Arveson says that an operator A acting on a separable Hilbert space H is diagonalizable:"If there is a (necessarily separable) σ-finite measure space ( X , μ ), a function f ∈ L ∞ ( X , μ ), and a unitary operator W : L 2 ( X , μ ) ↦ H such that W M f = A W",where M f is multiplication by f ∈ L ∞ ( X , μ ).Could someone elaborate on the notion of a separable σ-finite measure space and how it enters into the context of diagonalizable operators and the Spectral Theorem for (Normal) operators on a separable Hilbert space?

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Arveson says that an operator   A acting on a separable Hilbert space   H is diagonalizable:&quot;If there is a (necessarily separable)   σ-finite measure space   (  X  ,  μ  ), a function   f  ∈      L          ∞        (  X  ,  μ  ), and a unitary operator   W  :      L          2        (  X  ,  μ  )  ↦  H such that   W      M          f        =  A  W&quot;,where       M    f   is multiplication by   f  ∈      L          ∞        (  X  ,  μ  ).Could someone elaborate on the notion of a separable σ-finite measure space and how it enters into the context of diagonalizable operators and the Spectral Theorem for (Normal) operators on a separable Hilbert space?

asafand2c · Accepted Answer

Saying that a measure space is separable is not standard terminology. Here it just means that       L    2    (  X  ) is separable, which is obvious since it is isomorphic to   H.If   A is diagonalizable then it is unitarily equivalent to the normal operator       M    f  , so it is normal.To see the relation with the Spectral Theorem, given   Δ  ⊂  σ  (  A  ) Borel, define  F  (  Δ  )  =  W      M                  1                              f                          −              1                                (          Δ          )                          W    ∗    .Then one checks that   F is a spectral measure and that  A  =      ∫          σ      (      A      )        λ    d  F  (  λ  )  .

Arveson says that an operator A acting on a separable Hilbert space H is diagonalizable:

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