If $P(A \cup B) = 1$ then does $B = A^c$? Let A and B be two arbitrary events in a sample space S. Prove the following or provide a counterexample:
If $P(A \cup B) =  1$ then $B = A^c$

This is my original answer:
Since both $A, B \in S$ and $P(A \cup B) = 1$ then by definition $A \cup B = S$:
$$S = A \cup B$$
$$S \cap B^c = A \cup B \cap B^c$$
$$B^c = A$$
$$B = A^c$$

But I now realize that $P(A \cup B) = 1$ does not necessarily imply that $A \cup B = S$ so my proof doesn't work.
 A: Take any even $A$ such that $P(A)=1$. Take $B=A$. Then

*

*What is $P(A\cup B)$?

*Is $B=A^c$?

A: For one particularly straightforward counterexample, consider $A=B=S$.  We have here $P(A\cup B) = P(S\cup S) =P(S) = 1$ despite $B\neq A^c = \emptyset$
For a more exotic counterexample, let $N$ be a non-empty null set (an event who occurs with probability zero but is still technically in the sample space) who shares some element(s) with $A^c$ and let $B = A^c\setminus N$.
You make two major mistakes in your attempt.  The first is as you noted $P(A\cup B) = 1$ does not imply that $A\cup B = S$ due to the existence of null sets.
Your second major mistake is in going from $S=A\cup B$ to $S\cap B^c = A\cup B\cap B^c$.  You left off parentheses and evaluated as though parentheses were on the right as though $A\cup (B\cap B^c)$ leading to $A\cup (\emptyset)$ to $A$.  Rather, $S = (A\cup B)$ implies $S\cap B^c = (A\cup B)\cap B^c$ which leads to $B^c = A\cap B^c$.
Having corrected these two mistakes, we should have arrived at the conclusion that $P(A\cup B) = 1$ implies that $B^c$ is a subset of $A$ up to the difference of a nullset.  That is to say, $B^c\setminus A$ is a nullset.
For explicit example, consider the space $\{0,1,2,3,4\}$ where $P(\{1\})=P(\{2\})=P(\{3\})=P(\{4\})=\frac{1}{4}$ and $P(\{0\})=0$.  Letting $A=\{1,2,3\}$ and $B=\{2,3,4\}$ we do indeed have $P(A\cup B) = P(\{1,2,3,4\}) = 1$ and $B^c = \{0,1\}$ which although is not a subset of $\{1,2,3\}$ it does differ from a subset of $A$, namely $\{1\}$ by only a null-set.
A: If you consider the space of outcomes $\{a, b\}$ with probability $P(a) = 1$ and $P(b) = 0$.
Take the event $A = \{a\}$. Then $P(A) = 1$.
Take any event $B$ of the same space, then $P(A \cup B) = 1$.
Let $B = A$, then you have a counter example, because $B \neq A^c$
