I have this:Theorem. If E is any class of sets and if A is any subset of X, then S(E) ∩ A = S ( E ∩ A ) .Proof. Denote by C the class of all sets of the form B ∪ ( C − A ), where B ε S ( E ∩ A ) and C ε S ( E ) ;it is easy to verify that C is a σ-ring.We are using Paul Halmos's Measure Theory as our textbook. One of the proofs provided is incomplete. I don't like learning things without knowing the full proof. Is there full proof that shows why C is a σ-ring?I know I need to take two sets G , H in C and prove that G − H ∈ C But that means I have to show that ( B G ∪ ( C G − A ) ) − ( B H ∪ ( C H − A ) ) ∈ C , which is just too complicated to me. I am not sure how it is easy to verify.

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I have this:Theorem. If       E   is any class of sets and if       A   is any subset of X, then  S(E)     ∩      A    =      S    (    E       ∩      A    )  .Proof. Denote by       C   the class of all sets of the form B   ∪   (  C  −      A    ), where  B     ε         S    (    E         ∩          A        )       and     C     ε         S    (    E    )  ;it is easy to verify that       C   is a   σ-ring.We are using Paul Halmos&#039;s Measure Theory as our textbook. One of the proofs provided is incomplete. I don&#039;t like learning things without knowing the full proof. Is there full proof that shows why       C   is a   σ-ring?I know I need to take two sets   G  ,  H in       C   and prove that   G  −  H  ∈      C  But that means I have to show that   (      B    G    ∪  (      C    G    −  A  )  )  −  (      B    H    ∪  (      C    H    −  A  )  )  ∈      C  , which is just too complicated to me. I am not sure how it is easy to verify.

Sydnee Villegas · Accepted Answer

If we define       C   a bit differently it becomes easier to see that it is a   σ-ring. To give a better definition, let&#039;s aks the question: why do we even care about       C  ?The answer is that it is easy to see that       E    ⊂      C   and       C    ∩  A  =      S    (      E    ∩  A  ), and hence if it is a   σ-ring,       S    (      E    )  ⊂      C   and       S    (      E    )  ∩  A  ⊂      S    (      E    ∩  A  ).A better definition of       C   then, is       C    :=  {  C  ⊂  X  ∣  C  ∩  A  ∈      S    (      E    ∩  A  )  }.This class of sets truly is easy to verify is a   σ-ring. For let       C    1    ,      C    2    ∈      C  , then  (      C    1    −      C    2    )  ∩  A  =  (      C    1    ∩  A  )  −  (      C    2    ∩  A  )And for   {      C    n    }  ⊂      C   a countable collection            (        ⋃      C    n              )        ∩  A  =  ⋃    (      C    n    ∩  A  )

I have this: Theorem. If <mrow class="MJX-TeXAtom-ORD"> <mi class="MJX-tex-mathit" mathvar

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