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When source and target have continuous strictly increasing cumulative distributions, all transport maps are obtained by choosing a Lebesgue-measure-preserving map $S$ on the unit interval. The maps $F$ and $G$ turn the two measures into uniform measure. The choice $S=\mathrm{id}$ gives the monotone rearrangement.

The answer is the monotone rearrangementIt pushes $\mu$ to $\nu$ by the same cumulative-distribution argument as in part (a).

For the squared transport cost, the two-point swap difference isThus any optimal support has no crossed pairs: $x_1<x_2$ forces $y_1\le y_2$. This follows from the finite-cost converse proved above; every cost here is bounded. Conversely, an increasing graph is c-cyclically monotone for this cost. After removing the marginal terms $\sum x_i^2+\sum y_i^2$, the cycle inequalities say that pairing the sorted $x_i$ and $y_i$ maximizes $\sum x_i{y}_i$. Exchanging any inverted pairing increases that sum by the nonnegative product of the two differences, so repeated exchanges prove the inequality. Therefore the displayed increasing transport is optimal.

For uniqueness, let $\pi$ be any noncrossing coupling. The sets $A=\{x\le a\}$ and $B=\{y\le b\}$ cannot both have positive mass in $A\setminus B$ and $B\setminus A$: a point from each would give a strictly crossed pair. Consequently one of these differences has zero mass, andThis determines the joint distribution uniquely and is exactly the distribution of the common-quantile coupling $(F^{-1}(U),G^{-1}(U))$ for uniform $U$. Since $F$ is continuous and strictly increasing, that coupling is induced by $T=G^{-1}F$. Hence there is one optimal deterministic plan, with maps differing only on $\mu$-null sets; in fact it is the unique optimal coupling. This proves the one-dimensional quadratic transport uniqueness criterion directly.

The intended monotone rearrangement isHere $G^{-1}$ is the quantile function of $\nu$, agreeing with the ordinary inverse when $G$ is continuous and strictly increasing. With an atomless measure $\mu$, its cumulative distribution function $F$ is continuous, and the probability integral transform makes $F(X)$ uniform on $(0,1)$ for $X\sim \mu$. Thus $G^{-1}(F(X))\sim \nu$. The one-dimensional monotone rearrangement theorem says this transport map minimizes the cost $d(x-y)$ for convex continuous $d$, whenever the cost integrals are well defined. Values at exceptional endpoints may be chosen arbitrarily.

The printed assumptions omit an essential source condition. Invertibility of $G$ alone does not ensure an admissible transport map. For example, $\mu ={\delta }_0$ and $\nu$ a standard normal distribution satisfy the stated condition on $G$, but $T_{\#}{\delta }_0$ is always a Dirac measure. There is no solution to the Monge optimal transport problem in this example. The boxed answer therefore requires the additional assumption that $\mu$ is an atomless measure, or an equivalent condition making the displayed map admissible. For arbitrary sources the always admissible monotone transport plan is $(F^{-1},G^{-1}{)}_{\#}\mathcal{U}(0,1)$, which need not be induced by a map.

Interpret the invertibility assumption on $G$ as continuity and strict increase on the relevant range, so that $\nu$ is an atomless measure. Suppose a non-decreasing transport map $T$ with $T_{\#}\mu =\nu$ exists. Then $\mu$ is also an atomless measure: if $\mu (\{x\})>0$, the pushforward measure would give $\nu (\{T(x)\})\ge \mu (\{x\})>0$. Thus $F$ is continuous.

At any point $x$ where the non-decreasing representative is defined, the definition of a monotone function givesUsing the pushforward measure identity and continuity of the two cumulative distribution functions, we obtainConsequentlyThe monotone rearrangement in part (c) is optimal for the convex difference cost, so $T$ has the same cost and solves the Monge optimal transport problem. The meaningful uniqueness is up to a $\mu$-null set; arbitrary values away from the source do not affect transport or cost. Existence of the non-decreasing transport map supplies the source condition missing in part (c).

Write $d\mu =f(x)dx$, with $f\ge 0$. Applying polar coordinates and Fubini's theorem gives, for bounded nonnegative measurable $h$,Also $\tilde{\mu }(\{0\})=\mu (\{0\})=0$ by absolute continuity of measures. Extending $g$ by zero on the negative half-line proves

Both radial pushforward measures have compact support of a measure, so Brenier's theorem on the line supplies a unique optimal transport plan and a transport map $\tau$, unique $\tilde{\mu }$-almost everywhere, withIn one dimension, the gradient of a convex function is a nondecreasing function. Nonnegativity holds $\tilde{\mu }$-almost everywhere because ${\tau }_{\#}\tilde{\mu }=\tilde{\nu }$ is carried by $[0,\infty )$. To choose a nonnegative, nondecreasing representative everywhere, let $F,G$ be the radial cumulative distribution functions and use the monotone rearrangementAt $F(r)=0$ set $\tau (r)=0$, and at $F(r)=1$ use the largest point of the support of a measure $\tilde{\nu }$. These choices preserve monotonicity and boundedness. Since $F$ is continuous, $F(R)$ has the uniform distribution on $(0,1)$ when $R$ has law $\tilde{\mu }$, and the quantile function $G^{-1}$ sends this uniform distribution to $\tilde{\nu }$. Thus this representative has the required pushforward measure, even if the target has atoms.

If $X$ has a continuous cumulative distribution function $F$, then $F(X)$ has the uniform distribution on $(0,1)$. Combined with inverse transform sampling, this constructs the monotone rearrangement $G^{-1}(F(X))$ to a target with quantile function $G^{-1}$.