Undergrad Can a Finite Measure Space Have Uncountably Many Positive Measure Members?

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A finite measure space cannot contain uncountably many members in its σ-algebra with strictly positive measure. This is due to the properties of measures and the definition of finite measure spaces, which limit the total measure available. If there were uncountably many sets with positive measure, their total measure would exceed the finite measure of the space. The problem remains unsolved by participants, with the original poster providing a solution. The discussion emphasizes the constraints of measure theory in finite contexts.
Euge
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Here is this week's POTW:

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Prove that no finite measure space can have uncountably many members in its $\sigma$-algebra with strictly positive measure.

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No one answered this week’s problem. You can read my solution below.

Suppose $(X,M,\mu)$ is a finite measure space with uncountably many $A\in M$ such that $\mu(A) > 0$. There exists an $n \in \Bbb N$ such that uncountably many $A\in M$ with $\mu(A) > 1/n$. If $A$ is the countable disjoint union of sets $A_i\in M$ of measure greater than $1/n$, then $A\in M$ and $\mu(A) = \sum \mu(A_i) = \infty$. This is a contradiction.
 

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