Prooving $\mathbb{I}_{\{A \Delta B \}}=(\mathbb{I}_{\{ A\}}-\mathbb{I}_{\{ B\}})^2$

Question

Let $A$ and $B$ be two events from $\Omega, \mathcal{P}(\Omega),\mathbb{P})$. I need to show that next equal is true $$\mathbb{I}_{\{A \Delta B \}}=(\mathbb{I}_{\{ A\}}-\mathbb{I}_{\{ B\}})^2$$.

I think $A \Delta B=(A\cap \overline{B})\cup(B\cap \overline{A})$ Is it right? what to do next?

Answer 1

My notation for your $\mathbb I_{\{A\}}$ is $\mathbf1_A$.

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Since indicator functions only take values in $\{0,1\}$ for any $\omega\in \Omega$ the following statements are evidently equivalent:

• (1) $(\mathbf1_A-\mathbf1_B)^2(\omega)=1$
• (2) $(\mathbf1_A(\omega)-\mathbf1_B(\omega))^2=1$
• (3) $[\mathbf1_A(\omega)=1\wedge\mathbf1_B(\omega)=0]\vee[\mathbf1_A(\omega)=0\wedge\mathbf1_B(\omega)=1]$
• (4) $[\omega\in A\wedge \omega\in B^{\complement}]\vee[\omega\in A^{\complement}\wedge \omega\in B]$
• (5) $\omega\in (A\cap B^{\complement})\cup(A^{\complement}\cap B)=A\Delta B$

This proves that $(\mathbf1_A-\mathbf1_B)^2$ serves as indicator function of set $A\Delta B$.