Measure theory question: Countable sub-additivity

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SUMMARY

The discussion centers on the concept of countable sub-additivity in measure theory, specifically regarding the external measure denoted as ##m^*(x)##. It establishes that for sets ##E## and ##E_j##, the property states that if ##E=\bigcup_{j=0}^{\infty}E_j##, then ##m^*(E) \leq \sum_{j=0}^{\infty}m^*(E_j)## holds true. The confusion arises from the incorrect assumption that the reverse inequality ##\sum_{j=0}^{\infty} m^*(E_j)\leq m^*(E)## could also be valid, which is clarified as incorrect. The participant resolves their misunderstanding by recognizing that ##m^* \left (\bigcup_{j=0}^{\infty}E_j \right ) \neq \sum_{j=0}^{\infty} m^*(E_j)##.

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  • Study the properties of external measures in detail, focusing on countable additivity.
  • Explore the implications of monotonicity in measure theory.
  • Review examples of sets where countable sub-additivity applies.
  • Investigate the differences between external measures and other types of measures, such as Lebesgue measure.
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Mathematicians, students of analysis, and anyone studying measure theory who seeks to deepen their understanding of countable sub-additivity and external measures.

Cascabel
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I have a question on sub-additivity. For sets ##E## and ##E_j##, the property states that if

##E=\bigcup_{j=0}^{\infty}E_j##

then

##m^*(E) \leq \sum_{j=0}^{\infty}m^*(E_j)##, where ##m^*(x)## is the external measure of ##x##.

Since ##E\subset \bigcup_{j=0}^{\infty}E_j##, by set equality, the property seems to follow from monotonicity.

However, it is also true that, ##\bigcup_{j=0}^{\infty} E_j \subset E##, which seems to imply the reverse inequality, ##\sum_{j=0}^{\infty} m^*(E_j)\leq m^*(E)##, which is not true.

What's wrong?
 
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Sorry, figured it out.

Sorry, figured it out, ##m^* \left (\bigcup_{j=0}^{\infty}E_j \right ) \neq \sum_{j=0}^{\infty} m^*(E_j)##.
 

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