Fubini's Theorem

Let $(X, \mathcal{F}_1, \mu_1)$ and $(Y, \mathcal{F}_2, \mu_2)$ be two $\sigma$-finite measure space.

Produce Measure Space

Consider the product space $\Omega = X \times Y$.

Definition. A measurable rectangle is a product $A \times B$ for which $A \in \mathcal{F}_1$ and $B \in \mathcal{F}_2$.

Denote the collection of measurable rectangles as

We would like to construct a measure $\mu$ on $\sigma(\mathcal{S})$, the $\sigma$-field generated by $\mathcal{S}$, thus

1. For each $A \times B \in \mathcal{S}$, it's natural to define $\mu (A \times B) = \mu_1(A) \cdot \mu_2(B)$.

2. We prove that $\mathcal{S}$ is a semialgebra.

3. We show that $\mu$ is a measure on $\mathcal{S}$.

4. Finally, it follows by the Caratheodory’s Extension Theorem, $\mu$ has a unique extension on $\sigma(\mathcal{S})$.

Lemma 1. $\mathcal{S}$ is a semialgebra.

Proof.

• $\mathcal{S}$ is close under intersection: if $A \times B, C \times D \in \mathcal{S}$, then$$(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D).$$
• The complement of element in $\mathcal{S}$ is a finite disjoint union of elements in $\mathcal{S}$: if $A \times B \in \mathcal{S}$, since$$\Omega = (A \times B) + ( \bar A \times B ) + ( A \times \bar B ) + ( \bar A \times \bar B ),$$it holds that$$\overline{ A \times B } = ( \bar A \times B ) + ( A \times \bar B ) + ( \bar A \times \bar B ).$$

$\square$

Lemma 2. $\mu$ is a measure on $\mathcal{S}$.

Proof. It suffices to prove $\sigma$-additivity of $\mu$. Suppose $A \times B = +_{i \ge 1} (A_i \times B_i)$ is a countable disjoint union of elements in $\mathcal{S}$.

Fix an $x \in A$. For each $y \in B$, we have $(x, y) \in A \times B$. As the sets $\begin{Bmatrix} A_i \times B_i \end{Bmatrix}$ consists of a partition of $A \times B$, it follows that $(x, y)$ belongs to exactly one $A_i \times B_i$. Denote

Then $B = +_{i \in I(x) } B_i$.

As $\mu_2$ is a measure on $Y$, it is countably additive:

By monotone convergence theorem,

which is

$\square$

Theorem. $\mu$ has a unique extension $\bar \mu$ on $\sigma( \mathcal{S} )$, the $\sigma$-field generated by $\mathcal{S}$.

Proof. Note that $\mathcal{S}$ is a semialgebra and $\mu$ is a $\sigma$-finite measure. Applying Caratheodory’s Extension Theorem directly finishes the proof.

$\square$

Fubini's Theorem

Let $\begin{Bmatrix} a_i : i \in \mathbb{N} \end{Bmatrix}$ be a countably sequence of numbers. Then the series converges to the same limit under permutation, if

1. $a_i \ge 0, \forall i \in \mathbb{N}$. Note that $\sum_{i \ge 1} a_i$ can equal $\infty$.

2. $\sum_{i \ge 1} | a_i | < \infty$.

Integration is a generalization of discrete sum. Given a measure space $(\Omega, \sigma( \mathcal{S} ), \bar \mu)$, and a measure function $f$, the integral of $f$ exists, if

1. $f \ge 0$;
2. Or $\int |f| d\, \bar \mu < \infty$.

In case that $(\Omega, \sigma( \mathcal{S} ), \bar \mu)$ is a produce measure space, the Fubini's theorem states that if the double integral exists, it can be computed by iterated integrals.

Theorem (Fubini's Theorem). If $f \ge 0$ or $\int |f| d\, \bar \mu < \infty$, then there exists $A \in \mathcal{F}_1$, $B \in \mathcal{F}_2$, s.t., $\mu_1 (A) = \mu_2 (B) = 0$, and

$$\int_{X - A} \int_Y f(x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } f(x, y) d\, \bar \mu = \int_{Y - B} \int_X f(x, y) d\, \mu_1 \, d\, \mu_2.$$

Remark. For the case concerning $f \ge 0$, we have $A = B = \emptyset$, and the Theorem is called Tonelli’s theorem.

