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Motivation:

  • I have seen the interchange of limit/derivative and integral many times, but don't know how such operation makes sense.

  • I've always desired to remove this uncertainty by giving a detailed proof in which any argument is justified. In doing so, I've searched on the Internet and found that most proofs are either too long and need so many between lemmas (in this case, I feel that the main ideas are hidden), or omit important details (at least details not obvious for me).

  • Then I come across these $2$ lectures (here and here). I basically copy the whole argument in the former for the measurability part. For the part of repeated integral, I'm proud that I do not copy but adapt the idea in the latter lecture (which is for complete measure) to my case (which is $\sigma$-finite measure).

  • It takes me $3$ days to complete the proof. I'm very happy because I now understand the machinery that makes the theorem right. This effort also solidifies my understanding of monotone and dominated convergence theorems.

My question: My proof is very clear and detailed, so it's easy to follow. I hope somebody will take time help me verify it and point out anything unclear or incorrect.

Thank you so much for your help!


Related definitions of Bochner integrals can be found here. Let

  • $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be $\sigma$-finite measure spaces.

  • $\Sigma := \mathcal A \times \mathcal B$.

  • $\mathcal C := \mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$.

  • $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$.

  • $\mathcal S (X \times Y, \lambda, {\mathbb R}^+)$ the space of $\lambda$-simple functions from $X \times Y$ to ${\mathbb R}^+$.

  • $\mathcal L_0 (X \times Y, \lambda, \overline{\mathbb R}{}^+)$ the space of $\nu$-measurable functions $X \times Y$ to $\overline{\mathbb R}{}^+$.

  • $\mathcal L_1 (X \times Y, \lambda, \overline{\mathbb R}{}^+)$ the space of $\nu$-integrable functions $X \times Y$ to $\overline{\mathbb R}{}^+$.

Tonelli's theorem: Let

  • $f: X \times Y \to \overline{\mathbb R}{}^+$ measurable.

  • $f_x: Y \to \mathbb R, \, y \mapsto f(x, y)$ for all $x \in X$.

  • $f_y: X \to \mathbb R, \, x \mapsto f(x, y)$ for all $y \in Y$.

Then

(i) The maps $$\phi: X \ni x \mapsto \int_Y f_x \, \mathrm d \nu \quad \text{and} \quad \psi: Y \ni y \mapsto \int_X f_y \, \mathrm d \mu $$ are measurable.

(ii) $$\int_X \phi \, \mathrm d \mu = \int_{X \times Y} f \, \mathrm d \lambda = \int_Y \psi \, \mathrm d \nu.$$


Proof:

Lemma: Let $x \in X$ and $G \in \mathcal C$. The $x$-section $G_x$ defined by $G_x := \{y \in Y \mid (x,y) \in G\}$ is measurable.

Proof: Let $\mathcal D_x := \{G \in \mathcal C \mid G_x \text{ is measurable}\}$. Notice that $(G^c)_x =(G_x)^c$ and $(\bigcap_n G_n)_x = \bigcap_n (G_n)_x$. So $\mathcal D_x$ is a $\sigma$-algebra over $X \times Y$. For $A \times B \in \Sigma$, $(A \times B)_x = B$ if $x \in A$ and $\emptyset$ otherwise. So $\Sigma \subseteq \mathcal D_x$ and thus $\mathcal D_x = \mathcal C$.

We fix a non-decreasing sequence $(Y_n)$ of sets in $\mathcal B$ such that $\bigcup Y_n = Y$ and $\nu(Y_n) < \infty$. By symmetry, it's sufficient to prove for $\phi$.

We proceed to prove (i) for $f =1_G$ with $G \in \mathcal C$. We write $\phi_G$ instead of $\phi$. Then $\phi_G(x) = \int_Y 1_{G} (x,y) \mathrm d \nu(y) = \int_Y 1_{G_x} \mathrm d \nu = \nu(G_x)$, which is well-defined by our Lemma. Let $\mathcal D := \{G \in \mathcal C \mid \phi_G \text{ is measurable}\}$.

  • First, we assume $\nu$ is finite. If $G \in \mathcal D$, then $\phi_{G^\mathrm c} = \nu(Y) -\phi_G$ is measurable because $\nu(Y)<\infty$ and $\phi_G$ is measurable. If $(G_n)$ is a sequence of pairwise disjoint (p.w.d.) sets in $\mathcal D$, then $\phi_{\bigcup G_n} = \sum_n \phi_{G_n}$ is measurable. As a result, $\mathcal D$ is indeed a $\sigma$-algebra. If $G= A\times B \in \Sigma$, then $\phi_G = \nu(B) 1_A$ which is simple and thus measurable. Hence $\Sigma \subseteq \mathcal D$ and thus $\mathcal D = \mathcal C$.

