Characterizing subsets of integrable functions

Question

Let us consider the space $L^1(0,1;\mathbb{R})$ of real-valued, Lebesgue integrable functions defined on the interval $(0,1)$ (where we only distinguish functions which are not equal almost everywhere). Let us define the function $S \colon L^1(0,1;\mathbb{R}) \to 2^{2^{(0,1)}}$ given by the formula: \begin{equation} \forall f \in L^1(0,1;\mathbb{R}) \quad S(f) = \{ A \subseteq (0,1) \text{ measurable }\mid \int_A |f| \, \mathrm{d}x = \frac{1}{2} \lVert f \rVert_{L^1} \}. \end{equation}

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Is there any simple way to characterize all $g \in L^1(0,1;\mathbb{R})$ such that for previously given $f$ we have $S(f) = S(g)$?

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It is easy to see that we can multiply $f$ by any $h \in L^\infty(0,1;\mathbb{R})$ such that $|h|=c$ almost everywhere, where $c \ge0$, and we would have $S(fh ) = S(f)$, however, I am not sure whether this exhausts all possiblities.

Answer 1

Yes, the functions $g$ with $S(g)=S(f)$ are the functions $g=hf$. Indeed, suppose $f \ge 0$ and $\|f\|_{L^1}=1$, then for any measurable $A \subset (0,1)$, we can compute $\int_A |f|\,dx$ based on the value of $S(f)$. There is some $a_1 \in (0,1)$ such that $\int_{(0,a_1)}|f|\,dx=\frac{1}{2}$. There are also some $a_{11},a_{12} \in (0,1)$ such that $\int_{(0,a_{11}) \cup (a_1,a_{12})} |f|\,dx=\int_{(0,a_{11}) \cup (a_{12},1)} |f|\,dx=\frac{1}{2}$, so we have $\int_{(0,a_{11})}|f|\,dx=\frac{1}{4}$. Extending this reasoning, we can decompose $(0,1)$ into arbitrarily short subintervals whose measure we know (if there is an interval $E$ where $f=0$, then this will appear immediately since there would be sets $B \cup E, B \backslash E \in S(f)$ for each $B \in S(f)$).