Comparison with Discrete Sum

Let $X = Y = \mathbb{N}$, and $\mu_1 = \mu_2$ be counting measures. Let $\begin{Bmatrix} a_{i,j} : i, j \in \mathbb{N} \end{Bmatrix}$ be a sequency of numbers and define $f(i, j) = a_{i,j}$. Denote

By definition of integration, $\int_\Omega f(x, y) \, d\, \bar \mu$ exists if

1. Either $f \ge 0$. In this case

   $$\int_\Omega f(x, y) \, d\, \bar \mu = \sup \begin{Bmatrix} \sum_{ ( i,j ) \in S } a_{i,j} : \begin{array}{l} S \subseteq \mathbb{N} \times \mathbb{N} \\ |S| < \infty \end{array} \end{Bmatrix},$$

2. Or $\int_\Omega | f(x, y) | \, d\, \bar \mu < \infty$. In this case,

   $$\begin{aligned} &\int_\Omega f(x, y) \, d\, \bar \mu = \int_\Omega f_+ \, d\, \bar \mu - \int_\Omega f_- \, d\, \bar \mu \\ &= \sup \begin{Bmatrix} \sum_{ ( i,j ) \in S } a_{i,j}: \, \begin{matrix} S \subseteq E_1 \\ |S| < \infty \end{matrix} \end{Bmatrix} - \sup \begin{Bmatrix} \sum_{ ( i,j ) \in S } a_{i,j}: \, \begin{matrix} S \subseteq E_2 \\ |S| < \infty \end{matrix} \end{Bmatrix} . \end{aligned}$$

In the former case, the summation of elements in a chosen set $S$ grows in one direction. While in the later, the summation converges absolutely.

When $\int_\Omega f(x, y) \, d\, \bar \mu$ exists, the following iterated integrals exist:

Further, their values are equal.

$\square$

Proof of Fubini's Theorem

By symmetry, it suffices to prove that

Sketch of the Proof

Before proving that the iterated integral equals the double integral, we need to make sure that the first iterated integral is well-defined. Specifically,

• When $x$ is fixed, the function $f_x : Y \rightarrow \mathbb{R}$ defined by $f_x(y) \doteq f(x, y)$ is $\mathcal{F}_2$ measurable, and the integral $\int_Y f(x, y) \, d\, \mu_2$ exists.

• There is $\mathcal{F}_1$ measurable function $g : X \rightarrow \mathbb{R}$ such that $g(x) = \int_Y f(x, y) \, d\, \mu_2$ almost everywhere.

We will first establish the theorem for indicator functions $\mathbb{I}_E$ for sets $E \in \sigma( \mathcal{S} )$, which serve as basic building blocks for extending the theorem for more general functions.

Lemma. Given $E \in \sigma( \mathcal{S} )$, let $\mathbb{I}_E$ be the indicator for $(x, y) \in E$. Then

1. For each $x \in X$, the function $f_x(y) \doteq \mathbb{I}_E (x, y)$ is $\mathcal{F}_2$ measurable.

2. The function $g(x) \doteq \int_Y \mathbb{I}_E (x, y) d\, \mu_2$ is $\mathcal{F}_1$ measurable.

3. $$\int_X \int_Y \mathbb{I}_E (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } \mathbb{I}_E (x, y) d\, \bar \mu.$$

The proof of the lemma is deferred to the end of the article.

Proof of the Theorem

For now, we apply it to prove Theorem for the following three increasingly more general cases.

Case One. $f$ is a simple function on $(X \times Y, \sigma(\mathcal{S}))$.

Case Two. $f \ge 0$ and is $\sigma(\mathcal{S})$ measurable.

Case Three. $f$ is measurable and integrable.

Remark: this follows exactly the same order with which we define Lebesgue integrals.

Proof.

Case 1. Suppose that $f(x, y) = \sum_{i \in [n]} c_i \cdot \mathbb{I}_{E_i} (x, y)$ is a simple function on $(X \times Y, \sigma(\mathcal{S}))$.

