Let $(X,\mu)$ be a measure space and $$f = \sum_{j = 1}^m a_j \chi_{A_j} = \sum_{k = 1}^n b_k\chi_{B_k}$$ for $a_j,b_k \geqslant 0$, $A_j$, $B_k$ measurable. Now I want to show $$\sum_{j = 1}^m a_j \mu(A_j) = \sum_{k = 1}^n b_k \mu(B_k)$$ To do so, we define $A_{m + 1} := B_1,\dots,A_{m+n}:= B_n$ and $\mathfrak{D}$ to be the set of all elements of the form $$\bigcap_{i = 1}^{m + n} M_i$$ where $M_i \in\{A_i,A_i^c\}$. The set $\mathfrak{D}$ has following properties:
- The elements of $\mathfrak{D}$ are pairwise disjoint. Proof. This follows directly from the fact, that one contains elements of $A_i$ and the other of $A_i^c$.
- We have $$A_i = \bigcup_{Y \in \mathfrak{D}, Y \subseteq A_i} Y$$ for any $i = 1,\dots,m+n$. Proof. The inclusion $\supseteq$ is clear, since if it is in the union it must be in at least one $Y$, but $Y \subseteq A_i$ and so the element is in $A_i$.
However I do not see the inclusion $\subseteq $ in the second property. Any suggestions?