I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 29 on p.40 in Exercises 2B in this book.
Exercise 29
Give an example of a measurable space $(X,\mathcal{S})$ and a family $\{f_t\}_{t\in\mathbb{R}}$ such that each $f_ t$ is an $\mathcal{S}$-measurable function from $X$ to $[0,1]$, but the function $f:X\to [0,1]$ defined by $$f(x)=\sup\{f_t(x):t\in\mathbb{R}\}$$ is not $\mathcal{S}$-measurable.
[Compare this exercise to 2.53, where the index set is $\mathbb{Z}^+$ rather than $\mathbb{R}$.]
I was not able to solve Exercise 29.
I found aduh's answer.
It is not necessary to be $N=\mathbb{R}$ in aduh's answer.
I think the following solution is a special case of aduh's answer.
Is the following solution ok?
And I want to know another solution for Exercise 29.
Let $X:=\mathbb{C}.$
Let $\mathcal{S}:=\{E\subset X:E\text{ is countable or }X\setminus E\text{ is countable}\}.$
Then, $(X,\mathcal{S})$ is a measurable space. (See Example 2.28 on p.28 and Exercise 2 in Exercises 2B on p.38 in the book.)
Let $\{f_t\}_{t\in\mathbb{R}}$ be a family such that each $f_t$ is a function from $X$ to $[0,1]$ such that $f_t(x):=\chi_{\{t\}}(x)$ for every $x\in X.$
Let $t\in\mathbb{R}.$
Let $B\subset\mathbb{R}$ be a Borel set.
If $0\in B$ and $1\in B$, then $f_t^{-1}(B)=X\in\mathcal{S}.$
If $0\in B$ and $1\notin B$, then $f_t^{-1}(B)=X\setminus\{t\}\in\mathcal{S}.$
If $0\notin B$ and $1\in B$, then $f_t^{-1}(B)=\{t\}\in\mathcal{S}.$
If $0\notin B$ and $1\notin B$, then $f_t^{-1}(B)=\emptyset\in\mathcal{S}.$
So, $f_t$ is an $\mathcal{S}$-measurable function.
$\{f_t(x):t\in\mathbb{R}\}=\{0,1\}$ if $x\in\mathbb{R}.$
$\{f_t(x):t\in\mathbb{R}\}=\{0\}$ if $x\in X\setminus\mathbb{R}.$
So, $\sup\{f_t(x):t\in\mathbb{R}\}=1$ if $x\in\mathbb{R}.$
So, $\sup\{f_t(x):t\in\mathbb{R}\}=0$ if $x\in X\setminus\mathbb{R}.$
Therefore, $f(x):=\sup\{f_t(x):t\in\mathbb{R}\}=\chi_{\mathbb{R}}(x).$
$\{1\}$ is a Borel set and $f^{-1}(\{1\})=\chi_{\mathbb{R}}^{-1}(\{1\})=\mathbb{R}\notin\mathcal{S}.$
So, $f$ is not $\mathcal{S}$-measurable.