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.


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 $-C<f(x)<C$ 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.

  • $\begingroup$ 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? $\endgroup$ – Pedro Santos Apr 18 at 11:47
  • $\begingroup$ @PedroSantos I have edited my answer $\endgroup$ – Kavi Rama Murthy Apr 18 at 11:55
  • $\begingroup$ Ah yeah Thanks, yeah the sets are gonna have measure zero so those sets really arent gonna affect the integral , thanks !! $\endgroup$ – Pedro Santos Apr 18 at 11:56

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