Let μ, ν be finite measures on the non-degenerate compact interval [ a , b ] ⊆ R provided with the Borel σ-algebra. It is well-known that if μ ( B ) = ν ( B ) for every Borel set B belonging to a π-system that generates B ( [ a , b ] ) and that includes [ a , b ], then μ = ν. Is the same true if the equality is replaced by an inequality? In other words, suppose μ ( B ) ≤ ν ( B ) for every Borel set B belonging to a π-system that generates B ( [ a , b ] ) and that includes [ a , b ]. Is it the case that μ ≤ ν?

Question

Let   μ,   ν be finite measures on the non-degenerate compact interval   [  a  ,  b  ]  ⊆  R provided with the Borel   σ-algebra. It is well-known that if   μ  (  B  )  =  ν  (  B  ) for every Borel set   B belonging to a   π-system that generates   B  (  [  a  ,  b  ]  ) and that includes   [  a  ,  b  ], then   μ  =  ν. Is the same true if the equality is replaced by an inequality? In other words, suppose   μ  (  B  )  ≤  ν  (  B  ) for every Borel set   B belonging to a   π-system that generates   B  (  [  a  ,  b  ]  ) and that includes   [  a  ,  b  ]. Is it the case that   μ  ≤  ν?

Dreforganzv · Accepted Answer

No, consider the following   π system:  C  =  {  [  a  ,  t  )  ,  [  a  ,  b  ]    |    t  ∈  [  a  ,  b  ]  }define  μ  (  [  a  ,  t  )  =  t  −  a  +  100  ν  (  [  a  ,  t  )  )  =  t  −  a  +  50  μ  (  [  a  ,  b  ]  )  =  100  +  (  b  −  a  )  ν  (  [  a  ,  b  ]  )  =  100  +  (  b  −  a  )Note that   μ  (  C  )  ≥  ν  (  C  ) for every   C  ∈  C, but   μ  (  {  b  }  )  =  μ  (  [  a  ,  b  ]  )  −      lim    n    μ  (  [  a  ,  b  −  1  /  n  )  =  0 and   ν  (  {  b  }  )  =  50This is possible because we can&#039;t assure that   B  ∖  A  ∈  C for a   π- system

Let μ , ν be finite measures on the non-degenerate comp

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