Let
- $(E,\mathcal E,\lambda)$ be a measure space
- $I$ be a finite nonempty set
- $p,q_i$ be probability densities on $(E,\mathcal E,\lambda)$ for $i\in I$
- $\mu:=p\lambda$
- $w_i:E\to[0,1]$ be $\mathcal E$-measurable for $i\in I$
If $$\{p=0\}\subseteq\{q_i=0\}\subseteq\{w_i=0\}\;\;\;\text{for all }i\in I\tag1$$ and $$\bigcup_{i\in I}\{q_i>0\}=\{p>0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}\tag2,$$ are we able to simplify $$\sum_{i\in I}\mu\left(\left\{q_i>0\right\}\right)\int w_i\:{\rm d}\mu\tag3$$ or the same expression with $\{q_i>0\}$ replaced by $\{w_i>0\}$, i.e. $\sum_{i\in I}\mu\left(\left\{w_i>0\right\}\right)\int w_i\:{\rm d}\mu$?
Maybe there is a useful formula for $\mu\left(\left\{w_i>0\right\}\right)$ beyond the trivial inequality $\int w_i\:{\rm d}\mu\le\mu\left(\left\{w_i>0\right\}\right)$.