Is the probability of the union of events nondecreasing in the probability of the events? Can it be shown that the probability $P(A_1 \cup \dots \cup A_n)$ is nondecreasing in the probability of any event $A_i$?
This fact seems intuitive to me, independent of the fact whether $A_1, \dots, A_n$ are intersecting or not?
So, if for any $A_i$, the probability $P(A_i)$ weakly increases, will the probability $P(A_1 \cup \dots \cup A_n)$  weakly increase?
Is this correct?
 A: It is nondecreasing as a function of sets, i.e., if one of the events $A_i$ is replaced with a superset then the probability cannot decrease. But the probability $P(A_1\cup\ldots\cup A_n)$ is not even a function of the individual probabilities $P(A_i)$ as numbers, let alone a nondecreasing function of them.
Here is a counterexample in $[0,1]$ with the uniform probability measure. We have $P([0,\frac12]\cup[\frac12,1])>P([0,\frac12]\cup[0,\frac23])$ even though $P([\frac12,1])<P([0,\frac23]).$
(updated to answer question in comment)
In the special case where we are given independent $E_1,\ldots,E_m$ and every $A_i$ is an intersection of some of the $E_j$:
$$A_i=\bigcap_{j\in K_i}E_j,\ i=1,\ldots,n$$
let us demonstrate that $P[\bigcup_{i=1}^nA_i]$ depends on $P[E_k]$ nondecreasingly. Fix $k$ and let $I=\{i|1\leq i\leq n, k\in K_i\},$ $J=\{i|1\leq i\leq n, k\not\in K_i\}.$
For $i\in I$ write
$$A_i'=\bigcap_{j\in K_i\setminus\{k\}}E_j$$
Then we have
$$\eqalign{
\bigcup_{i=1}^nA_i&=\bigcup_{i\in I}A_i\cup\bigcup_{i\in J}A_i\\
&=\left(E_k\cap\bigcup_{i\in I}A_i'\right)\cup\bigcup_{i\in J}A_i\\
&=\left(E_k\cap\left(\bigcup_{i\in I}A_i'\setminus\bigcup_{i\in J}A_i\right)\right)\sqcup\bigcup_{i\in J}A_i\\
P\left[\bigcup_{i=1}^nA_i\right]&=P[E_k]P\left[\bigcup_{i\in I}A_i'\setminus\bigcup_{i\in J}A_i\right]+P\left[\bigcup_{i\in J}A_i\right]\\
}$$
where in the last line we have used independence, and the symbol $\sqcup$ denotes disjoint union. But the expression on the right is clearly nondecreasing in $P[E_k].$
A: Let $E_1,E_2,...,E_n$ measurables sets, and $B_i=A_i\cup E_i$.
Then, since $A_i\subseteq B_i$
$\begin{eqnarray}
&&P(B_1\cup B_2\cup...\cup B_n)-P(A_1\cup A_2\cup...\cup A_n)\\
&=&P(\bigcup B_i\backslash\bigcup A_i)\\
&=&P(\bigcup B_i\cap(\bigcup A_i)^c)\\
&=&P\left(\bigcup B_i\cap(\bigcap A_i^c)\right)\\
&=&P\left(\bigcup [B_i\cap(\bigcap A_i^c)]\right)\\
&=&P\left(\bigcup [E_i\cap(\bigcap A_i^c)]\right)\\
&=&P(\bigcup E_i\backslash(\bigcup A_i)^c)\\
&\le&P(\bigcup E_i)\\
&\le&P(E_1)+...+P(E_n)
\end{eqnarray}$
So, by the second and last line, $$0\le P(B_1\cup...\cup B_n)-P(A_1\cup...\cup A_n)\le P(E_1)+...+P(E_n).$$
This shows that $P$ is increasing and "continuos"
