$P( A\triangle B ) = 0 \Rightarrow P(A)=P(B)=P(A\cup B) + P(A \cap B) $

Question

I want to prove the following statement;

$$P( A\triangle B ) = 0 \Rightarrow P(A)=P(B)=P(A\cup B) \color{blue}{=} P(A \cap B) $$

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What I did is that

$$P(A\triangle B) =P((A\setminus B ) \cup (B\setminus A))=0$$

$$=P((A\cap B^c)\cup (B\cap A^c))=0$$

$$=P((A\cup B)\cap (A\cup A^c)\cap (B\cup B^c)\cap(B^c \cup B^c))=0$$

$$=P((A\cup B)\cap (B^c\cup A^c))=0$$

$$=P((A\cup B)\setminus (B\cap A))=0$$

Please help me showing this. thank you.

Answer 1

The probabilities have to look like this, from which everything follows.