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Let $c>0$. Let $X \subseteq (0,\infty)$ be a Lebesgue measurable set. Define $$ cX := \{ cx \mid x \in X \}. $$ Then $$ \int_{cX} \frac{dt}{t} = \int_{X} \frac{dt}{t}$$

Now I can prove this for $X$ an interval and, thus, any set generated by set operations on intervals. It is simply by using the Fundamental Theorem of Calculus and natural log $\ln$. But I'm not sure how to approach for general Lebesgue measurable set.

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    Use substitution aka transformation formula for integrals.2012-10-05
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    Have you done the analogous proof $\int_X dx = \int_{X+c}dx$2012-10-05
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    @martini Yes, I think that's a very clever way. But unfortunately I know no thoerem of that kind in the context of abstract measure theory. Do you know any? And maybe a reference book?2012-10-05
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    @GEdgar Yes, though not exactly your eqn but kinda simialr, namely, $\int_{X} f(x) dx = \int_{X+c} f(x-c) dx$. I know this holds by the translation invariance property of Lebesgue measure.2012-10-05
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    @julypraise Please use proper English.2012-10-06
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    What you consider here is the Lebesgue measure (or if you wish the Haar-measure) on the multiplicative group $\mathbb{R}^+$. It reflects $\log(cX) = \log c +\log X$.2012-10-06

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