# Characterization of $\mu$-measurable $\overline{\mathbb{R}}$-valued functions

Let $$(X, \mathcal{A}, \mu)$$ be a complete, $$\sigma$$-finite measure space and $$(E,|\cdot|)$$ a Banach space.

• We say $$f \in E^{X}$$ is $$\boldsymbol{\mu}\textbf{-simple}$$ if $$f(X)$$ is finite, $$f^{-1}(e) \in \mathcal{A}$$ for every $$e \in E,$$ and $$\mu\left(f^{-1}(E \backslash\{0\})\right)<\infty$$.

• Suppose $$f_n, f \in E^{X}$$ for $$n \in \mathbb{N} .$$ Then $$(f_n)_{n \in \mathbb N}$$ converges to $$f$$ $$\boldsymbol{\mu}\textbf{-almost everywhere}$$ if and only if there is a $$\mu$$-null set $$N$$ such that $$f_{n}(x) \rightarrow f(x)$$ for all $$x \in N^{c}$$.

• In the theory of integration, it is useful to consider not only real-valued functions but also maps into the extended number line $$\overline{\mathbb{R}}$$. Such maps are called $$\overline{\mathbb{R}}$$-valued functions.

• An $$\overline{\mathbb{R}}$$-valued function $$f: X \rightarrow \overline{\mathbb{R}}$$ is said to be $$\boldsymbol{\mu}\textbf{-measurable}$$ if $$\mathcal{A}$$ contains $$f^{-1}(-\infty), f^{-1}(\infty)$$, and $$f^{-1}(O)$$ for every open subset $$O$$ of $$\mathbb{R}$$.

After dicussing with @Thorgott, I came up with the following theorem. I've tried but to no avail. Could you please leave me some hints to finish it?

Theorem $$f: X \to \overline{\mathbb{R}}$$ is $$\mu$$-measurable if and only if there is a sequence of $$\mu$$-simple functions $$f_n: X \to \mathbb R$$ such that $$f_n \to f$$ $$\mu$$-almost everywhere.

• Do you have ideas for either direction? Or would you like help with both? What are some things you've tried? Also - which theorems of measure theory are you familiar with? – HallaSurvivor Jan 26 '20 at 19:46
• One direction is trivial. The other follows by splitting $f$ into a finite valued part and an indicator on the set where $f$ is $+\infty$ or $-\infty$. It is straightforward to construct a simple function converging to the latter ($n 1_{A_\infty \cap [-n,n]}$ for example if the space is $\mathbb{R}$, use the $\sigma$ finite part to do for general Banach space). – copper.hat Jan 26 '20 at 20:08
• Thank you so much @copper.hat! I will try your suggestion. – LE Anh Dung Jan 26 '20 at 20:20
• Hi @copper.hat! From your hints, I've figured a proof and posted it as an answer below. I'm not sure if the part I prove $f^{-1}(+\infty) \in \mathcal{A}$ (at the end of my proof) is correct or not. Could you please verify it? – LE Anh Dung Jan 30 '20 at 16:02

From @copper.hat's hints and my textbook, I've figured out a proof. I would be great if someone helps verify it. Thank you so much!

$$\textbf{My attempt}$$

$$\Longrightarrow$$

(i) We consider first the case $$\mu(X)<\infty$$. Let $$(a_k)_{k \in \mathbb N}$$ be an enumeration of $$\mathbb Q$$ and $$A_{k,n} = f^{-1} [ \mathbb B (a_k, 1/(n+1))]$$. Let $$A_{+} = f^{-1}(+\infty)$$ and $$A_{-} = f^{-1}(-\infty)$$. Then $$\{A_{k,n},A_{+},A_{-}\} \subseteq \mathcal A$$ for all $$(k,n) \in \mathbb N^2$$. The continuity of $$\mu$$ from above and the assumption $$\mu(X)<\infty$$ implies there are $$m_n$$ and $$B_n \in \mathcal A$$ such that $$B^c_n =A_+ \cup A_- \cup \bigcup_{k=0}^{m_n} A_{k,n} \quad \text{and} \quad \mu(B_n) < \frac{1}{2^{n+1}}$$

Now define $$\varphi_{n} \in {\mathbb R}^{X}$$ by $$\varphi_{n}(x) = \begin{cases} {a_{0}} & {\text {if} \quad x \in A_{0,n}} \\ {a_{k}} & {\text {if} \quad x \in A_{k, n} \setminus \bigcup_{p=0}^{k-1} A_{p, n} \quad \text {for} \quad 1 \le p \leq m_{n}} \\ n & {\text {if}} \quad x \in A_{+} \\ -n & {\text {if}} \quad x \in A_{-} \\ {0} & {\text {otherwise}} \end{cases}$$

Clearly, $$\varphi_{n}$$ is $$\mu$$-simple and $$\|\varphi_{n}(x) - f(x)\| < 1/(n+1)$$ for all $$x \in B_n^c$$. Define a decreasing sequence $$(C_n)_{n \in \mathbb N}$$ by $$C_n = \bigcup_{p=0}^{\infty} B_{n+p}$$. Then $$C_n^c \subseteq B_n^c$$ and $$\mu(C_n) \le \sum_{p=0}^\infty \mu(B_{n+p}) < 1/2^n$$. It therefore follows from the continuity of $$\mu$$ from above that $$C = \bigcap_{n=0}^\infty C_{n}$$ is $$\mu$$-null. We now set $$\psi_{n}(x) = \begin{cases} {\varphi_{n}(x)} & {\text {if} \quad x \in C_{n}^{c}} \\ n & {\text {if}} \quad x \in A_{+} \\ -n & {\text {if}} \quad x \in A_{-} \\ {0} & {\text {otherwise}}\end{cases}$$

Clearly, $$\psi_{n}$$ is $$\mu$$-simple. For $$x \in C^c$$, there exists $$n \in \mathbb N$$ such that $$x \in C_n^c$$. Then $$x \in C_{n+p}^c$$ for all $$p \in \mathbb N$$. So $$\|\psi_{n+p}(x) - f(x)\| =\|\varphi_{n+p}(x) - f(x)\| < 1/(n+p+1)$$ for all $$p \in \mathbb N$$. Hence $$\psi_{n} (x) \to f(x)$$ for all $$x \in A_+ \cup A_- \cup C^c$$.

