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Let's begin with a definition:

Let $U\subset \mathbb R^m$ be an open set and $K_i$ be a sequence of compact, J-measurable sets such that $U=\cup K_i$ and $K_i\subset\operatorname{int}K_{i+1}$ for every $i\in \mathbb N$. Given a continuous function $f:U\to \mathbb R$, we say the integral $\int_Uf(x)dx$ is convergent when for every such sequence $\{K_i\}$, there exists $\lim_{i\to \infty}\int_{K_i}f(x)dx$.

I want to prove the integral $\int_Uf(x)dx$ is convergent if and only if $\int_U|f(x)|dx$ is convergent.

I've already written down an attempt of solution but I got lost with the indices and the sups and infs. I need some hints how to proceed.

I need help.

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    This was just asked 25 minutes ago: http://math.stackexchange.com/questions/2118173/int-ufxdx-converge-iff-int-ufxdx-converge – 2017-01-28

2 Answers 2

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Hint: one direction is easy, just bound $f$ by $|f|$ to show the integral is finite. The other direction is a little trickier. Suppose that $\int |f|$ is not convergent. Then it must be that one of $\int f^+$ or $\int f^-$ is not convergent (else their sum would be convergent). Suppose $f^+$ is the bad one. Then Use a bad sequence of $K_i$ for $f^+$ and either a good sequence $K'_i$ for $f^-$ or a bad sequence that has $\int_{K'_i} f^-$ growing much slower than $\int_{K_i}f^+$ to get a single sequence $K_i \cup K'_i$ that has $\int_{K_i \cup K'_i} f$ going to $\infty$.

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    Thank you for your answer. Why if $\int |f|$ is not convergent, then $\int f^+$ or $\int f^-$ is not convergent? – 2017-01-28
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    Because if $\int f^+$ and $\int f^-$ are both convergent, then $\int |f|=\int f^+ -\int f^-$ is convergent too. – 2017-01-28
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    I would appreciate if you could explain a little bit which means "bad sequence" of $K_i$ for $f^+$? I'm struggling all day to get this idea. Thank you again – 2017-01-29
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    In the case that $\int f^+$ is not convergent, by definition there is a sequence of $K_i$s with $K_i \subset \mathrm{int}(K_{i+1})$ and $U = \cup_i K_i$ such that $\int_{K_i} f^+(dx)\,dx = \infty$. – 2017-01-29
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Suppose $f_+$ and $f_-$ the positive and negative parts of $f$, respectively. Then we can write $f = f_+ + f_-$ and $| f | = f_+ - f_-$. We know that if the integral of $f$ converges then its positive and negative parts converge as $| f |$ Converge. The return is analogous.