1. For fixed $x \in X$, according the lemma, the functions $f_x^{ (i) } (y) \doteq \mathbb{I}_{E_i} (x, y), i \in [n]$ are $\mathcal{F}_2$ measurable. It follows that function $f_x (y) = \sum_{i \in [n]} c_i \cdot f_x^{ (i) } (y)$ is $\mathcal{F}_2$ measurable.

2. Since $g^{(i)} (x) \doteq \int_Y \mathbb{I}_{E_i} (x, y) d\, \mu_2$ is $\mathcal{F}_1$ measurable, so is the function

3. Since $\int_X \int_Y \mathbb{I}_{E_i} (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } \mathbb{I}_{E_i} (x, y) d\, \bar \mu$,

Case Two. Suppose that $f \ge 0$ and is $\sigma(\mathcal{S})$ measurable. Then there exists simple, nonnegative functions $f^{ (n) } \uparrow f$.

1. For fixed $x \in X$. The functions $f_x^{ (n) } (y) \doteq f^{ (n) } (x, y) \uparrow f_x(y)$. Since they are $\mathcal{F}_2$ measurable, so is $f_x (y) \doteq f(x, y)$.

2. Let $g^{(n)} (x) \doteq \int_Y f^{ (n) } (x, y) d\, \mu_2$. Since $0 \le f_x^{ (n) } (y) \uparrow f_x(y)$, by monotone convergence theorem:

   $$g(x) = \int_Y f_x(y) d\, \mu_2 = \lim_n \int_Y f^{ (n) } (x, y) d\, \mu_2 = \lim_n g^{(n)} (x)$$

   It follows that $g^{(n)} (x) \uparrow g(x)$. Since they are $\mathcal{F}_1$ measurable, so is $g(x)$.

3. Since $\int_Y f^{ (n) } (x, y) d\, \mu_2 \uparrow \int_Y f(x, y) d\, \mu_2 \,$, $\int_X \int_Y f^{ (n) } (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } f^{ (n) } (x, y) d\, \bar \mu \,$, and $f^{(n)} (x, y) \uparrow f(x, y)$, by monotone convergence theorem,

Case Three. Suppose that $f(x, y)$ is a measurable and integrable function on $X \times Y$, i.e., $\int_{\Omega } |f(x, y)| d\, \bar \mu < \infty$. Write $f(x, y) = f^+(x, y) - f^-(x, y)$, such that $f^+(x, y), f^-(x, y) \ge 0$ are measurable and integrable.

1. For a fixed $x \in X$, since both $f^+_x(y) \doteq f^+(x, y), f^-_x(y) \doteq f^-(x, y)$ are $\mathcal{F}_2$ measurable, so is the function $f_x (y) \doteq f(x, y) = f^+(x, y) - f^-(x, y)$ is $\mathcal{F}_2$.

2. Since $f^+(x, y), f^-(x, y) \ge 0$ are measurable and integrable, we have

   $$\begin{aligned} \int_X \int_Y f^+(x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } f^+(x, y) d\, \bar \mu < \infty, \\ \int_X \int_Y f^-(x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } f^-(x, y) d\, \bar \mu < \infty. \end{aligned}$$

   It follows that $\exists A \subseteq \mathcal{F}_1$, s.t., $\mu_1(A) = 0$ and $g^+(x) \doteq \int_Y f^+(x, y) d\, \mu_2$ and $g^-(x) \doteq \int_Y f^-(x, y) d\, \mu_2$ are finite on $X \setminus A$. Define

   $$g(x) = \begin{cases} \int_Y f_x(y) d\, \mu_2 = g^+(x) - g^-(x), & \text{if } x \in X \setminus A \\ 0, & \text{if } x \in A \end{cases}.$$

   Then $g(x)$ is $\mathcal{F}_1$ measurable.

3. We have

$\square$

Proof of the Lemma

Lemma. Given $E \in \sigma( \mathcal{S} )$, let $\mathbb{I}_E$ be the indicator for $(x, y) \in E$. Then

1. For each $x \in X$, the function $f_x(y) \doteq \mathbb{I}_E (x, y)$ is $\mathcal{F}_2$ measurable.

2. The function $g(x) \doteq \int_Y \mathbb{I}_E (x, y) d\, \mu_2$ is $\mathcal{F}_1$ measurable.

3. $$\int_X \int_Y \mathbb{I}_E (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } \mathbb{I}_E (x, y) d\, \bar \mu.$$

There are various ways to prove this lemma. We present two proofs here.