  • Second, we assume $\nu$ is $\sigma$-finite. We define a sequence $(\nu_n)$ of finite measures on $Y$ by $\nu_n(B) := \nu(B \cap Y_n)$ for $B \in \mathcal B$. Let $\phi_{G,n}(x) := \nu_n(G_x)$ and $\mathcal D_n := \{G \in \mathcal C \mid \phi_{G, n} \text{ is measurable}\}$. As in the case $\nu$ is finite, we can prove that $\mathcal D_n = \mathcal C$. We have $G_x \cap Y_n \nearrow G_x \cap Y = G_x$, so $\nu_n (G_x) =\nu(G_x \cap Y_n) \nearrow \nu(G_x)$ by continuity from below of measure. This means $\phi_{G,n} \nearrow \phi_G$ and thus $\phi_{G}$ is measurable for $G \in \mathcal C$. Hence $\mathcal D = \mathcal C$.

By linearity of integrals, (i) also holds for $\mathcal S (X \times Y, \lambda, \mathbb R^+)$. If $f \in \mathcal L_0 (X \times Y, \lambda, \overline{\mathbb R}{}^+)$, then there is a non-decreasing sequence of $(f_n)$ in $\mathcal S (X \times Y, \lambda, \mathbb R^+)$ such that $f_n \to f$. It follows that (i) also holds for $f \in \mathcal L_0 (X \times Y, \lambda, \overline{\mathbb R}{}^+)$.

We proceed to prove (ii) for $f =1_G$ with $G=A \times B \in \Sigma$. As shown above, $\phi_G = \nu(B) 1_A$, so $\int_X \phi_{G} \, \mathrm d \mu = \mu(A)\nu(B)$. On the other hand, $\int_{X \times Y} 1_{G} \, \mathrm d \lambda = \lambda(A\times B) = \mu(A) \nu(B)$ by construction of product measure $\lambda$. Hence (ii) holds for this kind of $f$.

We proceed to prove (ii) for $f =1_G$ with $G = \bigcup_k A_k \times B_k$ with $A_k \times B_k \in \Sigma$. By this result, we can assume that $(A_k \times B_k)$ are p.w.d., which implies $\phi_{G} = \sum_{k} \phi_{A_k \times B_k}$ and $1_{G} = \sum_{k} 1_{A_k \times B_k}$. Clearly, $\sum_{k=1}^n \phi_{A_k \times B_k} \nearrow \phi_G$ and $\sum_{k=1}^n 1_{A_k \times B_k} \nearrow 1_{G}$ as $n \to \infty$. As such, \begin{align} \int_X \phi_{G} \mathrm d \mu &= \int_X \lim_n \sum_{k=1}^n \phi_{A_k \times B_k} \mathrm d \mu &&\overset{(1)}{=} \lim_n \int_X \sum_{k=1}^n \phi_{A_k \times B_k} \mathrm d \mu \\ &\overset{(2)}{=} \lim_n \sum_{k=1}^n \int_{X} \phi_{A_k \times B_k} \mathrm d \mu &&\overset{}{=} \lim_n \sum_{k=1}^n \int_{X \times Y} 1_{A_k \times B_k} \mathrm d \lambda \\ &\overset{(3)}{=} \lim_n \int_{X \times Y} \sum_{k=1}^n 1_{A_k \times B_k} \mathrm d \lambda &&\overset{(4)}{=} \int_{X \times Y} \lim_n \sum_{k=1}^n 1_{A_k \times B_k} \mathrm d \lambda\\ &= \int_{X \times Y} 1_G \mathrm d \lambda. \end{align}

Here we apply monotone convergence theorem (m.c.t) to obtain $(1)$ and $(4)$. The interchange of finite sum and integral in $(2)$ and $(3)$ is valid [even if all individual integrals are $\infty$] because all individual functions are non-negative. Hence (ii) also holds for this type of $G$.

We proceed to prove (ii) for $f =1_G$ with $G \in \mathcal C$ and $\lambda (G) < \infty$. By construction of product outer measure and the fact that $\lambda(G) < \infty$, there exists a sequence $(G_n)$ such that $G_n = \bigcup_m A^{(n)}_m \times B^{(n)}_m \supseteq G$, $A^{(n)}_m \times B^{(n)}_m \in \Sigma$, and $\lambda (G_n) \searrow \lambda(G)$. Without loss of generality, we assume $\lambda(G_1) < \infty$. Let $E_k := \bigcap_{n=1}^k G_n$ and $E := \bigcap_k E_k$. Then $G \subseteq E$, $\lambda(G) = \lambda(E)$, $E_1 = G_1$, and $E_k \searrow E$. By results (a) and (b), $E_k$ can be written as countable union of p.w.d. sets in $\Sigma$.

