Let $A,B⊂R^n$, if $ A $ is measurable and $m∗(A△B)=0$ show $B $is measurable and $m(B)=m(A)$ where $ m∗ $ denotes as outer measure

Question

It seems very trivial, but I don't know how to prove it.

Answer 1

Let $X$ be the universe we are working in.

Observe that $R\cap B\subseteq\left(R\cap A\right)\cup\left(A\bigtriangleup B\right)$ and $R\cap A\subseteq\left(R\cap B\right)\cup\left(A\bigtriangleup B\right)$ for any $R\subseteq X$.

An outer measure is subadditive and monotone, so:

$m^{*}\left(R\cap B\right)\leq m^{*}\left(R\cap A\right)+m^{*}\left(A\bigtriangleup B\right)=m^{*}\left(R\cap A\right)$

and:

$m^{*}\left(R\cap A\right)\leq m^{*}\left(R\cap B\right)+m^{*}\left(A\bigtriangleup B\right)=m^{*}\left(R\cap B\right)$

Proved is now that $m^{*}\left(R\cap A\right)=m^{*}\left(R\cap B\right)$ for any $R\subseteq X$.

Taking $R=X$ we find: $m^{*}\left(A\right)=m^{*}\left(B\right)$.

Observe that $A^{c}\bigtriangleup B^{c}=A\triangle B$ telling us that likewise we can find that $m^{*}\left(R\cap A^{c}\right)=m^{*}\left(R\cap B^{c}\right)$ for any $R\subseteq X$.

Making use of measurability of $A$ we find for any $R\subseteq X$:

$m^{*}\left(R\cap B\right)+m^{*}\left(R\cap B^{c}\right)=m^{*}\left(R\cap A\right)+m^{*}\left(R\cap A^{c}\right)=m^{*}\left(R\right)$

and conclude that $B$ is measurable.

So we end up with: $$m(A)=m^{*}(A)=m^{*}(B)=m(B)$$