12.4: Two kinds of absurdness idealism and dualism
This section is extracted from
$(\sharp):$  S. Ishikawa, "Measurement Theory in the Philosophy of Science" $\quad$ arXiv:1209.3483 [physics.histph] 2012 
Measurement theory (= quantum language ) has two kinds of absurdness. That is,
$(\sharp):$  $ {\mbox{ Two kinds of absurdness }} \left\{\begin{array}{ll} \mbox{idealism} {\cdots} \mbox{linguistic worldview} \\ {\mbox{ The limits of my language mean the limits of my world} } \\ \\ \mbox{dualism}\cdots {\mbox{Descartes=Kant philosophy}} \\ {\mbox{ The dualistic description for monistic phenomenon } } \end{array}\right. $ 
12.4.1: The linguistic interpretation  A spectator does not go up to the stage
Problem 12.13 [A spectator does not go up to the stage] Consider the elementary problem with two steps (a) and (b):
$(a):$ 
Consider an urn, in which 3 white balls and 2 black balls
are.
And consider the following trial:
 
(b): 
Then, calculate the probability that you have 2 white ball after (a)(i.e., three trials). 
Answer Put ${\mathbb N}_0$ $=\{0,1,2,\ldots\}$ with the counting measure. Assume that there are $m$ white balls and $n$ black balls in the urn. This situation is represented by a state $ (m,n) \in {\mathbb N}_0^2 $. We can define the dual causal operator ${\Phi^*}: {\cal M}_{+1}({\mathbb N}_0^2)$ $ \to {\cal M}_{+1}({\mathbb N}_0^2)$ such that
\begin{align} {\Phi^*}(\delta_{(m,n)}) = \left\{\begin{array}{ll} \frac{m}{m+n} \delta_{(m1,n)}+\frac{n}{m+n} \delta_{(m,n)} & \quad ( { \mbox{when} \;\; m \not= 0 \; )} \\ \delta_{(0,n)} & \quad {(\mbox{when } m = 0 \; )}. \end{array}\right. \tag{12.17} \end{align}
where $\delta_{(\cdot)}$ is the point measure. Let $T=\{0,1,2,3\}$ be discrete time. For each $t$ $\in T$, put $\Omega_t = {\mathbb N}_0^2$. Thus, we see:
\begin{align} & {[\Phi^*]}^3 (\delta_{(3,2)}) = {[\Phi^*]}^2 \left( \frac{3}{5}\delta_{(2,2)} + \frac{2}{5}\delta_{(3,2)} \right) \nonumber \\ = & {\Phi^*} \left( (\frac{3}{5} (\frac{2}{4} \delta_{(1,2)} +\frac{2}{4} \delta_{(2,2)} ) + \frac{2}{5}( \frac{3}{5} \delta_{(2,2)}+\frac{2}{5} \delta_{(3,2)} ) \right) \nonumber \\ = & {\Phi^*} \left( \frac{3}{10} \delta_{(1,2)} +\frac{27}{50} \delta_{(2,2)} + \frac{4}{25} \delta_{(3,2)} \right) \nonumber \\ = & \frac{3}{10} ( \frac{1}{3} \delta_{(0,2)}+\frac{2}{3} \delta_{(1,2)} ) +\frac{27}{50} ( \frac{2}{4} \delta_{(1,2)}+\frac{2}{4} \delta_{(2,2)} ) + \frac{4}{25} ( \frac{3}{5} \delta_{(2,2)}+\frac{2}{5} \delta_{(3,2)} ) \nonumber \\ = & \frac{1}{10} \delta_{(0,2)}+\frac{47}{100} \delta_{(1,2)} +\frac{183}{500} \delta_{(2,2)} + \frac{8}{125} \delta_{(3,2)} \tag{12.18} \end{align}
Define the observable ${\mathsf O} =({\mathbb N}_0,2^{{\mathbb N}_0}, F^)$ in $L^\infty (\Omega_3)$ such that
\begin{align} [F^{}(\Xi)](m,n) = \left\{\begin{array}{ll} 1 & \qquad (m,n ) \in \Xi \times {\mathbb N}_0 \subseteq \Omega_3 \\ 0 & \qquad (m,n ) \notin \Xi \times {\mathbb N}_0 \subseteq \Omega_3 \end{array}\right. \end{align}Therefore, the probability that a measured value "$2$" is obtained by the {{measurement}} ${\mathsf M}_{L^\infty ({\mathbb N}_0^2)}(\Phi^3{\mathsf O},$ $ S_{[(3,2)]})$ is given by
\begin{align} [\Phi^3 (F (\{2\}))](3,2) = \int_{\Omega_3} [F(\{2\})](\omega) ({[\Phi^*]}^3 (\delta_{(3,2)}) )(d \omega) = \frac{183}{500} \tag{12.19} \end{align}The above may be easy, but we should note that
$(c):$  the part (a) is related to causality, and the part (b) is related to measurement. 
$\fbox{Note 12.4}$ 
The part (a) is not related to
"probability".
That is because
The spirit of
measurement theory
says that

