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July 16th, 2017, 07:19 AM   #1
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On Step Functions - neat typeups of ideas

Hello all,

For a few weeks I've been working on this neat copy of my own attempt to prove some things about step functions. The image links below take you to snapshots of what is around two pages worth of lemmas and one small theorem at the end. I've included the links in order from top to bottom, but they overlap just in case. I have tried to be very precise and as such, appreciate any time anyone takes to give this stuff a look:

https://gyazo.com/d84aea07bc005c9817a1bc69098ef155

https://gyazo.com/72f6a772ec1c76202c0b604f8edbdff4

https://gyazo.com/97e29161ed025d4813d1ec79dea5930b

https://gyazo.com/94abc31cda51493b66a30368630730f3


Regards,

Magnitude.
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July 16th, 2017, 07:35 AM   #2
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This is pretty neat stuff. You prove accurately that the definition of the integral of a step function is well defined.

I have some pretty minor remarks, that might help you:
* An indicator function is usually written with \chi, thus $\chi_A$ and not $X_A$. (especially since you use $X_i$ to denote a set at some times)
* When you write a sum, you shouldn't write brackets if it's the sum over a product. So $\sum_i c_iA_i$ is fine, while $\sum_i (c_iA_i)$ is pretty tedious and not done. It's not that it's wrong, it's just that people will look funny if you write it.
* I'm not sure how your Proposition 2.3.4 deals with the step function $\chi_{[0,1]} + \chi_{[1,2]}$. You will need to take three disjoint intervals here: $[0,1)$, $\{1\}$ and $(1,2]$ and I don't think that actually happens.
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July 16th, 2017, 09:39 AM   #3
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Micrm@ass,

I think for your example of those two intervals, my method would see [0,1] and [1,2] get split into (-\infty,0) , [0,0] , (0,1) , [1,1] , (1,2) , [2,2] and [2,infty). Then, the unbounded intervals vanish as they are not a part of the two original intervals and the rest goes from there.
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July 16th, 2017, 10:34 AM   #4
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Quote:
Originally Posted by Magnitude View Post
Micrm@ass,

I think for your example of those two intervals, my method would see [0,1] and [1,2] get split into (-\infty,0) , [0,0] , (0,1) , [1,1] , (1,2) , [2,2] and [2,infty). Then, the unbounded intervals vanish as they are not a part of the two original intervals and the rest goes from there.
No, since there are only $n=3$ DISTINCT end points in my example, which in your proposition generates 2 bounded intervals instead of the needed 5.

So the issue I'm taking is with your sentence "Suppose the intervals $I_1$, ..., $I_r$ have $n$ distinct end points"
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July 16th, 2017, 12:49 PM   #5
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i see. But i am unsure of how to ammend this?
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July 16th, 2017, 01:31 PM   #6
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It's just an easy modification of your proof. Most of it is correct, you just need to account for certain overlaps.

Try to make a canonical form $$f = \sum c_i \chi_{(a_i, a_{i+1})} + \sum d_i \chi_{\{a_i\}}$$ So instead of arbitrary intervals, you reduce to open intervals plus certain values on the boundary.
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