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June 4th, 2017, 03:04 AM   #1
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Consistency of step functions

Hi,

In the images attached, I have attempted to prove that two equal step functions (which may not exist over the same intervals or have equal coefficients) evaluate to the same integral.

The first proof is that every step function can be represented as a step function over disjoint intervals. I call this the disjoint equivalent. Then, It is shown that any such equivalent may be partitioned further and that disjoint equivalent will have the same integral as the original function. Then, I take two step functions equal to one another, and represent them via a "common" equivalent ; one which can be obtained for both step functions using part 2. The result follows.

Not sure if this is correct, I understand it's a trivial thing to try to prove but of course we want an integral operator to be well defined, right?

M
Attached Images
File Type: jpg image1 (4).jpg (95.0 KB, 6 views)
File Type: jpg image2 (1).jpg (95.3 KB, 3 views)
File Type: jpg image3.jpg (94.3 KB, 4 views)
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June 17th, 2017, 05:14 AM   #2
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If they "may not exist over the same intervals or have equal coefficients" then what does it mean for two step functions to be "equal"?
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