# Proving statement about approximation of functions in completed measure spaces by functions belonging to the noncompleted space

Let $(X,\mathcal{A},\mu)$ be a measure space, $(X,\mathcal{A}_\mu,\overline{\mu})$ be its completion and $f: X \to [-\infty,\infty]$ $\mathcal{A}_\mu$-measurable. Then there exist $\mathcal{A}$-measurable functions $g,h: X \to [-\infty,\infty]$ with $g \leq f \leq h$ and $g = h$ $\mu$-a.e.

Proof. First assume $f \in \Sigma^+$. Then $$f = \sum_{i = 1}^n a_i \chi_{A_i}$$ where $a_i \leq 0$ and $A_i \in \mathcal{A}_\mu$ for $i = 1,\dots,n$. Since $\mathcal{A}_\mu$ is the completion of $\mathcal{A}$, there exist $E_i,F_i \in \mathcal{A}$ such that $$E_i \subseteq A_i \subseteq F_i \qquad \text{and} \qquad \mu(F_i \setminus E_i) = 0$$ for $i = 1,\dots,n$. Then $$g := \sum_{i = 1}^n a_i \chi_{E_i} \qquad \text{and} \qquad h := \sum_{i = 1}^n a_i \chi_{F_i}$$ have the desired properties. Now let $f$ be a $\mathcal{A}_\mu$-measurable nonnegative function. Then we find a sequence $(\varphi_n)_{n \in \mathbb{N}}$ of $\mathcal{A}_\mu$-measurable simple functions $0 \leq \varphi_n$ such that $\varphi_n \nearrow f$. By the first part we find sequences $(g_n)_{n\in\mathbb{N}}$ and $(h_n)_{n \in \mathbb{N}}$ of $\mathcal{A}$-measurable functions such that $$g_n \leq \varphi_n \leq h_n \qquad \text{and} \qquad g_n = h_n \>\mu\text{-a.e.}$$ for any $n \in \mathbb{N}$.

Now I am a bit unsure how to proceed. Intuitively, we just take $$g := \lim_{n \to \infty} g_n \qquad \text{and} \qquad h := \lim_{n \to \infty} h_n$$ but I am not sure if this works. I mean it is not clear that $g_n$ and $h_n$ converge pointwise. Any help?

Edit. Maybe better would be $$g := \limsup_{n \to \infty} g_n \qquad \text{and} \qquad h := \liminf_{n \to \infty} h_n$$

• The idea to use $\liminf$ and $\limsup$ works fine. – saz Jan 26 '17 at 7:01
• @saz Thanks a lot. Somehow I do not quite see yet, why then exactly $g = h$ holds $\mu$-a.e. (intuitively it is clear). Furthermore, I think there are counterexamples if we use for both $\limsup$ or $\liminf$ that this construction won't work. What are your thoughts about this? – TheGeekGreek Jan 26 '17 at 9:32

If we set

$$g:= \limsup_{n \to \infty} g_n \qquad \text{and} \qquad h := \liminf_{n \to \infty} h_n$$

(as you suggested), then both $g$ and $h$ are well-defined and $\mathcal{A}$-measurable. Moreover, since

$$N_n := \{g_n \neq h_n\}$$

satisfies $\mu(N_n)=0$ for all $n \geq 1$, we know that $N := \bigcup_n N_n \in \mathcal{A}$ is a $\mu$-null set and

$$\forall x \in X \backslash N, n \in \mathbb{N}: h_n(x) = g_n(x) = \varphi_n(x).$$

This implies

$$h(x) = g(x)=f(x) \qquad \text{for all x \in X \backslash N}.$$