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Problem: Let be $U\subset\mathbb{R}^m$ a open set. Show that $\int_{U}f(x)dx$ converge iff $\int_{U}|f(x)|dx$ converge.

It seems to be a simple matter, but I'm not having ideas to solve it. The settings are as follows:

Definition of integral improper: Given a continuous function (limited or not) it says that the improper integral $\int_{U}f(x)dx$ is convergent when, for all exhaustion $U=\cup K_i$, there is the limit $\lim_{i \rightarrow \infty}\int_{K_i}f(x)dx$ and its value is independent of the exhaustion taken.

Definition of Exhaustion: An exhaustion of $U$ is a sequence of J-measurable $K_i\subset U$ compact such that $U=\cup K_i$ and $K_i\subset int(K_{i+1}) \forall i\in\mathbb{N}$.

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    This is false: it cannot be proved. Indeed this statement implies that all convergent sequences are absolutely convergent (and this is false). – 2017-01-28
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    Suggestion: Consider the sets $U^{\pm} = \{x \in U : \pm f(x) \geq 0\}$, and show the integral of $f$ over $U$ converges iff the integral of $f$ over each of $U^{\pm}$ converges. (For one direction, I'd suggest the contrapositive: If the integrals $\int_{U^{\pm}} f(x)\, dx$ diverge, use this fact to pick exhaustions violating the definition of convergence of the improper integral.) – 2017-01-28
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    No, it is true: the key point that the sequence is not fixed, but for convergence "the value must be independent of the exhaustion taken". – 2017-01-28
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    If you need an idea to solve it: consider the functions $f_+$ and $f_-$, coinciding with $f$ on the positive and negative values respectively and 0 in all other points, so $f=f_++f_-$. Convergence of $|f|$ is equivalent to convergence of both $f_+$ and $f_-$. Statement readily follows from right to left. Other side: suppose $f_+$, for example, diverges, try to find some exhaustion that will exhaust the support of $f_+$ quickly and the support of $f_-$ very slowly, then $f$ will also diverge on this exhaustion. – 2017-01-28
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    Is $f$ a continuous function in the hypothesis? – 2017-01-28
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    You are assuming $\int_Uf$ convergente. This is hypothesis. – 2017-01-28
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    The going is immediate, is not it? If we assume $\int_U f$ convergent then $f^+$ or $f+-$ it is convergent and hence $\int_U |f|$ is also convergent. – 2017-01-28
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    I can not think of any exhaustion that solves my problem. – 2017-01-28
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    Anyone want to write to me? – 2017-01-28
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    I think I got something. I'll write. See if it's correct, please ... – 2017-01-29

1 Answers 1

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Is this solution?

Proof. Define $f_+,f_-$ parts positve and negative of $f$. Then, we have that $f=f_++f_-$ and $|f|=f_+-f_-$. Now, suppose that $\int_Uf(x)dx$ converge. Hence $\lim_{i\rightarrow \infty}\int_{K_i}f(x)dx=\lim_{i\rightarrow\infty}[\int_{K_i} f_+ + \int_{K_i}f_-]$ exist. Then, how $\int_{K_i} f_+,\int_{K_i}f_-$ converge, implies that $\int_{K_i}|f|$ converge too.

Is the reciprocal similar?

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    Your proof is incorrect. Convergence of the integral of $f$ does not imply convergence of the integrals of $f_+$ and $f_-$. – 2017-01-29
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    You have the correct solution? – 2017-01-29
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    Have you seen the proof that absolute convergence of series implies convergence? It's essentially the exact same thing. – 2017-01-29