Let μ is finite measure and μ ( A △ B ) = 0.Show that μ ( A ) = μ ( B )My work: μ ( A △ B ) = 0 = μ ( A | B ) + μ ( B | A ) = ( μ ( A ) − μ ( A ∩ B ) ) + ( μ ( B ) − μ ( B ∩ A ) )So μ ( A ) = − μ ( B ) + 2 μ ( B ∩ A )How is μ ( A ) = μ ( B )?

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Let   μ is finite measure and   μ  (  A  △  B  )  =  0.Show that   μ  (  A  )  =  μ  (  B  )My work:  μ  (  A  △  B  )  =  0  =  μ  (  A      |    B  )  +  μ  (  B      |    A  )  =  (  μ  (  A  )  −  μ  (  A  ∩  B  )  )  +  (  μ  (  B  )  −  μ  (  B  ∩  A  )  )So   μ  (  A  )  =  −  μ  (  B  )  +  2  μ  (  B  ∩  A  )How is   μ  (  A  )  =  μ  (  B  )?

Hailee Henderson · Accepted Answer

Since   μ is a measure, it must be non-negative. Now,   μ  (  A  ∖  B  )  +  μ  (  B  ∖  A  )  =  0. So, we must have   μ  (  A  ∖  B  )  =  μ  (  B  ∖  A  )  =  0.The rest is just our old set theory.

Yasmin Camacho · Accepted Answer

The measure   μ does not have to be finite. Note   A  Δ  B  =  {  x  :      χ    A    (  x  )  ≠      χ    B    (  x  )  }. Thus   μ  (  A  Δ  B  )  =  0 if and only if       χ    A    =      χ    B   a.e. If       χ    A    =      χ    B   a.e., then   μ  (  A  )  =  ∫      χ    A    =  ∫      χ    B    =  μ  (  B  ).

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