^{1/p}\). ☐

Since \(g\in L_q(S)\), for all \(\epsilon > 0\) there exists \(\delta>0\) such that if \(A\) is a measurable set with \(\mu A < \delta\) then

\[\int_A |g|^q \, d\mu < \epsilon.\]

Let \(B_0\subset S\) denote the set on which \(f_n\) does not converge to \(f\). Let \(B\subseteq S\) be such that \(f_n \rightarrow f\) uniformly in \(S\setminus B\), and such that \(\mu B< \delta\). (Such a set exists by Egoroff’s theorem since \(\mu S < \infty\).) Now throw the set \(B_0\) in with \(B\) (i.e. redefine \(B\) to be \(B\cup B_0\)). Then,

\[\begin{split}D_n &:= \int_S |f_n g - f g | \, d\mu = \int_S |f_n - f| |g | \, d\mu \\ &=\int_{S\setminus B} |f_n - f| |g | \, d\mu + \int_{B} |f_n - f| |g | \, d\mu \\ &\leq \| (f_n - f) \chi_{S\setminus B}\|_p \|g\|_q + \|f_n - f\|_p \|g \chi_B\|_q.\end{split}\]

The final inequality holds because \(f_n, f \in L_p\) implies \(|f_n - f| \in L_p\) and by Hölder’s inequality.

By Minkowski’s inequality, \(\|f_n - f\|_p \leq \|f_n\|_p + \|f\|_p\), so

\[D_n \leq \sup\{| f_n(x) - f(x)| : x\notin B\}\mu(S\setminus B)^{1/p} \|g\|_q + (\|f_n\|_p + \|f\|_p)\left(\int_B |g|^q\, d\mu\right)^{1/q}.\]

Now we are free to choose the \(\delta>0\) above so that

\[\left(\int_B |g|^q\, d\mu\right)^{1/q} < \frac{\epsilon}{2(M+\|f\|_p)}\]

holds whenever \(\mu B < \delta\). Also, \(f_n \rightarrow f\) uniformly on \(S\setminus B\), so let \(N\) be such that

\[\sup\{|f_N(x) - f(x)| : x ∉ B\} < \frac{ε}{2μ(S-B)^{1/p}\|g\|_q}.\]

Then \(D_N < ε/2 + ε/2\). ☐
The result is not true for \(p=1\) as the following example shows.

Let \(f_n = n χ_{[0,1/n]}, \, n=1, 2, \dots\).

First note that \(f_n → 0\) a.e. For if \(x ∈ (0,1]\), then there exists \(N>0\) such that \(1/N < x\) and thus \(f_n(x) = 0\) for all \(n≥ N\). Therefore, \(\{x∈ [0,1] : f_n(x) ↛ 0 \} = \{0\}\).

Now, if \(g\) is the constant function \(g = 1\), then \(∫ f_n g \, dμ = 1\) for all \(n=1, 2, \dots\). Therefore, \(∫ f_n g \, dμ → 1\), while \(∫ fg \, dμ = ∫ 0g\, dμ = 0\).

Finally, note that \(\|f_n\|_1 = n \, μ [0,\frac{1}{n}] = 1\) for \(n=1, 2, \dots\), so \(\{f_n\}\) satisfies the hypothesis \(\|f_n\|_1 ≤\) some constant \(M\). ☐

The Closed Graph Theorem If \(X\) and \(Y\) are Banach spaces and \(T : X \to Y\) is a linear mapping, then the graph \(Γ(T) := \{(x,y) ∈ X × Y: y = T x\}\) is closed if and only if \(T\) is bounded.

Proof. (⇒) Suppose \(\overline{Γ(T)} ⊆ Γ(T)\) (i.e., \(Γ(T)\) is closed).

Then, since \(T\) is linear, \(T(X)\) is a subspace of \(Y\), and \(\Gamma(T) = X \times T(X)\) is a subspace of the Banach space \(X \oplus Y\).

Endowed with the (usual) norm, \(\|(x, Tx)\| = \|x\| + \|Tx\|\), and the subspace \(\Gamma(T) = X \oplus T(X)\) is a Banach space (since we assume \(\Gamma(T)\) is closed, so \((x_n, Tx_n) \to (x, Tx) \in \Gamma(T)\)).