Proof One

Since $X$ and $Y$ are both $\sigma$-finite, there exists $A_k \in \mathcal{F}_1, \mu_1(A_k) < \infty, \forall n \in \mathbb{N}^+$, s.t., $A_k \uparrow X$, and $B_k \in \mathcal{F}_2, \mu_2(B_k) < \infty, \forall n \in \mathbb{N}^+$, s.t., $B_k \uparrow Y$. It follows that $\Omega_k \doteq A_k \times B_k \uparrow \Omega$, and that $\mu(\Omega_k) < \infty$.

Define

Claim. If $\mathcal{L}_k = \sigma(\mathcal{S})$, then the lemma holds.

Proof of the Claim. For each $E \in \sigma(\mathcal{S})$,

1. For fixed $x \in X$. The functions $f_x^{ (k) } (y) \doteq \mathbb{I}_{E \cap \Omega_k} (x, y) \uparrow f_x(y)$. Since they are $\mathcal{F}_2$ measurable, so is $f_x (y) \doteq f(x, y)$.

2. Let $g^{(k)} (x) \doteq \int_Y \mathbb{I}_{E \cap \Omega_k} (x, y) d\, \mu_2$ are increasing. Since $0 \le f_x^{ (n) } (y) \uparrow f_x(y)$, by monotone convergence theorem: by monotone convergence theorem:

   $$g(x) = \int_Y f_x(y) d\, \mu_2 = \lim_k \int_Y \mathbb{I}_{E \cap \Omega_k} (x, y) d\, \mu_2 = \lim_k g^{(k)} (x)$$

   It follows that $g^{(k)} (x) \uparrow g(x)$. Since they are $\mathcal{F}_1$ measurable, so is $g(x)$.

3. Since $\int_Y \mathbb{I}_{E \cap \Omega_k} (x, y) d\, \mu_2 \uparrow \int_Y \mathbb{I}_E (x, y) d\, \mu_2 \,$, $\int_X \int_Y \mathbb{I}_{E \cap \Omega_k} (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } \mathbb{I}_{E \cap \Omega_k} (x, y) d\, \bar \mu \,$, and $\mathbb{I}_{E \cap \Omega_k} (x, y) \uparrow \mathbb{I}_E (x, y)$, by monotone convergence theorem,

$\square$

It is left to prove that $\mathcal{L}_k = \sigma(\mathcal{S})$. The proof relies on the following fact.

Fact. If $\tilde{S}$ is an algebra, then the monotone class $\mathcal{M}(\tilde{S})$ generated by $\tilde{S}$ equals $\sigma(\tilde{S})$.

The proof consists of two steps:

• $\mathcal{L}_k$ contains the algebra $\tilde{S}$ generated by $S$.

• $\mathcal{L}_k$ is a monotone class.

In what follows, to ease the notations, we write $\mathcal{L}_k$ as $\mathcal{L}$ and assume that $\mu(\Omega) < \infty$.

$\mathcal{L}$ contains the algebra $\tilde{S}$ generated by $S$.

Let $E = +_i \big( A_i \times B_i \big)$ for a finite sequence of disjoint rectangles $A_i \times B_i$.

1. For each $x \in X$, the function $f_x(y) \doteq \mathbb{I}_E (x, y)$ is a simple function on $Y$ and there $\mathcal{F}_2$ measurable.

2. The function

   $$g(x) = \int_Y \mathbb{I}_E (x, y) d\, \mu_2 = \sum_i \mathbb{I}_{A_i} (x) \cdot \mu_2(B_i)$$

   is a simple function on $X$ and there $\mathcal{F}_1$ measurable.

$\mathcal{L}$ is a monotone class.

