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In a previous question I had $A \subset \bigcup_{k=1}^\infty R_k$ where $R_k$ in $\Bbb{R}^n$ are rectangles I then proceeded to use the following inequality $\left|\int_A f\right| \le \left|\int_{\bigcup_{k=1}^\infty R_k} f \right|$ which I am not really certain of. Does anyone know how to prove it? If its wrong what similar inequality should I use to prove the result here.

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user63697
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2 Answers2

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The inequality is not true in general (think of an $f$ that is positive on $A$ but such that it is negative outside $A$). But it does hold if $f\geq 0$. This is not an obstacle to you using it, because you would just have to split your function in its positive and negative part.

Martin Argerami
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  • I was thinking of using $| \int_A f| \le \int_A |f| \le \int_{ \bigcup_{k=1}^\infty R_k} |f|$ since $|f|>0$ and using $\sup |f|$ instead of $|\sup f|$ which doesn't change much – user63697 May 27 '13 at 16:29
  • @user63697: You mean $|f|\geq 0$, and the inequalities in your comment are correct. What do you mean by "using $\sup |f|$ instead of $|\sup f|$ which doesn't change much"? – Jonas Meyer May 27 '13 at 16:36
  • @JonasMeyer I'm referring to the linked question. – user63697 May 27 '13 at 16:41
  • @user63697: Thank you, that might help me guess, but doesn't really answer my question. My guess is that it will be helpful for you use the inequality $|\sup f|\leq\sup|f|$. "Doesn't change much" is vague, and presently meaningless to me out of context. – Jonas Meyer May 27 '13 at 16:44
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    @JonasMeyer I mean this inequality $|\int_{\bigcup_{k=1}^\infty R_k} f |\le |\sup f| \sum_{k=1}^\infty \mu(R_k)$ will be replaced by $\int_{\bigcup_{k=1}^\infty R_k} |f| \le \sup |f| \sum_{k=1}^\infty \mu(R_k)$ so instead of setting $|\sum_{k=1}^\infty \mu(R_k) < \epsilon / |\sup f|$ I set it to be less than $ \epsilon / \sup |f| $ . I mean it doesn't change much in the way the result is reached. – user63697 May 27 '13 at 16:49
  • @Jonas: yes, you are right. – Martin Argerami May 27 '13 at 18:30
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For $A \subset B$, $$ \left|\int_A f\right| \leq \left|\int_B f \right| $$ is true if either $f(x) \geq 0$ for all $x \in B$, or $f(x) \leq 0$ for all $x \in B$.

For a counter-example if this isn't the case, let $f(x) = x$, $A = [0,2]$, $B=[-2,2]$. Then $$ \left|\int_A f\right| = 2 > 0 = \left|\int_B f \right| \text{.} $$

In particular, you thus always have that $$ \int_A |f| \leq \int_B |f| \ $$ if $A \subset B$.

Jonas Meyer
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