Talk:Content (measure theory)

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This appears to be just the same thing as a "finitely-additive measure"; am I wrong in concluding that? -Chinju (talk) 00:40, 27 January 2008 (UTC)


 * I think you're right. Adammanifold (talk) 18:01, 5 November 2010 (UTC)

relationship between content (measure theory) and pre-measure?
hi guys,

so i saw the wiki for pre-measure and it says the reason it's not a measure is because it's not defined on a sigma-algebra.
 * that part is cool.

however, it seems to me that the content is more of a "precursor" to a measure than a pre-measure.
 * it seems content needn't be defined over a sigma algebra (hence a measure being a sigma-additive content), so wouldn't the content be more of a pre-measure since it's not in terms of ring theory, but sets?

in the case where the content is not defined over a sigma algebra, surely there is some sort of way we could tie it in with pre-measure?
 * i understand the latter is speaking in terms of rings, but so is Carathéodory's extension theorem, which is a measure-theoretic concept that could bridge these ideas together?

both content and pre-measure are sigma additive, too. surely i'm not off my rocker (I HOPE?! ;)) ? ty in advance for the chat guys. 174.3.155.181 (talk) 20:34, 7 July 2016 (UTC)


 * You are not off your rocker. content and pre-measure are similar concepts, the big difference being that a content is finitely additive and a pre-measure countably additive. --Mark viking (talk) 11:28, 20 July 2016 (UTC)

Content definition too broad?
I do not believe the stated definition of binary disjoint additivity implies finite additivity (which I assume one wants). For example, take the semiring $$\big\{\emptyset, \{1\}, \{2\}, \{3\}, \{1,2,3\}\big\}$$ and consider the content $$\mu(\emptyset) = 0,\quad \mu\{1\} = 1, \quad \mu\{2\} = 1, \quad \mu\{3\} = 1, \quad \mu\{1,2,3\} = 0.$$ This satisfies binary disjoint additivity, but not finite additivity. (And thus also not monotonicity as claimed in the article). --164.15.254.98 (talk) 10:06, 16 February 2023 (UTC)


 * Yep, I had the same concern, and this example seems to show we really do want finite disjoint additivity rather than just binary disjoint additivity. It's worth noting that this also contradicts the monotonicity claim under properties. But monotonicity will always hold for a content defined on a ring (in which case binary disjoint additivity does imply finite disjoint additivity), so the content you've defined cannot be extended to a content on the ring generated by your semiring (thus contradicting the page on Carathéodory's Extension Theorem).
 * I suggest the definition be changed on this page to include finite disjoint additivity rather than binary disjoint additivity. It could be noted that so long as the collection is closed under binary unions (for example, a ring), these two properties are equivalent. AJ LaMotta (talk) 17:01, 13 January 2024 (UTC)
 * Just made an edit to the page changing the definition from binary disjoint additivity to finite disjoint additivity. AJ LaMotta (talk) 17:13, 13 January 2024 (UTC)