Let B be a σ-algebra on a set X and let f : X → R ¯ be an extended real-valued function. Define the sets P = f − 1 ( − ∞ ) and N = f − 1 ( + ∞ ).Define a real-valued function g : X → R by g ( x ) = { f ( x ) , f ( x ) ∈ R 0 , o t h e r w i s e .Prove that f is B -measurable if and only if g is B -measurable and P , N ∈ B .So i tried to prove this like this: P , N ∈ B obviously. Now assume f is B -measurable. Let α ∈ R . Consider { x ∈ X | g ( x ) &gt; α }= { x ∈ X | f ( x ) &gt; α } ∖ P. In this circumstance P ∈ B , { x ∈ X | f ( x ) &gt; α } ∈ B so { x ∈ X | f ( x ) &gt; α } ∖ P ∈ B . So P c ∩ { x ∈ X | f ( x ) &gt; α } ∈ B . This is because, B is closed under finite intersection. ( ⇐ ) g is B -measurable. Then { x ∈ X | f ( x ) &gt; α } = { { x ∈ X | g ( x ) &gt; α } ∪ P , α &gt; 0 { x ∈ X | g ( x ) &gt; α } ∖ N , α ≤ 0 .Both of cases, observe that output sets belongs to B .But my professor said the proof was wrong. Where am I going wrong? Can you help me?

Question

Let       B   be a   σ-algebra on a set   X and let   f    :    X  →      R    ¯   be an extended real-valued function. Define the sets   P    =      f          −      1        (  −  ∞  ) and   N    =      f          −      1        (  +  ∞  ).Define a real-valued function   g    :    X  →      R   by  g  (  x  )  =      {                            f          (          x          )          ,                          f          (          x          )          ∈                      R                                                0          ,                          o          t          h          e          r          w          i          s          e                          .Prove that   f is       B  -measurable if and only if   g is       B  -measurable and   P  ,  N  ∈      B  .So i tried to prove this like this:  P  ,  N  ∈      B   obviously. Now assume   f is       B  -measurable. Let   α  ∈      R  . Consider  {  x  ∈  X      |    g  (  x  )  &amp;gt;  α  }=  {  x  ∈  X      |    f  (  x  )  &amp;gt;  α  }  ∖  P. In this circumstance   P  ∈      B  ,   {  x  ∈  X      |    f  (  x  )  &amp;gt;  α  }  ∈      B   so   {  x  ∈  X      |    f  (  x  )  &amp;gt;  α  }  ∖  P  ∈      B  . So       P    c    ∩  {  x  ∈  X      |    f  (  x  )  &amp;gt;  α  }  ∈      B  . This is because,       B   is closed under finite intersection.  (  ⇐  )  g is       B  -measurable. Then  {  x  ∈  X      |    f  (  x  )  &amp;gt;  α  }  =      {                            {          x          ∈          X                      |                    g          (          x          )          &amp;gt;          α          }          ∪          P          ,                          α          &amp;gt;          0                                      {          x          ∈          X                      |                    g          (          x          )          &amp;gt;          α          }          ∖          N          ,                          α          ≤          0                          .Both of cases, observe that output sets belongs to       B  .But my professor said the proof was wrong. Where am I going wrong? Can you help me?

America Barrera · Accepted Answer

{  x  ∈  X      |    g  (  x  )  &amp;gt;  α  }=  {  x  ∈  X      |    f  (  x  )  &amp;gt;  α  }  ∖  P is wrong. If   α  =  −  1 and   f  (  x  )  =  −  ∞ then   g  (  x  )  &amp;gt;  α. So   x is in LHS but not in RHS. You have to consider the cases   α  &amp;lt;  0 and   α  ≥  0 separately.

Let <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struck">B </mrow> be a

Answered question

Answer & Explanation

New Questions in Pre-Algebra