# The Unapologetic Mathematician

## Measurable Spaces, Measure Spaces, and Measurable Functions

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April 26, 2010 - Posted by | Analysis, Measure Theory

## 14 Comments »

1. is it enough $$mathcal{S}$$ be a $$\sigma ring$$ a probability book i’m studying says the space needs a $$\sigma – Field$$
Could you clarify please

thanks

Comment by cappa | April 27, 2010 | Reply

2. A $\sigma$-field is a different name for a $\sigma$-algebra. As I point out above, in many cases $\mathcal{S}$ actually is a $\sigma$-algebra (or -field). In this case, $X$ itself is in $\mathcal{S}$, and so it’s clear that every point in $X$ is in some measurable set.

However, we’re going to allow $\mathcal{S}$ to just be a $\sigma$-ring — $X$ itself might not be in $\mathcal{S}$ — so long as every point in $X$ is still in some measurable set.

Comment by John Armstrong | April 27, 2010 | Reply

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11. Is every measurable space is topological space

Comment by M. A. Hossain | September 27, 2011 | Reply

12. Not in any natural way. The obvious thing would be to try to use the $\sigma$-algebra of measurable sets as the collection of open sets in the topology, but it’s not necessarily closed under arbitrary unions.

Comment by John Armstrong | September 27, 2011 | Reply

13. Actually my query is as follows:
If T is a sigma-algebra In X, is T also a topology in X?

Comment by M. A. Hossain | September 28, 2011 | Reply

14. No, as I said; $T$ is not necessarily closed under arbitrary (uncountable) unions.

Comment by John Armstrong | September 28, 2011 | Reply