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There is a famous intuitive way to prove that $\pi$ is equal to $4$, by using square of perimeter $4$ with circle inscribed in it with radius of $1/2$. And now the question which arises to me is: What makes the difference between finding the length of the circle this way and finding the area under the graph of the function by summing the areas of rectangles with very small width ? What makes first wrong and second right?

And I know that there was a question about pi being equal to four but I think this is a bit different. And I think this question cannot be answered with some rigorous proof because it cannot be rigorously proved that area under the curve is really equal to Riemann integral or any other integral because you need to define area in the first place, which is made by Lebesgue measure which itself is an idea of somewhat "limitation". Correct me if I am wrong in any of my sentence. Thanks in advance.

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Юрій Ярош
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  • You can link to this question https://math.stackexchange.com/questions/12906/the-staircase-paradox-or-why-pi-ne4 – Yanko Sep 24 '18 at 11:16
  • In fact that linked question may contain an answer to your question. So please explain in what sense your question is different than the already existing one. – Yanko Sep 24 '18 at 11:18
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    @Yanko Your linked question is about debunking the $\pi = 4$ argument, while this question is about why, in the light of said debunking, we can still use Riemann integration. To me that's a related question, but different enough that it deserves its own existence. You might find an answer if you read each answer in the linked post looking for it, but I don't think that that by itself is enough to close this one. – Arthur Sep 24 '18 at 11:28
  • I think the basic idea (which I will not attempt to prove here) is that if you try to measure the perimeter of a circle as the limit of the perimeter of polygons whose vertices get closer and closer to the circle, you get different answers depending on how you choose your polygons, so it can't possibly be a valid method of measurement. On the other hand, if you try to measure the area of a circle in a similar way, you always get the same answer no matter how you choose your polygons, as long as their vertices approach the circle. So we take that limiting value to be the area of the circle. –  Sep 24 '18 at 11:38
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    @Yanko Actually, it contains a partial answer in the second paragraph of the accepted answer to the question you linked. – Юрій Ярош Sep 24 '18 at 11:47
  • In short, it comes down to whether or not the error in the approximation can be made small enough. – amd Sep 24 '18 at 19:28
  • @amd In the example with a circle the error of approximation doesn't change, but what this example tries to show is that the length of the square is essentially the same by kind of "imposition at infinity".So it tries to show that error is always zero. By the way, there is a example similar to the first, where the error of approximation is not constant, but it gives wrong result, but I don't know whether error tends to zero. You can look at it here on 5:10 https://www.youtube.com/watch?v=Rv0c7R8brjE. – Юрій Ярош Sep 25 '18 at 10:10
  • For both the "$\pi=4$" argument and the path shown at 5:10 in that video, the thing that is going to zero is the difference between the lengths of two consecutive "approximations". (In one case the difference goes to zero instantly, in the other it goes to zero in the limit.) But showing that your measurements converge to each other says nothing about the error of the measurements, that is, how far are they from the thing you're trying to measure. The error in each of these "approximations" begins large and remains large. – David K Sep 25 '18 at 12:04
  • @DavidK Ok. It seems reasonable, but I have one more question if you don`t mind. While I see why when defining area with e.g. Darboux Integral it is obvious that we get closer to the area by finding the lower and upper sums and that area is between them and that it should be equal to their sup and inf respectively, in the case sup is equal to inf, but how do we know that error of measurments is getting smaller when we do polygonal approximations of circumference ? – Юрій Ярош Oct 03 '18 at 09:32
  • That question seems much more difficult to me. We might just define length by polygonal approximation. We would of course insist that all vertices of the polygon are actually on the curve and not somewhere else, which immediately rules out the "staircase" measurement of the circle, but there are other concerns that must also be dealt with. I'd be interested in a good answer to that question too. – David K Oct 03 '18 at 11:55

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