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Fron Folland's Real Analysis

Let $(X,\mathcal{M},\mathcal{u})$ be a measure space with $\mathcal{u}(X) < \infty$, and let $(X,\mathcal{\overline{M}},\mathcal{\overline{u}})$ be its completion.

Suppose, $f:\:X \rightarrow \mathbb{R}$ is bounded. Then,

$f$ is $\mathcal{\overline{M}}$-measurable $\iff$ $\exists$ sequences $\{\phi_n\}$ and $\{\psi_n\}$ of $\mathcal{M}$-measurable simple functions such that $\phi_n \le f \le \psi_n$ and $\int(\psi_n - \phi_n)\,\mathrm{d}u < n^{-1}$. In this case, $\lim \int \phi_n\,\mathrm{d}u = \lim\int\psi_n\,\mathrm{d}u = \int f\,\mathrm{d}\overline{u}$

My work thus far:

$\implies$ Suppose $f$ is $\mathcal{\overline{M}}$-measurable. Then by properties of a complete measure space there exists a function $g$ that is $\mathcal{M}$-measureable such that $f = g \:\:\: \overline{u}\:a.e.$

Since $g$ is measurable, there exists a seq $\{ \phi_n\}$ of simple functions such that $0 \le \phi_1 \le \phi_2 \le \cdots \le g$

Here is where I get a little hazy. I want a function that converges to g from above. I feel like $\{\psi_n\}$ where $\psi_n = \sum_{j=0}^\infty \phi_j - \sum_{i=0}^n \phi_i$ will do the trick, but I'm not confident. Plus I am also wondering if I am missing something about the fact that $f = g$ a.e.

Any thoughts on how to proceed (or where to begin if I'm way off) are appreciated. Thanks!

P.S. there is a similar question here If a function is measurable with respect to the completion then it is equal to some measurable (with respect to the measure space) function a.e. but I have been unable to use it to make any headway.

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FAS
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  • http://math.stackexchange.com/questions/216199/sequences-of-simple-functions-converging-to-f – user62089 Apr 02 '13 at 10:12
  • Thanks, this got me thinking about my understanding of simple functions, and led me to http://math.stackexchange.com/questions/142968/how-to-approximate-simple-function-using-step-function , which has been useful. I think I will have an improvement to my work above soon. – FAS Apr 02 '13 at 17:31

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Hopefully it is okay to answer this here as I figure it out.

$\Longrightarrow $ Suppose $ f $ is $ \mathcal{\overline{M}} $ - measurable. Then by properties of a complete measure space there exists a function $g$ that is $\mathcal{M}$-measureable such that $f = g \:\:\:\:u -a.e.$

For every $n \in \mathbb{N} $ partion $\mathbb{R_+}$ into $(0,n] \cup (n,\infty)$. Partion $(0,n]$ into:

$$I_{n,k} = \left(\frac{k}{2n},\frac{k+1}{2n}\right], \:\:\: k = 0,1,2,...,2n^2-1$$

and let

$$\mathcal{M_{n,k}} = g^{-1}(I_{n,k}) = \left\{x \in X \:|\: \frac{k}{2n} \lt |g(x)| \le \frac{k+1}{2n} \right\} \quad and \quad \mathcal{M_{n,2n^2}} = g^{-1}\left( (n,\infty)\right)$$

where $\mathcal{M_{n,k}} \in \mathcal{M}$ since $g$ is measurable. Then if $\phi_n = \sum \frac{k}{2n}\chi_\mathcal{M_{n,k}}$ and $\psi_n = \sum \frac{k+1}{2n}\chi_\mathcal{M_{n,k}}$

$$|\phi_n| \le |g(x)| \le |\psi_n| \quad\quad |\psi_n - \phi_n| < \frac{1}{n}, \forall n$$

If everything up till now is correct, my question is do I use the fact that $f$ is bounded to use Dominated Convergence and show that $\int|\psi_n - \phi_n| \lt \frac{1}{n}$?

FAS
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  • @ZevChonoles is it okay to post answer like this even though incomplete? I could move this back to the original question but that also seems like an awkward place to put it. Thanks. – FAS Apr 02 '13 at 19:02