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Lets create a function that only takes irrational numbers, $\overline {\mathbb{Q}}$. Lets have $f(x)=c \cdot x$, but x can only be irrational. How would one go about finding the measure, loosley speaking the area under the graph, of the function, within a bounded region that is. Is it $0$ or something else? My intuition brings me to believe the measure of any function defined over the irrationals is just equal to the integral, again loosely speaking, since the cardinality of irrationals is greater than the cardinality of the rationals. If my guess is correct, that should imply that any bounded function defined over $ \mathbb{Q}$ has a measure of $0$.

Asaf Karagila
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Zach466920
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1 Answers1

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$$\mathbb Q \text{ and } \overline {\mathbb Q}$$ are Lebesgue measurable sets. One can define the measurable function

$$\overline f(x)=\begin{cases}cx,& \text{ if } x \in \overline{\mathbb Q}\\ cx,&\text{ if } x\in \mathbb Q.\end{cases}$$

Then, for a bounded measurable $A$:

$$\int_A \overline fd\lambda=\int_{A\setminus\mathbb Q}\ \ cx\ d\lambda +\int_{A \cap\ \mathbb Q}cx\ d\lambda=\ \int_Af d\lambda.$$

Because, considering $\{r_i, i=1,2,...\}$, an ordering of the rationals, we have

$$\int_{A \cap\mathbb Q}cx\ d\lambda=\sum _{r_i \in A \cap\mathbb Q}c\ r_i\lambda(\{r_i\})=0.$$

zoli
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  • Thankyou, I interpreted $\lambda$ as the measure density function, is this correct? Could you elaborate a bit on how you define the function $\lambda(n)$. It seems you are using the cardinality of the rationals to evaluate the integral to $0$. Is this correct? – Zach466920 Mar 31 '15 at 17:12
  • $\lambda$ is not a density function. $\lambda$ is the extension of the simplest "measure" defined on intervals by their length. Measure theory (https://en.wikipedia.org/wiki/Measure_(mathematics)) is about the ways that this extension is possible on the real line or even more abstract sets. Otherwise: yes. The basis of the argumentation is related to the cardinality of the rationals. – zoli Mar 31 '15 at 17:15
  • Ok, that's fine. So $\lambda({r_i})$ approaches $0$ as the measure, length, of the interval becomes 0 for rationals. Could you point me in the right direction to prove the summation over $t_i$ where $t_i$ is an element of the irrationals doesn't evaluate to $0$? – Zach466920 Mar 31 '15 at 17:24
  • Summation over an uncountable (index) set is not defined in measure theory. I think that the whole measure theory was developed by Borel and Lebesgue and others because people wanted to avoid such paradoxical problems. This, however, cannot be further explained via comments. Please, consult the wiki page I suggested above. – zoli Mar 31 '15 at 17:33
  • I've read that, that's why I was questioning your summation... – Zach466920 Mar 31 '15 at 17:38
  • Are you still questioning my summation? My summation is a humble one: it is over a countable set of zeros; $c\times r_i \times 0$ is zero for all $i$. And we have only countable many rationals in A. – zoli Mar 31 '15 at 17:41
  • That does make sense, but your answer seems to implictly say otherwise since the integral value and the integral over the rationals are known...and implies the integral over the irrationals is equal to the integral over the reals. – Zach466920 Mar 31 '15 at 17:53