(ii) We next consider the case $$\mu(X)=\infty$$. Because $$\mu$$ is $$\sigma$$-finite, there is a sequence $$(A_k)_{k \in \mathbb N}$$ of pairwise disjoint subsets in $$\mathcal{A}$$ such that $$\bigcup_{k=0}^\infty A_{k}=X$$ and $$\mu (A_{k}) < \infty$$. As in (i), for each $$A_k$$, there is a sequence $$(\psi^k_{n})_{n \in \mathbb N}$$ of $$\mu$$-simple functions and a $$\mu$$-null set $$C_k$$ such that $$\psi^k_{n} (x) \to f(x)$$ for all $$x \in A_k \setminus C_k$$. Moreover, $$C=\bigcup_{k=0}^\infty C_{k}$$ is $$\mu$$-null. We define a sequence $$(\psi_{n})_{n \in \mathbb N}$$ by $$\psi_{n}(x) = \begin{cases} {\psi^k_{n}(x)} & {\text {if} \quad x \in \bigcup_{k=0}^n A_{k}} \\ {0} & {\text {otherwise}}\end{cases}$$

Clearly, $$(\psi_{n})_{n \in \mathbb N}$$ is a sequence of $$\mu$$-simple functions such that $$\psi_{n} (x) \to f(x)$$ for all $$x \in \bigcap_{k=0}^\infty C_k^c$$.

$$\Longleftarrow$$

Assume there exist a sequence $$(\psi_{n})_{n \in \mathbb N}$$ of $$\mu$$-simple functions and a $$\mu$$-null set $$N$$ such that $$\psi_{n} (x) \to f(x)$$ for all $$x \in N^c$$.

Let $$O$$ be open in $$\mathbb R$$. We define a sequence $$(O_k)_{k \in \mathbb N^*}$$ by $$O_{k} = \{y \in O \mid d(y, O^{c})>1 / k \}$$. Then $$O_{k}$$ is open and $$\overline{O}_{k} \subseteq O$$. Let $$x \in N^{c}$$. We have $$x \in O \iff \exists k \in \mathbb{N}^{*}: x \in O_k$$. Therefore, $$f(x) \in O$$ if and only if there exists $$(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}$$ such that $$\forall n \ge m_k: \varphi_{n}(x) \in O_{k}$$. Consequently, $$x \in f^{-1}(O)$$ if and only if there exists $$(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}$$ such that $$\forall n \ge m_k: x \in \varphi^{-1}_{n}(O_{k})$$. As a result, $$f^{-1}(O) \cap N^{c} = \left ( \bigcup_{(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}} \bigcap_{n \ge m_k} \varphi_{n}^{-1} (O_{k}) \right ) \cap N^{c} = \bigcup_{(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}} \bigcap_{n \ge m_k} \left ( \varphi_{n}^{-1} (O_{k}) \cap N^{c}\right )$$

Because $$\varphi_{n}$$ is $$\mu$$-simple, $$\varphi_{n}^{-1}(O_{k}) \in \mathcal{A}$$ for all $$(n,k) \in \mathbb{N} \times \mathbb{N}^{*}$$. Hence $$f^{-1}(O) \cap N^{c} \in \mathcal{A}$$. Furthermore, the completeness of $$\mu$$ implies $$f^{-1}(O) \cap N$$ is a $$\mu$$-null set. Altogether, we obtain $$f^{-1}(O)=\left(f^{-1}(O) \cap N\right) \cup\left(f^{-1}(O) \cap N^{c}\right) \in \mathcal{A}$$

Let $$x \in N^c$$. We have $$f(x) = +\infty \iff \forall M \in \mathbb N, \exists N \in \mathbb N,\forall n \ge N: \varphi_n(x) \ge M$$. Consequently, $$x \in f^{-1}(+\infty) \iff \forall M \in \mathbb N, \exists N \in \mathbb N,\forall n \ge N: x \in \varphi^{-1}_n ([M, \infty))$$. As a result, \begin{aligned} f^{-1}(+\infty) \cap N^c &= \left( \bigcap_{M=0}^\infty \bigcup_{N=0}^\infty \bigcap_{n=N}^\infty \varphi^{-1}_n ([M, \infty)) \right) \cap N^c \\ &= \bigcap_{M=0}^\infty \bigcup_{N=0}^\infty \bigcap_{n=N}^\infty \left( \varphi^{-1}_n ([M, \infty)) \cap N^c \right) \end{aligned}

Because $$\varphi_{n}$$ is $$\mu$$-simple, $$\varphi_{n}^{-1}([M, \infty)) \in \mathcal{A}$$ for all $$(n,M) \in \mathbb{N} \times \mathbb{N}$$. Hence $$f^{-1}(+\infty) \cap N^{c} \in \mathcal{A}$$. Furthermore, the completeness of $$\mu$$ implies $$f^{-1}(+\infty) \cap N$$ is a $$\mu$$-null set. Altogether, we obtain $$f^{-1}(+\infty)=\left(f^{-1}(+\infty) \cap N\right) \cup\left(f^{-1}(+\infty) \cap N^{c}\right) \in \mathcal{A}$$

With similar reasoning, we have $$f^{-1}(-\infty) \in \mathcal{A}$$.