From Definition 1 in Math 245A Note 1:
Given two Boolean algebras $\mathcal{B}$, $\mathcal{B}'$ on ${X}$, we say that ${{\mathcal B}'}$ is finer than, a sub-algebra of, or a refinement of ${{\mathcal B}}$, or that ${{\mathcal B}}$ is coarser than or a coarsening of ${{\mathcal B}'}$, if ${{\mathcal B} \subset {\mathcal B}'}$.
Also, we are given:
Definition 10 (Integral of simple functions) An (unsigned) simple function ${f: X \rightarrow [0,+\infty]}$ on a measurable space ${(X,{\mathcal B})}$ ($\mathcal B$ finite) is a measurable function that takes on finitely many values ${a_1,\ldots,a_k}$. Note that such a function is then automatically measurable with respect to at least one finite sub-${\sigma}$-algebra ${{\mathcal B}'}$ of ${{\mathcal B}}$, namely the ${\sigma}$-algebra ${{\mathcal B}'}$ generated by the preimages ${f^{-1}(\{a_1\}),\ldots,f^{-1}(\{a_k\})}$ of ${a_1,\ldots,a_k}$. We then define the simple integral ${\hbox{Simp} \int_X f\ d\mu}$ by the formula
$\displaystyle \hbox{Simp} \int_X f\ d\mu := \hbox{Simp} \int_X f\ d\mu\downharpoonright_{{\mathcal B}'}$,
where ${\mu\downharpoonright_{{\mathcal B}'}: {\mathcal B}' \rightarrow [0,+\infty]}$ is the restriction of ${\mu: {\mathcal B} \rightarrow [0,+\infty]}$ to ${{\mathcal B}'}$.
Question: The second definition implies that the sub-$\sigma$-algebra $\mathcal{B}'$ is the coarser algebra that is contained in the finer algebra $\mathcal{B}$ (hence it makes sense to talk about the "restriction" of $\mu$ to $\mathcal{B}')$. Yet from Definition 1, if $\mathcal{B}'$ is a sub-algebra of $\mathcal{B}$, then it is the other way round that $\mathcal{B} \subset \mathcal{B}'$. Where did I get it wrong here?
