# Lebesgue integral of a measurable function squeezed between two integrable functions

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 $$\lim_{n\rightarrow \infty} {\int g_n} = \int g$$?

• 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. Nov 19, 2015 at 21:43
• Yep it's $f$ and $h$. Made the necessary corrections. Thanks Nov 19, 2015 at 21:46
• 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? Nov 19, 2015 at 21:50

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 $$\phi$$.
• 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? May 16, 2023 at 0:22
• If $f \leq g$ then $\int f \leq \int g$. May 16, 2023 at 0:59
• 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. May 16, 2023 at 1:35
• 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. May 16, 2023 at 12:02