Consider the continuous projection mappings, \(\pi_1, \pi_2\), defined on \(\Gamma(T)\) by \(\pi_1(x, Tx) = x\) and \(\pi_2(x, Tx) = Tx\).

Then \(\pi_1\) is a continuous bijection of \(\Gamma(X)\) onto \(X\). Therefore, by the inverse mapping theorem, \(\pi^{-1}\) is continuous.

Since \(\pi_1\colon \Gamma(T) \to X\) and \(\pi_2\colon \Gamma(T) \to T(X)\) and \(\pi_1^{-1}\colon X \to \Gamma(T)\), we can write \(T = \pi_2 \circ \pi_1^{-1}\) as a mapping from \(X\) to \(\Gamma(T)\) to \(T(X) \subseteq Y\).

Since \(\pi_1^{-1}\) and \(\pi_2\) are both continuous, so is \(T\).

(⇐) Suppose \(T\) is bounded (equivalently, continuous).

Let \((x,y)\) be a limit point of \(\Gamma(T)\) in \(X \oplus Y\). That is, \(\exists \{(x_n, Tx_n)\} \subseteq X \times Y\) such that \(\|(x_n, T x_n) - (x,y)\| \to 0\).

To see that \((x,y) \in \Gamma(T)\) (so \(\Gamma(T)\) is closed), observe that \(x_n \to x\) implies \(T x_n \to T x\), , since \(T\) is continuous.

Therefore, \(T x = y\), so \((x,y) = (x, Tx) \in \Gamma(T)\), as desired.

First we recall some well-known facts about normed linear spaces.

Fact 1. If \(X^∗\) is separable, then \(X\) is separable.

Fact 2. \(ℓ_1\) is a separable.

Fact 3. \(ℓ_∞\) is not separable.

Fact 4. \(ℓ_1^∗ ≅ ℓ_∞\).

Fact 5. If \(H\) is a separable Hilbert space, then \(H^{∗∗} ≅ H\).

We will use these facts to prove that \(ℓ_1\) is not a Hilbert space (so \(ℓ_1\) is a counterexample that disproves assertion a.)

If \(ℓ_1\) is a Hilbert space, then \(ℓ_∞^∗ ≅ ℓ_1\), by Fact 5. Therefore, \(ℓ_∞^∗\) is separable since \(ℓ_1\) is (Fact 2). But then Fact 1 implies \(ℓ_∞\) is separable, which contradicts Fact 3. Therefore, \(ℓ_∞^∗ ≇ ℓ_1\), so \(ℓ_1\) is not a Hilbert space.
Claim. \(X = (C[0,1], \| \; \|_1)\) is not a Banach space.

Proof. For \(n = 4, 5, 6, \dots\), let

\[\begin{split}f_n(x) = \begin{cases} 0, & 0≤ x < \frac{1}{4}-\frac{1}{n},\\ 1 - \frac{n}{4} + nx, & \frac{1}{4}-\frac{1}{n} ≤ x < \frac{1}{4},\\ 1, & \frac{1}{4} ≤ x < \frac{3}{4}\\ 1 + \frac{3n}{4} - nx, & \frac{3}{4}≤ x < \frac{3}{4}+\frac{1}{n}\\ 0, & \frac{3}{4}+\frac{1}{n}\leq x ≤ 1,\\ \end{cases}\end{split}\]

Thus \(f_n ∈ C[0,1]\) is constant except on the interval \(A_n := [\frac{1}{4}-\frac{1}{m}, \frac{1}{4}]\), where it is linearly increasing from 0 to 1, and the interval \(B_n := [\frac{3}{4}, \frac{3}{4}+\frac{1}{m}]\), where it is linearly decreasing from 1 to 0.

We show that \(\{f_n\}\) is a Cauchy sequence in \(X = (C[0,1], \| \; \|_1)\) that does not converge to a function in \(C[0,1]\).

For all \(n > m ≥ 4\), \(f_n\) and \(f_m\) are zero outside the interval \([\frac{1}{4}-\frac{1}{m}, \frac{3}{4}+\frac{1}{m}]\) and equal to \(1\) in the interval \([\frac{1}{4}, \frac{3}{4}]\).

