I am trying to answer the following question:Suppose a function f has a measurable domain and is continuous except at a finite number of points. Is f necessarily measurable?And I found this answer online:"Let E denote the (Lebesgue) measurable domain of f and define E 0 = { x ∈ E | f is not continuous at x }Since E 0 is finite, m ( E 0 ) = 0 and f is measurable on E 0 . By proposition 3, f is measurable on E ∼ E 0 as it is continuous on this set. We conclude from proposition 5(ii) that f is measurable on E."Why f restricted to the set of its discontinuities E 0 is measurable? by what theorem, proposition or logical justification?

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I am trying to answer the following question:Suppose a function   f has a measurable domain and is continuous except at a finite number of points. Is   f necessarily measurable?And I found this answer online:&quot;Let   E denote the (Lebesgue) measurable domain of   f and define      E    0    =  {  x  ∈  E      |    f   is not continuous at   x  }Since       E    0   is finite,   m  (      E    0    )  =  0 and   f is measurable on       E    0  . By proposition 3,   f is measurable on   E  ∼      E    0   as it is continuous on this set. We conclude from proposition 5(ii) that   f is measurable on   E.&quot;Why   f restricted to the set of its discontinuities       E    0   is measurable? by what theorem, proposition or logical justification?

xxsailojaixxv5 · Accepted Answer

Since       E    0   is finite and therefore discrete, the restriction of the   σ-algebra of Borel sets to       E    0   is       P    (      E    0    ). Therefore, any function defined on       E    0   is measurable.EditIf   (  X  ,      A    ) is a measurable space (ie   X is a set and       A   is a   σ-algebra on   X) and   Y  ⊂  X is a subset, then   Y is canonically a measurable space   (  Y  ,      A              |        Y    ) where the restriction of the   σ-algebra       A   to the subset   Y is defined as :      A              |        Y    =  {  Y  ∩  W  :  W  ∈      A    }If   X is a topological space and       A    =      B    (  X  ) is its Borel   σ-algebra, then       A              |        Y    =      B    (  Y  ) (ie the restriction of a Borel algebra is the Borel algebra of the topological subspace).

I am trying to answer the following question: Suppose a function f has a measurable domain

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