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 $\|a\|:=\sup\_{\epsilon>0}|a\_\epsilon|.$



Measurability of a net
I assume "strongly measurable" is [in the sense of Bochner](https://en.wikipedia.org/wiki/Bochner_measurable_function). I define *nonnegative* measurable functions $f\_\epsilon$ for my example. See below$^\*$ for a modification with *positive* measurable functions.




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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.




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$^\*$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.




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Related counterexample for family $(f\_n)\_{n \in \mathbb N}$.  

Now define $f\_n(x)$ using the [Rademacher funtions](https://en.wikipedia.org/wiki/Rademacher_system) $r\_n$:
$$
f\_n(x) = \begin{cases}
1+r\_n(x),\quad&\text{ if } 0 < x < 1
\\
0,\quad&\text{ otherwise}
\end{cases}
$$
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.


