# Events that are Equal with Probability 1 Proof Question

I have a question regarding two equal events with probability 1. On page 21 of Papoulis' "Probability, Random Variables, and Stochastic Processes", 4th edition, there is an equation:

$$P(A\bar{B}) + P(\bar{A}B) = 0 \iff P(A) = P(B) = P(AB)$$

I am able to prove the relation in the reverse direction, but how do I prove it in the forward direction?

$$P(A\bar{B}) + P(\bar{A}B) = 0 \implies P(A) = P(B) = P(AB)$$

My work:

$$P(A\bar{B}) + P(\bar{A}B) = 0 \implies P(A\bar{B}) + P(\bar{A}B) + P(AB) = P(AB)$$

$$\implies P(A\bar{B}) + P(B) = P(AB) \implies P(A) + P(B) = 2P(AB)$$

For the last step, I added $P(AB)$ to both sides. I am stuck at this point. Can I have some hints on how to proceed? Thank you in advance for your help.

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See here – Dilip Sarwate Oct 15 '11 at 22:27

$P(A\bar{B}) + P(\bar{A}B) = 0 \Rightarrow P(A\bar{B}) = 0$ and $P(\bar{A}B) = 0$. Do you see how that helps?

Added: Now that the OP has seen how to proceed, let's finish the argument for the sake of completeness.

Since $P(A\bar{B}) = 0$, the fact that $P(A) = P(AB) + P(A\bar{B})$ implies $P(A) = P(AB)$. Similarly, $P(\bar{A}B) = 0 \Rightarrow P(B) = P(AB)$. Thus $P(A) = P(B) = P(AB)$.

In a case like this, where there are only two sets $A$ and $B$ involved, often it helps the understanding to draw the Venn diagram.

For example, the image below probably makes it more clear why $P(A\bar{B}) + P(\bar{A}B) = 0 \iff P(A) = P(B) = P(AB)$.

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Yes, thanks. So, $P(A) + P(B) = 0 \implies P(A) = P(B) = 0$, since $P(A) \geq 0$ and $P(B) \geq 0$. – jrand Oct 15 '11 at 23:00