If $\cup_iA_i=A\subseteq\{\sum_iw_i=1\}$, can we show that $\sum_i\left(\int w_i\:{\rm d}μ\right)\int_{A_i}g\:{\rm d}μ=\int_Ag\:{\rm d}μ$?

Question

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 $i\in I$
• $g:E\to[0,\infty)$ be $\mathcal E$-measurable with $\{p=0\}\subseteq\{g=0\}$

If $$\{p=0\}\subseteq\{q_i=0\}\subseteq\{w_i=0\}\;\;\;\text{for all }i\in I\tag1$$ and $$\bigcup_{i\in I}\{q_i>0\}=\{p>0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}\tag2,$$ I'd like to show that $$\sum_{i\in I}\left(\int w_i\:{\rm d}\mu\right)\int_{\{\:q_i\:>\:0\:\}}g\:{\rm d}\mu=\int g\:{\rm d}\mu.\tag3$$

If $|I|=1$, the claim is easy to prove.

Are we able to find a (weak) additional assumption for the general case? The only thing I'm not willing to assume is $\{q_i=0\}\subseteq\{p=0\}$ for all $i\in I$ (but this assumption would be sufficient).