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I'm stariting to study probability and some really interesting questions starting to bother me. Let's consider the unit circle $C$ and $D$ - the circle with radius $\frac{1}{2}$. I know that the probability of randomly picked point $p$ of C to be in $D$ too is $\frac{\text{area of } D}{\text{area of } C} = \frac{1}{4}$, while the probability $p$ to lie over any particular line through $C$ is $0$. People say that this is because the line does not have area, while the circle has. So I guess this is the case with curves as well, but if the curve is very thick (like $\sin \frac{1}{x}$ near $0$) is the probability of point to lie over such curve still $0$?

brick
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    It is still measure zero, so, yes, the probability is zero. – Michael Burr Apr 05 '15 at 15:49
  • It would be clearer in this context if you said disk or ball rather than circle. To mathematicians the circle is usually the boundary of the disk. – Ian Apr 05 '15 at 16:06
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    It is interesting that you meantioned $\sin\left(\frac1x\right)$. You should see this question. – Eff Apr 05 '15 at 16:24

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Interesting question! If the curve is very very thick, we can get non-zero probability. For example there is a curve that goes through every point of the unit square. (Please see Wikipedia, Space Filling Curve.) But $y=\sin(1/x)$ is not thick enough.

Remark: One should give at least an informal argument that our curve is not thick enough. Let $\epsilon\gt 0$ be small positive. The part of the curve from $-\epsilon/4$ to $\epsilon/4$ has measure $\le$ the area of the rectangle with base $\epsilon/2$ and height $2$, so has measure $\le \epsilon$. And the part of the curve with $|x|\gt \epsilon/4$ is well-behaved, "thin."

André Nicolas
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  • What do you mean by 'measure' of the curve? – brick Apr 05 '15 at 18:22
  • I am using the term informally, though it can be made formal. The curve has (two-dimensional) Lebesgue measure $0$. Roughly speaking this means that for any $\epsilon\gt 0$ we can find a region of area $\le \epsilon$ which completely contains the curve. Somewhat more precisely, there is a collection $D_1,D_2,\dots$ of open disks such that the unuion of the $D_i$ contains the curve and the sum of the areas of the disks is less than $\epsilon$. – André Nicolas Apr 05 '15 at 18:28