[1] Let $f$ be non negative and $\int_\Omega fd\mu$ is finite then f is finite valued almost everywhere.
[2]Let $E_1=(-\infty,-2) \bigcup (2,\infty)$ $E_2=(-\infty,-3)\bigcup(3,\infty)$ and $E_3=(-4,4)$
Define $F=\chi_E -\chi_E + \chi_E $( ie.$E_1-E_2+E_3)$
Then prove that $\int_\Omega fd\mu$ exists although $\mu(E_1)-\mu(E_2)+\mu(E_3)$ is undefined
Note here that $\Omega= \mathbb{R} , \Sigma=$ Borel class, $\mu$ is the lebesgue measure on $\Sigma$
Everything is as defined in the H.L ROyden
MY ATTEMPT
[1] First I write $ A=(x \in \Omega: f(x) =\infty) $ and suppose $ \mu (A)>0$
For each $n \geq0$ the function $n.\chi_A$ is a simple function satisfying $0\leq n. \chi_A\leq f$
Hence $0\leq n. \chi_A=\int_\Omega n.\chi_A \leq \int_\Omega fd\mu$ for each $n\geq0$
This implies that $\int_\Omega fd\mu =\infty$
Is this correct for part one?
[2]Also for part two do I have to find the union of all and then integrate? I'm not sure exactly how to proceed.
The measure of an interval is it's length but I don't see how I can find the length of an open interval from infinity. There exits an infinite amount of open intervals and there are so many open covers.
What is the thinking behind beginning this question?
[1] correct (except what is remarked by @francescop21 in a comment).
[2]
I preassume that you forgot the indices and meant: $f=\chi_{E_1}-\chi_{E_2}+\chi_{E_3}$.
If so then note that $\chi_{E_1}-\chi_{E_2}=\chi_{[-3,-2)}+\chi_{(2,3]}$ so that $f=\chi_{[-3,-2)}+\chi_{(2,3]}+\chi_{(-4,4)}$ which is a nonnegative measurable function.
That is enough for the existence of $\int fd\mu$.
If moreover $\mu$ is the Lebesguemeasure then $f$ is also integrable with:$$\int fd\mu=\lambda([-3,-2))+\lambda((-2,-3])+\lambda((-4,4))=1+1+8=10$$