So the only region where these functions may differ is \(A_m ∪ B_m\), and over this region we have \(\sup \{|f_n(x) - f_m(x)| : x ∈ A_m ∪ B_m\} ≤ 1\). Therefore,

\[\|f_n - f_m\|_1 ≤ \sup \{|f_n(x) - f_m(x)| : x ∈ A_m ∪ B_m\} ⋅ m(A_m ∪ B_m) ≤ m(A_m \cup B_m) ≤ \frac{1}{m},\]

which tends to 0 as \(m\) tends to infinity. Therefore, \(\{f_n\}\) is a Cauchy sequence in \(X\).

On the other hand, \(f_n\) converges to the characteristic function \(χ_{[1/4,3/4]}\) which is not continuous, so \(\lim_{n → ∞} f_n ∉ C[0,1]\).
We adopt the convention that the term separable applies only to infinite Hilbert spaces.

Claim. Every separable Hilbert space is isometrically isomorphic to \(\ell_2\).

Proof. Let \(ℋ\) be a separable Hilbert space and let \(\{x_n\}\) be a countable dense set in \(ℋ\).

By throwing away elements if necessary, we can assume \(\{x_n\}\) is a set of linearly independent vectors whose span is dense in \(ℋ\).

Applying the Gram-Schmidt procedure to \(\{x_n\}\), we can assume it is a set of orthonormal vectors whose span is dense in \(ℋ\).

Define the mapping \(U: ℋ → ℂ^ω\) (from \(ℋ\) to the sequence space \(ℂ^ω\)) as follows:

\[U x = \{⟨x, x_n⟩\}_{n=0}^∞\]

Then \(U\) is a unitary operator from \(ℋ\) onto \(ℓ_2\).

To see this, let \(x∈ ℋ\). We first show \(U x ∈ ℓ_2\).

Observe that \(x = ∑_{n=0}^∞ ⟨ x, x_n⟩ x_n\), so

(44)¶\[\|x\|^2_{ℋ} = ⟨ x, ∑_{n=0}^∞ ⟨ x, x_n⟩ x_n⟩= ∑_{n=0}^∞ \overline{⟨ x, x_n⟩} ⟨ x, x_n⟩ = ∑_{n=0}^∞ |⟨ x, x_n⟩| = \|U x\|_{ℓ_2}^2.\]

Since \(x ∈ ℋ\), \(\|x\|_2 < ∞\), so \(\|Ux\|_2 < ∞\). Thus, \(U x ∈ ℓ_2\) and (44) shows \(U\) is unitary.

To complete the proof, we must show that \(U\) maps \(ℋ\) onto \(ℓ_2\).

Fix \(y = \{y_i\}_{i=0}^∞ ∈ ℓ_2\). Define \(x = ∑_{i=0}^∞ y_i x_i\). Then,

\[\{⟨ x, x_n⟩\}_{i=0}^∞ = U x = U(∑_{i=0}^∞ y_i x_i) = ∑_{i=0}^∞ y_i Ux_i = \{y_i\}_{i=0}^∞.\]

The last equality holds since \(U x_i = \{⟨ x_i, x_j⟩\}_{j=0}^∞\) which is the function from \(ω\) to \(ℂ\) that takes \(i\) to 1 and every other \(j ∈ ω\) to 0.

We have thus shown that there exists \(x ∈ ℋ\) such that \(U x = y\), as desired. ☐

See Fubini’s theorem in the appendix.
(Note: It’s likely that this solution supplies many more of the technical details than most examiners would expect one to produce while sitting for a timed examination.)
1. In order to apply Fubini’s theorem to the function \(F(x,y):= f(x-y)g(y)\), we should first observe that \(F(x,y)\) is \((ℬ(ℝ) \otimes ℬ(ℝ))\)-measurable, assuming \(f\) and \(g\) are \(ℬ(ℝ)\)-measurable.
  
  The details are not hard, but it’s instructive to work through them at least once in a lifetime.
  
  The function \(φ(x,y) = x-y\) is a continuous mapping from \(ℝ^2\) to \(ℝ\), so it is a Borel measurable function.
  
