the proof of the measurability of a sequence of monotone functions.

Question

I understand that $E_{kn}$ is disjoint and the union of $E_{kn}$ is equal to $X$. But, I don't understand that the next sentence "If we define $\varphi_n$ to be equal to $k2^{-n}$ on $E_{kn}$, then $\varphi_n$ belongs to $M(X, \mathbf{X})$". Could you elaborate on this?

Answer 1

The whole construction of $\phi_n$ boils down to:$$\phi_n(x):=\min\left(2^{-n}\lfloor2^nf(x))\rfloor,n\right)$$ It is immediate that for every $n$ the image of the function $\phi_n$ is finite.

Then if $I$ denotes the image of $\phi_n$ for every element $p\in I$ the set $\phi^{-1}(\{p\})$ is measurable.

We have:$$\phi_n=\sum_{p\in I}p1_{\phi^{-1}(\{p\})}$$where $I$ is finite and the sets $\phi^{-1}(\{p\})$ are measurable, are disjoint and cover $X$.

Note that the inverse image of every set wrt $\phi$ is a finite union of measurable sets of the form $\phi^{-1}(\{p\})$ hence is measurable.

This makes clear that such a function is measurable.