# Disintegration of von Neumann algebra

Let $B=\int_{Y}^{\oplus}B_yd_y$ be the direct integral decomposition of vNa $B$ into factors and if $P=\int_{Y}^{\oplus} p_ydy$ is a projection in $B$ and $p_y$ is equivalent to a projection $q_y$ in $B_y$ for all or a.e. $y\in Y$. Is there a good choice of $q_y$ up to equivalence such that $Q=\int_{Y}^{\oplus} q_ydy$ makes sense and $P$ and $Q$ are equivalent in $B$?

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I don't think the question makes sense as is. Why not choose $q_y = p_y$? –  Jesse Peterson Oct 17 '11 at 21:40
@Jesse: I presume that when the OP says “good choice of $q_y$ up to equivalence” he means the choice of a representative of the equivalence class of a section with respect to sets of measure 0. –  Dmitri Pavlov Oct 18 '11 at 7:47
That's certainly one interpretation. I wonder if the OP would mind clarifying what is meant by "good", "choice" (since the $q_y$ are already defined), and "up to equivalence"? –  Jesse Peterson Oct 18 '11 at 12:46
For each $y\in Y$ we can find a $q_y$ which is equivalent to $p_y$ (we don't know whether $q_y$ equals $p_y$ or not, let's say they are different), but $y\mapsto q_y$ is not necessarily measurable in the sense that $Q=\int_Y^{\oplus}q_ydy$ makes sense. So I want to find $q'_y$ which is equivalent to $q_y$ in $B_y$ and $y\mapsto q'_y$ is measurable. I need a kind of measurable selection theorem. Thanks for your attention :) –  m07kl Oct 18 '11 at 15:18
@m07kl: So what is wrong with choosing $q_y' = p_y$? –  Jesse Peterson Oct 18 '11 at 15:35