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I am trying to prove this statement that seems very easy but i am struggling with it. If $B-A$ is measurable then $B^c\cup A$ is also measurable but that does not help really. I am pretty sure the end goal is to manipulate this to get $A$ by itself which would imply it is measurable but i am failing to see how to do that.Just a hint would be appreciated.

Sorfosh
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$B=A \setminus (A\setminus B)=A\cap(A\setminus B)^c=(A\cup(A\setminus B)^c)^c$, so $B$ is written as unions and complements of measurables. Hence it is measurable as well.

Matteo Spadetto
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    Sigh... that was so easy. I kept thinking $A \cap B$ is nothing helpful but it is $B$... Thank you. – Sorfosh Oct 02 '19 at 17:16