expectation of the weighted sum of positive random variables Let $X_{i}(\omega)$, for $i=1,\dots,k$, and $Y(\omega)$ be the random variables defined on the same probability space $(\Omega, F,P)$, such that $E[X_{i}] \leq E[Y]$ for every $i$ and both of them are positive with probability $1$. Next, let $A_{i}$ be disjoint elements of the sigma-algebra $F$ such that $\cup_{i}A_{i} = \Omega$.
Is the following correct:
$$
\sum_{i}^{k}E[X_{i}I\{A_{i}\}] \leq E[Y]?
$$
 A: No. Consider the following: $k=2$, $\Omega = \{1,2\}$, $P(1)=P(2) = 1/2$, and $A_1 = \{1\}$, $A_2=\{2\}$ with
$$
X_i = \frac{1}{50}+\mathbb{1}_{\{i\}},\qquad i\in\{1,2\} 
$$
and $Y=1$ (constant r.v.). Then $\mathbb{E} X_i = \frac{1}{50}+\frac{1}{2} \leq \mathbb{E} Y$ for $i\in\{1,2\}$, yet
$$
\sum_{i=1}^2 \mathbb{E}[ X_i \mathbb{1}_{A_i} ]
= 2\cdot\frac{51}{100}  > 1 = \mathbb{E} Y
$$ 
(The $1/50$ is simply an annoying technicality to make the $X_i$'s positive with probability $1$. Remove it if you only want non-negative -- same idea, simpler answer.)
A: Let $k\geq2$, let $P(A_i)>0$ for $i\in\{1,\dots,k\}$ and let $c>0$.
Let $X_i=c+1_{A_i}$ and let $Y=c+1_A$ such that $\mathsf P(A_i)\leq\mathsf P(A)<1$ for every $i\in\{1,\dots,k\}$.
For instance you can take $A=A_{i_0}$ where $\mathsf P(A_{i_0})$ is maximal among the $\mathsf P(A_i)$.
Then $Y$ and the $X_i$ are  positive and:  $$\mathsf EX_i=c+\mathsf P(A_i)\leq c+\mathsf P(A)=\mathsf EY$$ for every $i\in\{1,\dots,k\}$.
But also we have: $$\sum_{i=1}^k\mathsf E[X_i1_{A_i}]=\sum_{i=1}^k(c+1)\mathsf P(A_i)=c+1>c+\mathsf P(A)=\mathsf EY$$
