Measurability of a net

Question

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 $\|a\|:=\sup_{\epsilon>0}|a_\epsilon|.$

Answer 1

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.

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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 $r_n$: $$ f_n(x) = \begin{cases} 1+r_n(x),\quad&\text{ if } 0 < x < 1 \\ 0,\quad&\text{ otherwise} \end{cases} $$ 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.