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/Part 4: Measure Theory and Integration/Chapter 15: Measurable Functions

15.5 Measurable Functions into Metric Spaces

Proposition 15.5.1. Let $(X, \cm)$, $(Z, \cn)$ be measurable spaces, $\seqf{Y_j}$ separable metrisable topological spaces, $F: \prod_{j = 1}^{n} Y_{j} \to Z$ be a $(\cb_{\prod_{j = 1}^n Y_j}, \cn)$-measurable function. For any $\seqf{f_j}$ where for each $1 \le j \le n$, $f_{j}: X \to Y_{j}$ is $(\cm, \cb_{Y_j})$-measurable, the composition

\[X \to Z \quad x \mapsto F(f_{1}(x), \cdots, f_{n}(x))\]

is $(\cm, \cn)$-measurable.

Proof. By Proposition 15.2.3, $\cb_{\prod_{j = 1}^n Y_j}= \bigotimes_{j = 1}^{n} \cb_{Y_j}$, so

\[X \to \prod_{j = 1}^{n} Y_{j} \quad x \mapsto (f_{1}(x), \cdots, f_{n}(x))\]

is $(\cm, \cb_{\prod_{j = 1}^n Y_j})$-measurable. Therefore the composition is $(\cm, \cn)$-measurable.$\square$

Proposition 15.5.2. Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, $f, g: X \to Y$ be $(\cm, \cb_{Y})$-measurable functions, then the following functions are measurable:

  1. For any metric $d$ on $Y$, $x \mapsto d(f(x), f(y))$.

  2. If $Y$ is a TVS over $K \in \RC$ and $\lambda \in K$, $\lambda f + g$. In particular, $\bracs{f = g}\in \cm$.

  3. If $Y \in \RC$, $fg$.

Proof. By Proposition 15.5.1.$\square$

Proposition 15.5.3. Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, and $f: X \to Y$ be a function, then the following are equivalent:

  1. $f$ is $(\cm, \cb_{Y})$-measurable.

  2. For each $\phi \in C(X; [0, 1])$, $\phi \circ f$ is $(\cm, \cb_{\real})$-measurable.

Proof. (2) $\Rightarrow$ (1): For each $U \subset X$ open, the function

\[d_{U^c}: X \to [0, 1] \quad x \mapsto d(x, U^{c}) \wedge 1\]

is continuous by Proposition 7.2.2. By Proposition 7.2.4, $\bracsn{d_{U^c} > 0}= U$. Thus $\bracs{f \in U}= \bracsn{d_{U^c} \circ f > 0}$.$\square$

Proposition 15.5.4. Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, and $\seq{f_n}$ be $(\cm, \cb_{Y})$-measurable functions, then:

  1. If $Y$ is completely metrisable, then $\bracsn{\limv{n}f_n \text{ exists}}\in \cm$.

  2. If $f = \limv{n}f_{n}$ exists, then it is $(\cm, \cb_{Y})$-measurable.

Proof. (1): Let $d$ be a complete metric on $Y$, then for any $x \in X$, $\limv{n}f_{n}(x)$ exists if and only if $\seq{f_n(x)}$ is Cauchy. In which case,

\[\bracs{\limv{n}f_n \text{ exists}}= \bigcap_{k \in \natp}\bigcup_{N \in \natp}\bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracsn{d(f_m, f_n) < 1/k}\]

is measurable by Proposition 15.5.2.

(2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_{n}$ is $(\cm, \cb_{\real})$-measurable by Proposition 15.3.2. Thus $f$ is $(\cm, \cb_{\real})$-measurable by Proposition 15.5.3.$\square$

Proposition 15.5.5. Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^{Y}$This mapping is typically obtained as slices of the level sets of a continuous function $Y \times Y \to \real$. such that

  1. For each $y \in Y$, $y \in \ol{N(y)^o}$.

  2. $\bigcap_{y \in Y}N(y) \ne \emptyset$.

