Value of a measure theoretic integral over a domain shrinking to a single set

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SUMMARY

The equality f(x)=\lim\limits_{\Omega\rightarrow\{x\}} \frac{1}{\mu(\Omega)}\int_\Omega f d\mu holds under specific conditions related to the Lebesgue differentiation theorem. The discussion emphasizes that for a sequence of sets \Omega_i shrinking to a single point x, the limit \lim\limits_{i\rightarrow\infty} y_i must equal f(x) if the function f is Lebesgue integrable over the domain and the measure \mu is appropriately defined. The convergence of the integral averages y_i to the function value at the point x is contingent on the properties of the measure and the behavior of the function near x.

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  • Understanding of Lebesgue integration
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  • Knowledge of the Lebesgue differentiation theorem
  • Basic calculus and limits
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  • Explore properties of Lebesgue integrable functions
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Mathematicians, students of analysis, and researchers in measure theory who are interested in the properties of integrals and their convergence behavior in relation to pointwise evaluations.

peb78
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Hi. Under what conditions does the following equality hold?

[itex]f(x)=\lim\limits_{\Omega\rightarrow\{x\}} \frac{1}{\mu(\Omega)}\int_\Omega f d\mu[/itex]

where [itex]\mu[/itex] is some measure. Being a little more careful, let [itex]\Omega_i[/itex] be a sequence of sets such that [itex]\Omega_{i+1}\subseteq\Omega_i[/itex] and

[itex]\bigcap\limits_{i=1}^{\infty} \Omega_i=\{x\}[/itex].

Then, define the consider the sequence [itex]\{y_i\}_{i=1}^\infty[/itex] where

[itex]y_i=\frac{1}{\mu(\Omega_i)}\int_{\Omega_i} f d\mu[/itex]

Under what conditions does [itex]\lim\limits_{i\rightarrow\infty} y_i=f(x)[/itex]?
 
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That's exactly what I'm looking for. Thanks!
 

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