Case One $E_n \in \mathcal{L}, n \in \mathbb{N}^+$, s.t., $E_n \uparrow E$

1. For each $x \in X$, we have $f_x^{ (n) } (y) = \mathbb{I}_{E_n} (x, y) \uparrow f_x(y)$. Since the $f_x^{ (n) } (y)$ $\mathcal{F}_2$ measurable, so is the function $f_x(y)$.

2. Let $g^{(n)} (x) \doteq \int_Y f^{ (n) } (x, y) d\, \mu_2$. Since $0 \le f_x^{ (n) } (y) \uparrow f_x(y)$, by monotone convergence theorem:

   $$g(x) = \int_Y f_x(y) d\, \mu_2 = \lim_n \int_Y f_x^{ (n) } (y) d\, \mu_2 = \lim_n g^{(n)} (x)$$

   It follows that $g^{(n)} (x) \uparrow g(x)$. Since they are $\mathcal{F}_1$ measurable, so is $g(x)$.

3. Since $\int_Y f^{ (n) } (x, y) d\, \mu_2 \uparrow \int_Y f(x, y) d\, \mu_2 \,$, $\int_X \int_Y f^{ (n) } (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } f^{ (n) } (x, y) d\, \bar \mu \,$, and $f^{(n)} (x, y) \uparrow f(x, y)$, by monotone convergence theorem,

Case Two $E_n \in \mathcal{L}, n \in \mathbb{N}^+$, s.t., $E_n \downarrow E$

1. For each $x \in X$, we have $f_x^{ (n) } (y) = \mathbb{I}_{E_n} (x, y) \downarrow f_x(y)$. Since the $f_x^{ (n) } (y)$ are $\mathcal{F}_2$ measurable, so is the function $f_x(y)$.

2. Since $f_x^{ (n) } (y) = \mathbb{I}_{E_n} (x, y) \downarrow f_x(y)$ and $f_x^{ (n) } (y) \le 1$, $0 \le 1 - f_x^{ (n) } (y) \uparrow 1 - f_x(y)$. By monotone convergence theorem:

   $$\begin{aligned} \mu_2(Y) - \lim_n g^{(n)} (x) &= \lim_n \int_Y \left( 1 - f_x^{ (n) } (y) \right) d\, \mu_2 \\ &= \int_Y \left( 1 - f_x(y) \right) d\, \mu_2 \\ &= \mu_2(Y) - g(x). \end{aligned}$$

   Since $\mu_2(Y) < \infty$, $\lim_n g^{(n)} (x) = g(x)$. Since they are $\mathcal{F}_1$ measurable, so is $g(x)$.

3. Since $0 \le \mu_2(Y) - \int_Y f^{ (n) } (x, y) d\, \mu_2 \uparrow \mu_2(Y) - \int_Y f(x, y) d\, \mu_2 \,$, $\int_X \int_Y f^{ (n) } (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } f^{ (n) } (x, y) d\, \bar \mu \,$, and $0 \le 1 - f^{(n)} (x, y) \uparrow 1 - f(x, y)$, by monotone convergence theorem,

   $$\begin{aligned} \mu(\Omega) - \int_X \int_Y f(x, y) d\, \mu_2 \, d\, \mu_1 &= \int_X \left( \mu_2(Y) - \int_Y f(x, y) d\, \mu_2 \right) \, d\, \mu_1 \\ &= \lim_n \int_X \left( \mu_2(Y) - \int_Y f^{ (n) } (x, y) d\, \mu_2 \right) \, d\, \mu_1 \\ &= \mu(\Omega) - \lim_n \int_{\Omega } f^{ (n) } (x, y) d\, \bar \mu \\ &= \lim_n \int_{\Omega } \left( 1 - f^{ (n) } (x, y) \right) d\, \bar \mu \\ &= \int_{\Omega } \lim_n \left(1 - f^{ (n) } (x, y) \right) d\, \bar \mu \\ &= \mu(\Omega) - \int_{\Omega } f (x, y) d\, \bar \mu. \end{aligned}$$

   Observing that $\mu(\Omega) < \infty$ finishes the proof.

