23.6 Localisable Measures

Proof. (1): $X \in \mathcal{S}$.

(2): Since $\mu$ is finite, for any $E \in \ce$, $\mu(S \cap E) = \mu(E)$ if and only if $\mu(E \setminus S) = 0$.

(3): Firstly, for any $S, T \in \mathcal{S}$ and $E \in \ce$,

so $S \cap T \in \mathcal{S}$. Now let $\seq{S_n}\subset X$ and $E \in \ce$, then since $\mu$ is finite,

by continuity from above. By (2), $\bigcap_{n \in \natp}S_{n} \in \mathcal{S}$.

(4): Let $M = \inf\bracs{\mu(T)|T \in \mathcal{S}}$ and $\seq{S_n}\subset \mathcal{S}$ such that $\limv{n}\mu(S_{n}) = M$, then by continuity from above,

By (3), $\bigcap_{n \in \natp}S_{n} \in \mathcal{S}$, therefore the minimum is achieved.

(5): Let $R \in \mathcal{S}$. By (3), $S \cap R \in \mathcal{S}$ with $\mu(S \cap R) = \inf\bracs{\mu(T)|T \in \mathcal{S}}= \mu(S)$, so

and $S$ is the essential supremum of $\ce$.$\square$

Proof. Let $\ce \subset \cm$ and $\seqi{A}\subset \cm$ be a decomposition of $X$.

For each $i \in I$, let $\ce_{i} = \bracs{E \cap A_i|E \in \ce}$. By Lemma 23.6.4, there exists an essential upper bound $S_{i} \in \cm$ of $\ce_{i}$ contained in $A_{i}$ with respect to the restricted measure $\mu|_{A_i}$. In other words,

1. (i)

   $S_{i}$ is an essential upper bound of $\ce_{i}$.

2. (ii)

   For any essential upper bound $T \in \cm$ of $\ce_{i}$ with $T \subset A_{i}$, $\mu(S_{i} \setminus T) = 0$.

Now, let $S = \bigsqcup_{i \in I}S_{i}$, then since $X = \bigsqcup_{i \in I}A_{i}$ and $S_{i} \subset A_{i}$ for all $i \in I$, $S \cap A_{i} = S_{i} \in \cm$ for all $i \in I$, so $S \in \cm$. For any $E \in \ce$, $E \cap A_{i} \in \ce_{i}$. Passing through the decomposition, (i) implies that,

and $S$ is an essential upper bound of $\ce$.

Finally, let $T \in \cm$ be an essential upper bound of $\ce$, then $T \cap A_{i}$ is an essential upper bound of $\ce_{i}$ for all $i \in I$. The decomposition and (ii) then shows that

therefore $S$ is an essential supremum of $\ce$.$\square$

Proof. (1): Let $A, B \in \cf$, then

so $F_{B}^{c}$ is an essential upper bound of $\bracs{E_A}_{A \in \cf}$. Since $E$ is an essential supremum of $\bracs{E_A}_{A \in \cf}$, $\mu(E \setminus F_{B}^{c}) = \mu(E \cap F_{B}) = 0$.

(2): For any $B \in \cf$, $\mu(F_{B} \setminus E^{c}) = \mu(F_{B} \cap E) = \mu(E \cap F_{B}) =0$. Thus $E^{c}$ is an essential upper bound of $\bracs{F_B}_{B \in \cf}$. Given that $F$ is an essential supremum of $\bracsn{F_B}_{B \in \cf}$, $\mu(F \cap E) = \mu(F \setminus E^{c}) = 0$.$\square$

Proof. First suppose that $Y$ is finite. For each $y \in Y$, let $P(y)$ be an essential supremum of $\bracs{f_A^{-1}(y)|A \in \cf}$. By Lemma 23.6.6, for any $x, y \in Y$ with $x \ne y$, $\mu(P(x) \cap P(y)) = 0$. After modification by null sets, assume without loss of generality that $X = \bigsqcup_{y \in Y}P(y)$.

For each $x \in X$, let $f(x) \in Y$ be the unique element of $Y$ such that $x \in P(f(x))$, then:

1. (1)

   For each $y \in Y$, $f^{-1}(y) = P(y)$, so $f \in \mathcal{L}^{0}(X; Y)$ is measurable.

2. (2)

   Let $A \in \cf$, then for each $y \in Y$, $\mu(f_{A}^{-1}(y) \setminus P(y)) = 0$. On the other hand,

   \begin{align*}\mu\braks{(P(y) \cap A) \setminus f_A^{-1}(y)}&\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap f_{A}^{-1}(z) ) \\&\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap P(z)) \\&+ \sum_{z \in Y \setminus \bracs{y}}\mu(f_{A}^{-1}(z) \setminus P(z)) \\&= 0\end{align*}

   
   

   so $f|_{A} = f_{A}$ almost everywhere on $A$.

3. (U)

   For all $A \in \cf$, $f|_{A} = g|_{A}$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere.

