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The question says it. Let $A = \{\,x \,|\, 0 < x \le 10\,\}$. What would be the average of all the items in this set? How do you prove it? UPDATE $x$ belongs to the set of real numbers.My thoughts:Is it possible to find the average of the items of $A = {x|0 <= x <= 10}$ where $x$ belongs to the set of real numbers?If it is, why couldn’t we use the same approach for my question?

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    What are your thoughts? – Christoph Dec 17 '18 at 20:53
  • To what set do the $x$'s belong? If it's $\mathbb N$, then you will be able to add and average the items. But if it's $\mathbb Q$ or $\mathbb R$, you will have to change your question since these the interval for these 2 sets has an infinite number of elements. – MasB Dec 17 '18 at 20:59
  • $x$ belongs to the set of real numbers –  Dec 17 '18 at 21:02
  • For a closed interval we have the average value function from calculus. But for open/non closed intervals this is interesting – T. Fo Dec 17 '18 at 21:05
  • Yes, it's possible. What kind of class do you need it for? Calculus? Probability theory? – Botond Dec 17 '18 at 21:11
  • It’s for probability theory. –  Dec 17 '18 at 22:36

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Considering $(0,10]$ or $[0,10]$ as subspaces of $\mathbb R$ with the Lebesgue measure $\mu$, the question of "average" is a measure theoretic (or probability theoretic) one. While average of a finite set $X\subset\mathbb R$ may be defined as $$ \frac{1}{|X|} \sum_{x\in X} x, $$ we can define the average of a set $X\subset\mathbb R$ of finite measure $\mu(X)<\infty$ as the Lebesgue integral $$ \frac{1}{\mu(X)} \int_X x\,\mathrm d\mu(x). $$ When $X=(0,10]$ or $X=[0,10]$ you have $\mu(X)=10$ and obtain the average $$ \frac{1}{10} \int_0^{10} x\,\mathrm dx =5. $$

Christoph
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