Measurability of a net Let $(f_\epsilon)_{\epsilon>0}$ be a family of positive measurable functions on $L_p(\mathbb R)$ where $1<p<\infty.$ Assume that the pointwise supremum $f^*(x)=\sup_{\epsilon>0}|f_\epsilon(x)|$ is in $L_p(\mathbb R).$ Define $F:\mathbb R\to \ell_{(0,\infty)}^\infty$ defined by $F(x)=(f_{\epsilon}(x))_{\epsilon>0}.$ Can we show that $F$ is strongly measurable? Here $\ell_{(0,\infty)}^\infty:=\{a:(0,\infty)\to \mathbb C:\sup_{\epsilon>0}|a_\epsilon|<\infty\}$ we define $\|a\|:=\sup_{\epsilon>0}|a_\epsilon|.$
 A: I assume "strongly measurable" is in the sense of Bochner.  I define nonnegative measurable functions $f_\epsilon$ for my example.  See below$^*$ for a modification with positive measurable functions.

Let $f_\epsilon$ be defined by
$$
f_\epsilon(x) = \begin{cases}
1,\quad&\text{ if }0<x<\epsilon<1
\\
0,\quad&\text{ otherwise}
\end{cases}
$$
so $f^*(x) = \sup_{\epsilon > 0}f_\epsilon(x) = \mathbf1_{(0,1)}(x)$ and therefore $f^* \in L_p(\mathbb R)$.
Define $F : \mathbb R \to \ell_{(0,\infty)}^\infty$ by
$F(x) = (f_\epsilon(x))_{\epsilon \in (0,\infty)}$.
Note: for $0<x<y<1$ we have
$$
\|F(y) - F(x)\|_\infty 
= \sup_{\epsilon\in(0,\infty)} |f_\epsilon(y) - f_\epsilon(x)|
\ge |f_y(y) - f_y(x)| = 1 .
$$
So the range of $F$ (even if we omit a set of $x$ with measure zero) is nonseparable.  Thus $F$ is not strongly measurable.

$^*$The above example has nonnegative functions $f_\epsilon$.  For a similar example with positive measurable functions, we may do this:
choose a fixed positive measurable function $g \in L^p$ and consider $g+f_\epsilon$.  Then for
$G(x) = ((g+f_\epsilon)(x))_{\epsilon \in (0,\infty)}$.  We have
$$
\|G(y) - G(x)\|_\infty = \|F(y) - F(x)\|_\infty
$$
so we get the same conclusion that $G$ is not strongly measurable.

Related counterexample for family $(f_n)_{n \in \mathbb N}$.
Now define $f_n(x)$ using the Rademacher funtions $r_n$:
$$
f_n(x) = \begin{cases}
1+r_n(x),\quad&\text{ if } 0 < x < 1
\\
0,\quad&\text{ otherwise}
\end{cases}
$$
Then $f^*(x) = \sup_m f_n(x) = 2\mathbf1_{(0,1)}(x)$ so $f^*\in L_p(\mathbb R)$.
Define $F : \mathbb R \to \ell_\infty(\mathbb N)$ by
$F(x) = (f_n(x))_{n \in \mathbb N}$.
If $0<x<y<1$ then
$$
\|F(y) - F(x)\|_\infty = 2 .
$$
So we get the same conclusion, $F$ is not Bochner measurable.
