Essentially bounded functions and simple functions

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Discussion Overview

The discussion revolves around the proof that essentially bounded functions can be expressed as uniform limits of simple functions, specifically under the conditions of sigma-finite and positive measure. The focus includes aspects of measurability and the construction of simple functions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to prove that essentially bounded functions are uniform limits of simple functions, emphasizing the need for sigma-finite and positive measure.
  • Another participant clarifies that the function in question is also measurable, which is crucial for the proof.
  • A different participant suggests that describing the limit in terms of unions and intersections of measurable sets is key to demonstrating measurability, and mentions a potential reference from Wikipedia for a proof.
  • Another contribution proposes dividing the range of the function into small intervals and constructing a simple function that approximates the original function within a specified error margin, noting that the limit is uniform almost everywhere due to the essential boundedness.

Areas of Agreement / Disagreement

Participants express various approaches to the problem, but there is no consensus on a single method or proof. Multiple competing views and techniques are presented without resolution.

Contextual Notes

Participants do not fully resolve the assumptions required for the proof, such as the specifics of measurability and the nature of the intervals used in the construction of simple functions.

Shaji D R
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How to prove that essentially bounded functions are uniform limit of simple functions. Here measure is sigma finite and positive.
 
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I need help. I forgot to indicate that the function is measurable also.
 
Trick is usually to describe limit in terms of unions, intersections of measurable sets. I mean this in order to show that the limit is measurable. for the rest, partition your domain in "enough" (compact) pieces for the vertical intervals [n, n+1). I think I remember Wikipedia had a proof.
 
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assume your function is bounded and divide up the range into small intervals. for each interval [a,b] take the functionm tohave value a on the inverse image of that interval...this gives you a simple function whichn lies within |b-a| of your function on that set...of course the limit is only uniform a.e. since the function is only essentially bounded and not bounded.
 

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