 # It is my understanding that &#x03C3;<!-- σ --> -algebras are necessary to satisfy certain desir Semaj Christian 2022-06-25 Answered
It is my understanding that $\sigma$-algebras are necessary to satisfy certain desired properties of a measure, and that these conditions are mutually inconsistent if we consider arbitrary open sets of $\mathbb{R}$. However, I have also read that power set of a set is necessarily a $\sigma$-algebra, so how come we cannot use the power set of $\mathbb{R}$ to define a measure space? I feel I am missing something fundamental here.
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There are disjoint subsets $A$ and $B$ of $\mathbb{R}$ such that
$\mu \left(A\cup B\right)<\mu \left(A\right)+\mu \left(B\right)$
violating the additivity condition of a measure μ induced from the "length" function on intervals. The collection of (Lebesgue) measurable sets is strictly smaller than the powerset.
EDIT: It is pointed out in the comments that the powerset is an adequate sigma-algebra for other measures like "atomic" measures on $\mathbb{R}$. Thus your intuition is not far off.

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