Suppose $f \leq g\leq h $ where both $f $ and $h$ are integrable over $E$ and $g$ is measurable in E. Is monotonicity enough to show that $g $ is also integrable?
I was thinking of
Integral comparison test which says that if g is dominated by a nonnegative integrable function h then g is integrable. However this test requires g to be of finite measure which is not in the hypothesis. But is the finiteness of g implied by its being bounded below (f ) and above (h ) by integrable functions? I know that the test also requires h to be nonnegative but since h is integrable iff int |h| is integrable and hleq |h| then g must really be dominated by nonnegative integrable function (|hl ) Is my logic correct?
I am also thinking of using the Lebesgue Dominated Convergence Theorem. Can i just assume to have a sequence of measurable functions $g_n $ converging pointwise to $g $ so that i can have the $\lim_{n\rightarrow \infty} {\int g_n} = \int g $?
Decompose the functions into positive and negative parts respectively $f= f_+ + f_-$, $h=h_+ + h_-$. Now see that $$|g| \leq \max(|f_-|, |h_+|).$$ From integrability of $f$ and $h$, the right hand side is finite, so $g$ is integrable.
To dispel any doubts about the fact that for measurable function $\phi$ such that $0\leq \phi \leq \psi$ for some integrable function $\psi$ implies the integrability of $\phi$, here is a short argument:
Assume that $\phi$ is measurable. Writing down the definition of Lebesgue integral for $\phi$ we bound it pointwise by $\psi$. The construction of Lebesgue integral is such that it is a monotone increasing sequence with respect to partitions of the domain, which by the above argument we have just bounded. Bounded increasing sequence has a finite limit which proves the integrability of $\phi$.