$\square$

Proof Two

Lemma. Given $E \in \sigma( \mathcal{S} )$, let $f(x, y) = \mathbb{I}_E (x, y)$ be the indicator for $(x, y) \in E$. Then

1. For each $x \in X$, the function $f_x(y) \doteq \mathbb{I}_E (x, y)$ is $\mathcal{F}_2$ measurable.

2. The function $g(x) \doteq \int_Y \mathbb{I}_E (x, y) d\, \mu_2$ is $\mathcal{F}_1$ measurable.

3. $$\int_X \int_Y \mathbb{I}_E (x, y) d\, \mu_2 \, d\, \mu_1 = \int_{\Omega } \mathbb{I}_E (x, y) d\, \bar \mu.$$

Proof of (1).

Definition. Let $E_x \doteq \begin{Bmatrix} y : (x, y) \in E \end{Bmatrix}$ be the cross-section at $x$.

It suffices to show that

Claim. For each $x \in X$ and $E \in \sigma(\mathcal{S})$, $E_x \in \mathcal{F}_2$.

Proof of the Claim. Fix an $x \in X$. Define the function $T_x : Y \rightarrow X \times Y$ by $T_x (y) \doteq (x, y).$ Then for each $E = A \times B \in \mathcal{S}$,

Therefore $T_x^{-1} (E) = E_x \in \mathcal{F}_2$. Since $\mathcal{S}$ generates $\sigma(\mathcal{S})$, we conclude that $T_x$ is measurable $\mathcal{F}_2 / \sigma(\mathcal{S})$.

$\square$

Proof of (2) and (3).

Define the set $\mathcal{L} \subseteq \sigma(\mathcal{S})$ as the collection of $E$ for which the condition (2) and (3) hold.

1. Clearly it holds that $\mathcal{S} \subseteq \mathcal{L}$;
2. We show that $\mathcal{L}$ is a $\lambda$ system.

Since $\mathcal{S}$ is a $\pi$ system, via Dynkin's $\pi-\lambda$ theorem, $\mathcal{L}$ is a $\sigma$-algebra, and hence $\mathcal{L} = \sigma(\mathcal{S})$.

We will now check that $\mathcal{L}$ is a $\lambda$-system. With the same technique applied in previous proof, we can suppose without loss of generality that $\mu(\Omega) < \infty$.

1. $\Omega \in \mathcal{L}$.

2. If $E \in \mathcal{L}$, then $\bar E \in \mathcal{L}$. First, observe that $\bar E_x = \Omega_x - E_x = B - E_x \in \mathcal{F}_2$. Since $\mu_2(B) < \infty$,

   $$g_{\bar E} (x) = \mu_2 ( \bar E_x ) = \mu_2 (\Omega_x - E_x) = \mu_2 (B) - \mu_2 (E_x).$$

   Since $g_E(x) = \mu_2 (E_x)$ is $\mathcal{F}_1$ measurable, so is $g_{\bar E}(x)$. Integrating over $x$ gives

   $$\begin{aligned} \int_X \mu_2 ( \bar E_x ) \, d\, \mu_1 &= \int_X \mu_2 (B) \, d\, \mu_1 - \int_X \mu_2 (E_x) \, d\, \mu_1 \\ &= \mu (\Omega) - \mu(E). \end{aligned}$$

   Since $\mu(\Omega) < \infty$, we have $\mu(\bar E) = \mu (\Omega) - \mu(E) = \int_X \mu_2 ( \bar E_x ) \, d\, \mu_1$.

3. If $E_n \in \mathcal{L}, n \in \mathbb{N}^+$ is a disjoint sequence of sets, then $\cup_{n \ge 1} E_n \in \mathcal{L}$. Define $g_{E_n} (x) \doteq \mu_2 ( (E_n)_x )$. Then $g(x) \doteq \sum_{n \ge 1} g_{E_n} (x)$ is $\mathcal{F}_1$ measurable, since $g = \sup_{n \ge 1} \sum_{k \in [n]} g_{E_k}$. By monotone convergence theorem and that $\mu$ is a measure, it holds that

   $$\int_X g \, d\, \mu_1 = \int_X \sum_{n \ge 1} g_{E_n} \, d\, \mu_1 = \sum_{n \ge 1} \int_X g_{E_n} \, d\, \mu_1 = \sum_{n \ge 1} \mu (E_n) = \mu (E).$$

   Therefore, $E \in \mathcal{L}$.