Proving the continuity of probability ${A_n}$ is a sequence of increasing events. Then I want to prove that $P(\cup A_n)$ is the least upper bound of the sequence $P(A_n)$. I am able to prove that it is an upper bound, but stuck in proving that it is indeed "least" upper bound.
BTW, I don't want the usual proof of "continuity of probability" from the probability axioms where we break the union into disjoint sets.  
 A: The trick is to take the $A_n$ and "disjointize" them as follows.
Let $B_1=A_1$, $B_2=A_2\setminus A_1$, $B_3=A_3\setminus A_2$, and generally $B_n=A_n\setminus A_{n-1}$. Since $A_1\subset A_2\subset \cdots$ then the $B_n$ are disjoint and so we get that $\cup_{n=1}^j B_n=\cup_{n=1}^j A_n=A_j$, for $1\leq j\leq\infty$ where $A_\infty=\cup_{n=1}^\infty A_n$ by definition. Since the $B_n$ are disjoint then we have that $P(A_\infty)=P(\cup_{n=1}^\infty B_n)=\sum_{n=1}^\infty P(B_n)$. Let $b_j=\sum_{n=1}^j B_n$. Then notice that $P(A_j)=b_j$ for $1\leq j\leq\infty$ and $b_j\rightarrow b_\infty$. Thus it just reduces to show that if you have a sequence of montone increasing reals ($b_j$) that converge to a point $(b_\infty)$ then the limit is the sup of the sequence.
A: If you have a sequence of events $\{A_{n}\}$ then, it is true that $A_{n} \subseteq \cup_{k=n}^{\infty}A_{k}$. So, using measure property we have:
$$P(A_{n}) \leq P(\cup_{k=n}^{\infty}A_{k})$$
There you have that is the least upper bound for the sequence $P(A_{n})$.
