how to show that if $\int_{*}f=\int^{*}f$ then $f$ is measurable and $\int f=\int_{*}f=\int^{*}f$. I am working with this:
well im try like this:
Let $E_{\alpha}=\{x\in X: |f(x)|\leq \alpha\}$ with $\alpha\in [0,\infty]$. I want to show that $E_{\alpha}\in \Sigma$ the sigma algebra in $X$. Im not sure, i need to compare $\alpha$ with the all extended simple function, whatever let $\epsilon>0$ and $L:=\int_{*}f$ then by properties of $\sup$ exist $\psi$ extended simple function such that $L<\int \psi +\epsilon$, in this moment i believe that i need to compare $\alpha$ with $\psi$ (im not sure)
If $\alpha \geq \psi$ or $\psi\geq \alpha$.
when $\alpha \geq \psi$ and $\psi\leq f$ then $\psi \leq \min\{\alpha,f\}=f$ imply that can separate the set $E_{\alpha}$ right? with measurable sets, but im confuse how to continued, please give me any hint, fot this and the other statement, please, thank you
Well, im try to do like Jose27 said before,
Let $ L=\inf \{\int\phi:\phi\in L^{*}\; extended\; simple, f\leq \phi\; a.e\}$
$ U=\sup \{\int\phi:\phi\in L^{*}\; extended\; simple, \phi\leq f\; a.e\}$
By hypothesis $U=L$ then for $\epsilon>0$ exist $\psi,\phi$ extended simple function such that $\psi-\epsilon<f<\phi+\epsilon$ then by monotony on the integrals we have
$\int \psi-\epsilon<\int f<\int \phi+\epsilon < \infty$ (If $f$ is integrable then f is measurable?)
If i take $\epsilon=\frac{1}{n}$ and use $\lim$ when $n\rightarrow \infty$ finish the first statement? Thank you
Im still with problem!!