Let $p(\cdot) :[0;1] \to [1;\infty)$ be a measurable function.
Define $p^{*}(x)$ as the decreasing rearrangement of the measurable function $p(\cdot)$.
Hence we know that $p(\cdot)$ and $ p^{*}(x)$ are equimeasurable to each other. ( Since $p(\cdot)$ is positive-valued).
Let $p(\cdot\cdot) : [0;1]^2 \to [1; \infty)$ be a measurable function.
Define $p^{*}(x,y)$ as the decreasing rearrangement of the measurable function $p(\cdot\cdot)$.
Thereby we know that $p(\cdot\cdot)$ and $p^{*}(x,y)$ are equimeasurable to each other.(Since $p(\cdot\cdot)$ is positive-valued).
Define $\bar p(x,y) = p^{*}(x) + p^{*}(y)$ .
There is the question:
Will $\bar p(x,y)$ be equimeasurable to $p(\cdot\cdot)$ ?
I'm trying to figure out something from these definitions:
Note that :
$(1)$ Nonnegative functions $f$ and $g$ are called equimeasurable if $\eta_f = \eta_g$ , i.e.,
$$m\{f \gt y \} = m\{g \gt y \} .$$
$(2)$ Functions $|f|$ and $f^{*}$ are equimeasurable.
Any help would be appreciated.