Mathematics Asked by Alan Simonin on January 17, 2021
This is a question in 2 steps :
Let $H$ be a vector field of bounded functions $Omega rightarrow mathbb{R}$. Assume that $H$ contains the constant functions and has the following property : If $(f_n)$ is an increasing and bounded function series of positive functions of $H$ then $f:=limf_n in H$. Let $C$ be a subset of $H$ stable under multiplication. Then $H$ contains all bounded $sigma(C)$-measurable functions i.e. $L_b^0(sigma(C)) subset H$
Now, if we take a bounded $sigma(C)$-measurable function, it is also a function in $H$. But the properties of $H$ are too vague and not very useful for us, right ? It does not have any property related to measurability.
I know that this theorem is used to prove that if two probability functions are equal on a generating part, stable under $cap$ then they are equal on the whole set as well. But how can we see that on this theorem ?
If $A$ is an algebra on a set $Omega$, the the smallest monotone class containing $A$ is equal to the smallest sigma-algebra containing $A$
I can see that the property on $H$ saying that if $(f_n)$ is an increasing and bounded function series of positive functions of $H$ then $f:=limf_n in H$ is written monotone class all over it, but do we have that the smallest monotone class containing $C$ is equal to the smallest sigma-algebra containing $C$ ? Is that the point of the functional monotone class theorem ?
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