Essentially bounded functions and simple functions

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SUMMARY

This discussion focuses on proving that essentially bounded functions are the uniform limit of simple functions under the conditions of sigma-finite and positive measures. The key approach involves using measurable sets to describe limits through unions and intersections, ensuring the limit remains measurable. The method includes partitioning the domain into compact pieces and dividing the range into small intervals, allowing the construction of simple functions that approximate the essentially bounded function within a specified tolerance. The conclusion emphasizes that the uniform limit holds almost everywhere due to the nature of essential boundedness.

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  • Understanding of measurable functions and sigma-finite measures
  • Familiarity with the concept of simple functions in measure theory
  • Knowledge of limits and uniform convergence in functional analysis
  • Experience with partitioning domains and constructing approximations
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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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