  3. For any $y_{0} \in Y$, $\bracs{y \in Y|y_0 \in N(y)}\in \cb_{Y}$.

Then, for any $f: X \to Y$, the following are equivalent:

  1. $f$ is $(\cm, \cb_{Y})$-measurable.

  2. There exists a sequence $\seq{f_n}$ of $(\cm, \cb_{Y})$-measurable simple functions such that

    1. For each $x \in X$ and $n \in \natp$, $f_{n}(x) \in N(f(x))$.

    2. $f_{n} \to f$ pointwise.

  3. There exists a sequence $\seq{f_n}$ of $(\cm, \cb_{Y})$-measurable simple functions such that $f_{n} \to f$ pointwise.

Proof. (1) $\Rightarrow$ (2): Let $\seq{y_n}\subset Y$ be a dense subset of $Y$. Assume without loss of generality that $y_{1} \in \bigcap_{y \in Y}N(y)$. For each $N \in \nat$ and $x \in X$, let

\[C(N, x) = \bracs{1 \le n \le N|y_n \in N(f(x))}\]

By assumption (b), $1 \in C(N, x) \ne \emptyset$. Let

\[k(N, x) = \min\bracs{n \in C(N, x) \bigg | d(f(x), y_n) = \min_{m \in C(N, x)}d(f(x), y_m)}\]

then for any $k \in \natp$,

\[\bracs{x \in X|k(n, x) \le k}= \bigcup_{j = 1}^{k}\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}\]

For each $1 \le m \le N$, $y \mapsto d(y, y_{m})$ is continuous. Thus $x \mapsto d(f(x), y_{m})$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_{m})$ are $(\cm, \cb_{\real})$-measurable by Proposition 15.3.2 and assumption (c). By Proposition 15.5.2,

\[\bracs{x \in X \bigg | d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}\in \cm\]

for each $1 \le j \le k$. Combining this with assumption (c) shows that $\bracs{x \in X|k(n, x) \le k}\in \cm$.

Let $f_{N}(x) = y_{k(N, x)}$, then for each $N \in \natp$, $f_{N}(x) \subset \bracs{y_n|1 \le n \le N}$, so $f_{N}(x)$ is finitely valued. Since $x \mapsto k{(N, x)}$ is $(\cm, 2^{\nat})$-measurable, $f_{N}$ is $(\cm, \cb_{Y})$-measurable.

Fix $x \in X$, then

\[f(x) \in \ol{N^o(f(x))}= \ol{\bracs{y_n| n \in \natp, y_n \in N^o(f(x))}}\]

by assumption (a) and Definition 4.5.5. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_{N} \in N^{o}(f(x))$ and $d(f(x), y_{N}) < \eps$. In which case, $d(f(x), f_{n}(x)) < \eps$ for all $n \ge N$. Therefore $f_{n} \to f$ pointwise.

(3) $\Rightarrow$ (1): By Proposition 15.5.4.$\square$

Proposition 15.5.6. Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_{E})$ be a separable normed space, and $f: X \to E$, then the following are equivalent:

  1. $f$ is $(\cm, \cb_{E})$-measurable.

  2. There exists simple functions $\seq{f_n}$ such that $\abs{f_n}\le \abs{f}$ for all $n \in \natp$, and $f_{n} \to f$ pointwise.

Proof. Let

\[N: E \to 2^{E} \quad y \mapsto B_{E}(0, \norm{y}_{E})\]

then

  1. $y \in \ol{B_E(0, \norm{y}_E)}$.

  2. $0 \in \bigcap_{y \in E}N(y)$.

  3. For any fixed $y_{0} \in E$,

    \[\bracs{y \in E|y_0 \in N(y)}= \bracs{y \in E|\norm{y_0}_E \le \norm{y}_E}\in \cb_{E}\]

By Proposition 15.5.5, (1) and (2) are equivalent.$\square$

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