Therefore $f$ is the desired function.

Now suppose that $Y$ is an arbitrary separable metrisable space. By Lemma 26.6.4, there exists $\seq{I_n}\subset Y^{Y}$ such that:

1. (i)

   $I_{n} \to \text{Id}$ pointwise as $n \to \infty$.

2. (ii)

   For each $n \in \natp$, $I_{n}(Y)$ is finite and Borel measurable.

For each $n \in \natp$, let $f_{A, n}= I_{n} \circ f_{A}$, then

1. (a)

   For each $A \in \cf$, $f_{A, n}\in \mathcal{L}^{0}(A; Y)$.

2. (b)

   For each $A, B \in \cf$, $f_{A, n}|_{A \cap B}= f_{B, n}|_{A \cap B}$ almost everywhere.

3. (c)

   For each $A \in \cf$, $f_{A, n}(A) \subset I_{n}(Y)$.

By the finite case, there exists $f_{n}: X \to Y$ such that:

1. (1)

   $f_{n} \in \mathcal{L}^{0}(X; Y)$.

2. (2)

   For each $A \in \cf$, $f_{n}|_{A} = f_{A, n}$ almost everywhere.

Since $Y$ is Polish, Proposition 26.5.4 implies that $\bracsn{\limv{n}f_n \text{ exists}}\in \cm$. For each $A \in \cf$, $f_{A, n}\to f_{A}$ pointwise by (i). For each$n \in \natp$, $f_{A, n}= f_{n}|_{A}$ almost everywhere on $A$ by (2). Thus

As $\cf$ is a scaffold for $\mu$, $\mu\bracsn{\limv{n}f_n \text{ does not exist}}= 0$. By (1) and Proposition 26.5.4, there exists $f \in \mathcal{L}^{0}(X; Y)$ such that $f = \limv{n}f_{n}$ almost everywhere. In which case,

1. (1)

   $f \in \mathcal{L}^{0}(X; Y)$.

2. (2)

   For each $A \in \cf$, $f|_{A} = \limv{n}f_{n}|_{A} = \limv{n}f_{A, n}= f_{A}$ almost everywhere.

3. (U)

   For all $A \in \cf$, $f|_{A} = g|_{A}$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere.

$\square$

Proof. Let $\seqi{f}\subset L^{\infty}(X; \real)$ and $M \in \real$ such that $f_{i} \le M$ almost everywhere for all $i \in I$.

Fix $A \in \cm$ with $\mu(A) < \infty$, and let

then since $f_{i} \le M$ almost everywhere for all $i \in I$, $\mathcal{S}_{A} \ne \emptyset$, and $m_{A} = \inf_{g \in \mathcal{S}_A}\int g d\mu \in \real$.

Let $\seq{g_{A, n}}\subset \mathcal{S}_{A}$ such that $\seq{g_{A, n}}$ is decreasing pointwise and $\limv{n}\int_{A} g_{A, n}d\mu \downto m_{A}$. Take $g_{A} = \limv{n}g_{A, n}$, then by the Dominated Convergence Theorem, $\int g_{A} d\mu = m_{A}$.

For each $i \in I$, since $g_{A, n}\ge f_{i}|_{A}$ almost everywhere for all $n \in \natp$, $g_{A} \ge f_{i}|_{A}$ almost everywhere as well. Thus $g_{A} \in \mathcal{S}_{A}$. For any $h \in \mathcal{S}_{A}$, $g_{A} \wedge h \in \mathcal{S}_{A}$ with

Thus $g_{A} \wedge h = g_{A}$ almost everywhere, so $g_{A} \le h$ almost everywhere, and $g_{A}$ is an essential supremum of $\bracsn{f_i|_A}_{i \in I}$.

Now, let $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$, then $\one_{A \cap B}g_{B} + \one_{A \setminus B}M \in \mathcal{S}_{A}$, and

Thus $g_{A} \wedge (\one_{A \cap B}g_{B} + \one_{A \setminus B}M) = g_{A}$ almost everywhere, so $g_{A}|_{A \cap B}\le g_{B}|_{A \cap B}$ almost everywhere. As the argument is symmetric, $g_{A}|_{A \cap B}= g_{B}|_{A \cap B}$ almost everywhere.

By the gluing lemma for measurable functions, there exists a measurable function $g:X \to \real$ such that $g|_{A} = g_{A}$ for all $A \in \cm$ with $\mu(A) < \infty$.

Let $h \in L^{\infty}(X; \real)$ with $h \ge f_{i}$ almost everywhere for all $i \in I$, then for any $A \in \cm$ with $\mu(A) < \infty$,

As $\mu$ is semifinite, $\mu(\bracs{h < g}) = 0$. Finally, since $g_{A} \le M$ almost everywhere for all $A \in \cm$ with $\mu(A) < \infty$, $g \le M$ almost everywhere. Therefore $g \in L^{\infty}(X; \real)$ is indeed the essential supremum of $\seqi{f}$.$\square$