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For a test prep question:

Let $g(x)=\tanh (x)$, and let $\mu$ be the measure generated by $g$. Which subsets of the reals are $\mu$-measurable? Are polynomials integrable? And finally, is $|\sinh|$ integrable?

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Defining the measure of a set $E$ as the integral of $g'(x)$ over that set. Then, I think every set will be measurable as long as its characteristic function is measurable too, in the Lebesgue measure. I don't know how to relate the two measures at hand now. – FPP Sep 13 '12 at 19:41
And I don't think the polynomials are integrable, since they diverge to infinity or minus infinity. – FPP Sep 13 '12 at 19:42
Since $\int f(x) d \tanh(x) = \int f(x) \mathbb{sech}^2(x) dx$ and $\int |f(x) |\mathbb{sech}^2(x) dx \leq 2\int |f(x)| e^{-2x}dx$, it follows that all polynomials are integrable. It also follows from this that $\sinh$ is integrable. – copper.hat Sep 13 '12 at 19:48
up vote 1 down vote accepted

Measurable sets are the same as the usual case because $\tanh$ is strictly increasing. Integration with function $f$ becomes $\int f(x)\tanh'(x) dx = \int \frac{f(x)}{\cosh^2(x)} dx$. Substituting a polynomial or $|\sinh|$ for $f$ will make the integral finite, so polynomials and $|\sinh|$ are integrable. (The domain is $(-\infty, \infty)$.)

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Ah, nice, thanks. Another question... How could you find out the value of the integral of $\sinh (x)$? – FPP Sep 13 '12 at 20:05
$\int_0^\infty \frac{\sinh}{\cosh^2(x)} dx = \int_0^{\infty} \operatorname{sech}(x)\tanh(x)dx = -\operatorname{sech}(x)|_0^{\infty} = 1$. – Tunococ Sep 13 '12 at 20:09

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