Im not sure about the above question. Im guessing that there is none, else the question would probably not be asked that way, but i can't really pinpoint where the contradiction lies.
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1What is $\Omega$, or what are you assuming about it? There certainly is in some cases... – Robert Israel Jul 20 '15 at 20:27
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Perhaps it's standard notation, but to me, unknown. What is $\Omega$? – ZenoCosini Jul 20 '15 at 20:28
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Im sorry, in our class $\Omega$ is a non-empty, arbitrary set. In this case there should be an $\mathbb{R}$. – InspectorPing Jul 20 '15 at 20:34
2 Answers
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Hint:
Strict monotonicity implies injectivity on the domain.
ZenoCosini
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oh, I wrote it just wrong. Strictly monoton only if one to one. sorry. – user251257 Jul 20 '15 at 20:43
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And what is the connection between being injective and being integrable? – Mariano Suárez-Álvarez Jul 20 '15 at 21:09
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Hint: Assume without lost of generality $f:\mathbb R\to[0,\infty)$ is strictly increasing. Then, we have: $\int_{\mathbb R} f(x) \; \mathrm dx \ge \int_0^\infty f(x) \; \mathrm dx \ge \int_0^\infty f(0) \; \mathrm dx$.
Solution:
From $0\le f(-1) < f(0)$ follows $\int_0^\infty f(0) \mathrm dx = \infty$. Thus, $f$ is not integrable.
user251257
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