  • As shown above, $\int_X \phi_{E_k} \mathrm d \mu = \int_{X \times Y} 1_{E_k} \mathrm d \lambda$ for all $k$. Notice that $E_x = (\bigcap_{k} E_k)_x = \bigcap_{k} (E_k)_x$ implies $(E_k)_x \searrow E_x$. By continuity from above of measure, $\phi_{E_k} = \nu((E_k)_x) \searrow \nu(E_x) =\phi_{E}$ and $1_{E_k} \searrow 1_E$. We get \begin{align} \int_X \phi_{E} \mathrm d \mu &= \int_X \lim_k \phi_{E_k} \mathrm d \mu &&\overset{(5)}{=} \lim_k \int_X \phi_{E_k} \mathrm d \mu \\ &= \lim_k \int_{X \times Y} 1_{E_k} \mathrm d \lambda &&\overset{(6)}{=} \int_{X \times Y} \lim_k 1_{E_k} \mathrm d \lambda \\ &= \int_{X \times Y} 1_{E} \mathrm d \lambda &&= \lambda(E). \end{align}

  • It follows from $E_k \subseteq E_1$ that $0 \le \phi_{E_k} \le \phi_{E_1}$ and $0 \le 1_{E_k} \le 1_{E_1}$ for all $k$. Also, $\int_X \phi_{E_1} \mathrm d \mu = \int_{X \times Y} 1_{E_1} \mathrm d \lambda = \lambda(E_1)= \lambda(G_1) < \infty$. Here we apply dominated convergence theorem (d.c.t.) to obtain $(5)$ and $(6)$.

  • Let $F := E \setminus G$. We have $\lambda(E) = \lambda (G) + \lambda(F)$ and $\lambda(E) = \lambda(G) < \infty$, so $\lambda (F) =0$. With the same reasoning, there is a sequence $H_1 \supseteq H_{2} \supseteq \cdots \supseteq H_k \supseteq \cdots$ such that $H_k$ is a countable union of p.w.d sets in $\Sigma$ and that $H := \bigcap_k H_k \supseteq F$, $\lambda(F) = \lambda (H)=0$, $\phi_{H_k} \searrow \phi_H$, and $1_{H_k} \searrow 1_H$. By twice applications of d.c.t again, we get $\int_X \phi_{H} \mathrm d \mu = \int_{X \times Y} 1_{H} \mathrm d \lambda = \lambda(H) = 0$. This means $\nu(H_x) =\phi_H (x) = 0$ for $\mu$-a.e $x \in X$. On the other hand, $F_x \subseteq H_x$ is measurable by our Lemma, so $\phi_F (x) =\nu (F_x)=0$ for $\mu$-a.e $x \in X$. Hence $\int_X \phi_{F} \mathrm d \mu =0$.

  • It follows from $G \cup F = E$ and $G \cap F =\emptyset$ that $\phi_G +\phi_F = \phi_E$ and thus $\int_X ( \phi_{G} + \phi_F) \mathrm d \mu = \int_X \phi_{E} \mathrm d \mu$. Notice that $\phi_G, \phi_F \ge0$, so $\int_X \phi_{G} \mathrm d \mu + \int_X \phi_{F} \mathrm d \mu = \int_X \phi_{E} \mathrm d \mu$ [even if both individual integrals are $\infty$] and thus $\int_X \phi_{G} \mathrm d \mu = \int_X \phi_{E} \mathrm d \mu$. It follows from $\int_X \phi_{E} \mathrm d \mu = \lambda (G)$ that (ii) holds for $G \in \mathcal C$ with $\lambda (G) < \infty$.

By linearity of integrals, (ii) also holds for $\mathcal S (X \times Y, \lambda, \mathbb R^+)$. If $f \in \mathcal L_0 (X \times Y, \lambda, \overline{\mathbb R}{}^+)$, then there is a non-decreasing sequence of $(f_n)$ in $\mathcal S (X \times Y, \lambda, \mathbb R^+)$ such that $f_n \to f$. By m.c.t., (ii) also holds for $f \in \mathcal L_0 (X \times Y, \lambda, \overline{\mathbb R}{}^+)$.

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  • $\begingroup$ Some condition must be missing from the statement of the lemma. $\endgroup$
    – Berci
    Oct 23, 2021 at 20:22
  • $\begingroup$ What do your diagonal arrows mean? I’m sure this is basic notation but I’ve not really studied the subject $\endgroup$
    – FShrike
    Oct 23, 2021 at 20:24
  • 1
    $\begingroup$ @FShrike The notation $f_n \nearrow f$ usually means $f_n$ converges pointwise to $f$ and $f_{n} \le f_{n+1}$ for all $n$, i.e., $(f_n)$ is a non-decreasing sequence. $\endgroup$
    – Akira
    Oct 23, 2021 at 20:27
  • $\begingroup$ @Berci I'm unable to figure it out. Can you elaborate on those conditions. Your explanation possibly sheds light on this related question. $\endgroup$
    – Akira
    Oct 23, 2021 at 20:29
  • $\begingroup$ I mean as it's written, there's no condition on $G\subseteq X\times Y$, but you state its $x$-sections and $G$ itself are measurable. I'm sure you wanted to put a condition. (Or maybe I misinterpreted something..) $\endgroup$
    – Berci
    Oct 24, 2021 at 7:24

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