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?


Let $x\in A_i$. For every $j\neq i$ we either have $x\in A_j$ or $A_j^c$. Hence, making the right choices of $M_j$ will give you $x \in \bigcap_{k=1}^{m+n} M_k$. Thus

$$ x\in \bigcup_{Y\in \mathcal{D}, Y\subseteq A_i} Y $$

As $x\in A_i$ arbitrary, we conclude

$$ A_i \subseteq \bigcup_{Y\in \mathcal{D}, Y\subseteq A_i} Y $$

  • $\begingroup$ Thanks. This was too easy. $\endgroup$ – TheGeekGreek Sep 25 '16 at 10:56

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