12.4.2:In the beginning was the wordsFit feet to shoes
Remark 12.14 [The confusion between measurement and causality ( Continued from Example 2.31)] Recall Example 2.31 [The measurement of "cold or hot" for water]. Consider the measurement ${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}_{{{{{c}}}}{{{{h}}}}},$ $ S_{[\omega]} )$ where $\omega=5 \mbox{°C}$. Then we say that
$(a):$  By the { {{measurement}}} ${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}_{{{{{c}}}}{{{{h}}}}}, S_{[ \omega(=5)]} )$, the probability that a measured value $x(\in X =\{{{{{c}}}}, {{{{h}}}}\})$ belongs to a set $ \left[\begin{array}{cc} {} \emptyset (={\text {empty set}}) \\ \{ \mbox{c}\} {} \\ \{ \mbox{h} \} \\ \{ \mbox{c} ,\mbox{h}\} \end{array}\right] $ is equal to $ \left[\begin{array}{cc} {} 0 \\ {} [F(\{ c \})] (5)=1 \\ {} [F(\{ h \})](5) =0 \\ {} 1 \end{array}\right] $ 
$\quad$  $\qquad$ "5°C" is the cause and "cold" is a result. 
$(b):$  $\qquad \qquad $ $ \underset{\mbox{(cause)}}{\fbox{5 °C}} \longrightarrow \underset{\mbox{(result)}}{\fbox{cold}} $ 
$\fbox{Note 12.5}$ 
However,
from the different point of view,
the above (b)
can be justified as follows.
Define the
dual causal operator
$
{\Phi^*}
:
{\cal M}([0, 100])
\to
{\cal M}(\{{{{{c}}}}, {{{{h}}}}\})$
by
\begin{align}
&
[{\Phi^*}
\delta_\omega
](D)
=
f_{ {{{{c}}}} }(\omega)
\cdot
\delta_{\mbox{ C}}
(D)
+
f_{{{{{h}}}} }(\omega)
\cdot
\delta_{\mbox{ H}}(D)
\qquad
(\forall \omega \in [0,100],\;\;
\forall D \subseteq \{{{{{c}}}}, {{{{h}}}}\})
\end{align}
Then,
the (b) can be regarded as "causality".
That is,

Remark 12.15 [Mixed measurement and causality] Reconsider Problem 9.5 (urn problem:mixed measurement). That is, consider a state space $\Omega=\{\omega_1, \omega_2 \}$, and define the observable ${\mathsf O} = ( \{ {{w}}, {{b}} \}, 2^{\{ {{w}}, {{b}} \} } , F)$ in $L^\infty (\Omega)$ in Problem 9.5. Define the mixed state by $\rho^m =p \delta_{\omega_1} +(1p) \delta_{\omega_2}$. Then the probability that a measured value $x$ $(\in \{ {{w}} , {{b}} \})$ is obtained by the mixed measurement ${\mathsf M}_{L^\infty(\Omega)}({\mathsf O}, S_{[{}\ast{}] }(\rho^m) )$ is given by
\begin{align} P(\{ x \}) &= \int_\Omega [F(\{ x \})]( \omega) \rho^m(d \omega) = p [F(\{ x \})](\omega_1) + (1p) [F(\{ x \})](\omega_2) \nonumber \\ &= \left\{\begin{array}{ll} 0.8 p + 0.4 (1p) \quad & (\mbox{when }x={{w}}{}\; ) \\ 0.2 p + 0.6 (1p)) \quad & (\mbox{when }x={{b}}{}\; ) \end{array}\right. \tag{12.20} \end{align}Now, define a new state space $\Omega_0$ by $\Omega_0=\{\omega_0\}$. And define the dual (nondeterministic) causal operator ${\Phi^*}: {\cal M}_{+1}(\Omega_0)$ $ \to {\cal M}_{+1}(\Omega)$ by ${\Phi^*}(\delta_{\omega_0})$ $ =p \delta_{\omega_1} +(1p) \delta_{\omega_2}$. Thus, we have the (nondeterministic) causal operator ${\Phi}: L^\infty (\Omega)$ $ \to L^\infty (\Omega_0)$. Here, consider a pure measurement ${\mathsf M}_{L^\infty (\Omega_0)}(\Phi{\mathsf O}, S_{[\omega_0]})$. Then, the probability that a measured value $x$ $(\in \{ {{w}} , {{b}} \})$ is obtained by the measurement is given by
\begin{align} P(\{ x \}) &= [\Phi (F (\{ x \}))](\omega_0) = \int_\Omega [F(\{ x \})]( \omega) \rho^m (d \omega) \\ &= \left\{\begin{array}{ll} 0.8 p + 0.4 (1p) \quad & (\mbox{when }x={{w}}{}\; ) \\ 0.2 p + 0.6 (1p)) \quad & (\mbox{when }x={{b}}{}\; ) \end{array}\right. \end{align}which is equal to the (12.20). Therefore, the mixed measurement ${\mathsf M}_{L^\infty (\Omega)}({\mathsf O}, S_{[{}\ast{}] }(\nu_0) )$ can be regarded as the pure measurement ${\mathsf M}_{L^\infty (\Omega_0)}(\Phi{\mathsf O}, S_{[\omega_0]})$.
$\fbox{Note 12.6}$ 

$\fbox{Note 12.7}$ 
In the book
"The astonishing hypothesis"
by F. Click (the most noted for being a codiscoverer of the structure of the DNA molecule in 1953 with James Watson)),
Dr. Click said that
It should be note that this (a) and the dualism do not contradict. That is because quantum language says:
