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June 4th, 2017, 02:04 AM   #1
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Joined: Jan 2016
From: United Kingdom

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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
 image1 (4).jpg (95.0 KB, 6 views) image2 (1).jpg (95.3 KB, 3 views) image3.jpg (94.3 KB, 4 views)

 June 17th, 2017, 04:14 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,240 Thanks: 884 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"?

 Tags consistency, functions, step

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