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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

  1. 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?

  2. 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 $?

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  • $\begingroup$ It is long ago that I learned this stuff, but can't you argue that $g-f$ is measurable and $0 \le g-f \le h-f$ ? Then $g-f$ is integrable and therefore $g = (g-f) +f $ is integrable. – And you probably meant "both f and h are integrable" in the first sentence of your question. $\endgroup$
    – Martin R
    Nov 19, 2015 at 21:43
  • $\begingroup$ Yep it's $f $ and $h $. Made the necessary corrections. Thanks $\endgroup$ Nov 19, 2015 at 21:46
  • $\begingroup$ How would it follow that $g-f $ is integrable? Sorry i am pretty lost. The blurry part is how does it follow immediately since it is still a need to show that $g $ is integrable? $\endgroup$ Nov 19, 2015 at 21:50

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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 $\phi$.

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  • $\begingroup$ Thanks. So you're using this fact: Suppose that $f$ and $g$ are measurable functions such that $0\le f \le g$ and $\int g \lt \infty$. Then we have $\int f \lt \infty$. How this fact can be proved? $\endgroup$
    – S.H.W
    May 16, 2023 at 0:22
  • $\begingroup$ If $f \leq g$ then $\int f \leq \int g$. $\endgroup$ May 16, 2023 at 0:59
  • $\begingroup$ I think there is a subtlety here. We know that $g$ is integrable and $0\le f \le g$ but we don't want to assume that $f$ is integrable as well. In fact we are trying to show that $\int f$ exists. I'm looking for a proof of this fact and by combining it with your answer the proof is complete. $\endgroup$
    – S.H.W
    May 16, 2023 at 1:35
  • $\begingroup$ You assume that $f$ is measurable. Now you can write down the very definition of lebesgue integral for $f$ and bound it pointwise by $g$. 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, therefore this sequence has a finite limit. $\endgroup$ May 16, 2023 at 12:02
  • $\begingroup$ You're right, thanks. I think it would be nice if you add the proof to your answer, if you like. After the time limit is passed, I will award the bounty. $\endgroup$
    – S.H.W
    May 16, 2023 at 14:47

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