  Consider the composition \((f ∘ φ)(x,y)= f(x-y)\). If \(U⊆ ℝ\) is open, then \((f ∘ φ)^{-1}(U)= φ^{-1}f^{-1}(U)\).
  
  Assume for now that \(f\) is a Borel function, so \(f^{-1}(U) ∈ ℬ(ℝ)\).
  
  Let \(Ω := \{V ⊆ ℝ : φ^{-1}(V) ∈ ℬ(ℝ) ⊗ ℬ(ℝ)\}\).
  
  Clearly \(Ω\) contains the open sets. In fact, it’s not hard to show that \(Ω\) is a σ-algebra, so it must contain \(ℬ(ℝ)\).
  
  Therefore, \(f^{-1}(U) ∈ Ω\) for every open set \(U ⊆ ℝ\).
  
  We have thus observed that if \(f: ℝ → ℝ\) is Borel and \(φ: ℝ^2 → ℝ\) is continuous, then \(f ∘ φ\) is Borel.
  
  Next, let \(g: ℝ → ℝ\) be a function and note that \(g(y) = (g ∘ π_2)(x,y)\), where \(π_2(x,y) = y\) is the (continuous) second projection.
  
  Thus, if \(g\) is Borel, then \(g ∘ π_2: ℝ^2 → ℝ\) is Borel.
  
  Finally, we note that \(f(x-y)g(y)\) is the product \(f(x-y)g(y) = (f ∘ φ)(x,y) (g ∘ π_2)(x,y)\) of two Borel functions, which is also Borel, as expected.
  
  The final technical detail needed to make this observation useful in the present context is the fact that for every Lebesgue measurable function \(f\) there exists a Borel function \(f̃\) such that \(f̃ = f\) a.e. (See, e.g., Chapter 9 of Rudin [Rud87].)
  
  So we assume without loss of generality that the functions involved in this problem are Borel.
  
  Now that we have completed the tedious task of showing that \(f(x-y)g(y)\) is a measurable function on \(ℝ^2\), we can apply Fubini’s theorem, as follows:
  
  (45)¶\[\begin{split}∬ |f(x-y) g(y)| \, dx\, dy &= ∫ |g(y)| \left(∫ |f(x-y)|\, dx\right)\, dy &\\ &= ∫ |g(y)| \left(∫ |f(x)|\, dx\right)\, dy & (∵ \text{ invariance of Leb. meas.}) \\ &= ∫ |g(y)| \, dy ∫ |f(x)|\, dx < ∞ & (∵ f, g ∈ L_1(ℝ)).\end{split}\]
  
  Therefore, Fubini’s theorem part b. implies \(f(x-y)g(y) ∈ L_1(dx × dy)\), so Fubini’s theorem (Part c) implies
  
  \[F(x_0,y) = f(x_0-y) g(y) ∈ L_1(dy) \text{ for almost every } x_0 ∈ ℝ,\]
  
  and
  
  \[h(x) = ∫ f(x-t) g(t) \, dt ∈ L_1(dx).\]
2. Having proved that \(h\) exists for almost every \(x ∈ ℝ\), we now we show \(\|h\|_1 ≤ \|f\|_1 \|g\|_1\).
  
  This follows immediately from Fubini’s theorem (Part a). Indeed,
  
  \[\begin{split}∫ |h| \, dm &= ∫ \left| ∫ f(x-y) g(y)\, dy\right|\, dx \\ & ≤ ∬ |f(x-y) g(y)| \, dy \, dx\\ & = ∬ | f(x-y) g(y)|\, dx\, dy\\ & = \|f\|_1 \|g\|_1,\end{split}\]
  
  where the last equality follows from (45).

See the Radon-Nikodym theorem in the appendix of theorems.
(coming soon)

Footnotes

1: Since \(f\) is an extended real valued function, we must not forget to include \(A_\infty\), without which the union in (39) would not be all of \([a,b]\).
2: Disclaimer: you should check the first proof carefully for yourself and decide whether you believe it.
3: Judging from Parts a and b of this problem, the third proof may be closer to what the examiners had in mind.