# Equivalence with measurable functions and spaces

i have been reading Follands Real Analysis Book and i got stuck with one exercise. It says that if $$\lambda(X)$$ is finite and $$(X,M,\lambda)$$ is a measure space and $$(X, \overline{M}, \overline{\lambda})$$ its completion. Suppose $$f$$ is bounded, then it is $$\overline{M}$$-mensurable iff theres exists sequences of $$M$$ simple mensurable functions $${\phi_n}$$ and $$\psi_n$$ such that $$\phi_n \leq f \leq \psi_n$$ and $$\int \psi_n - \phi_n \leq 1/n$$.

I think i can do one of the implications cause if there exists sequence of functions like that we will get by the dominated convergence theorem that $$f$$ will be $$M-mensurable$$ and so it will be $$\overline{M} -mensurable$$ and we can swap the limits with the integral. Is this correct ?? Or i need to complete it more??

But i have no idea how to do the other one, my idea was that if $$f$$ is mensurable in the extension the there exists simple functions that converge to $$f$$ and $$-f$$ but how do i know that these simple functions are $$M-mensurable$$?? , any help is welcomed.

## 1 Answer

If $$f$$ is bounded and $$M-$$ measurable there exist simple functions $$g_n$$ increasing to $$f^{+}$$ and simple functions $$h_n$$ decreasing to $$f^{-}$$. Hence $$\phi_n \equiv g_n-h_n \leq f$$ and $$\int (f-\phi_n) \to 0$$. By applying this to $$-f$$ we can find the $$\psi_n$$'s. Once you get $$\int (\psi_n-\phi_n) \to 0$$ you can go to a subseqeunce to get the inequality $$\int (\psi_n-\phi_n) \leq \frac 1 n$$ for all $$n$$.

Now suppose $$f$$ is bounded and $$\overset {-}M$$ measurable. There exists an $$M-$$ measurable function $$g$$ such that $$f=g$$ almost everywhere. Suppose $$f(x)=g(x)$$ for $$x \notin E$$ where $$E$$ has measure $$0$$. Suppose $$-C for all $$x$$. Now choose $$\phi_n$$ and $$\psi_n$$ as above with $$f$$ changed to $$g$$ and then define $$\Phi_n=\phi_n-I_E C,\Psi_n=\psi_n+I_E C$$. Then $$\Phi_n$$ and $$\Psi_n$$ are simple functions satisfying the required properties.

• Yes that was my idea , but my question is when i find these sequence of simple functions they are defined in the completed measure how do i know that they are also going to work with respect to $M$? There could be sets where they are defined that arent in $M$ right? – Pedro Santos Apr 18 at 11:47
• @PedroSantos I have edited my answer – Kavi Rama Murthy Apr 18 at 11:55
• Ah yeah Thanks, yeah the sets are gonna have measure zero so those sets really arent gonna affect the integral , thanks !! – Pedro Santos Apr 18 at 11:56