One character of code

(a) Shears preserve area

$S_f$ is well defined on $\mathbb{T}^2$ because $f$ is 1-periodic, so the formula does not depend on the representative of $y$, and replacing $x$ by $x+1$ changes the output by $(1,0)\equiv(0,0)$. It is a bijection with inverse $(x,y)\mapsto(x - f(y),\ y)$, well defined for the same reason. Likewise $R_g$ is a bijection with inverse $(x,y)\mapsto(x,\ y - g(x))$. Both maps and their inverses are Borel measurable, being compositions of the Borel map $(x,y)\mapsto(x, f(y), y)$ with continuous addition.

Let $B$ be a Borel set and, for $y\in[0,1)$, let $B^y = \{x\in\mathbb{R}/\mathbb{Z} : (x,y)\in B\}$ be its horizontal slice. The horizontal slice of $S_f^{-1}(B) = \{(x,y) : (x + f(y), y)\in B\}$ at height $y$ is $B^y - f(y)$, a translate of $B^y$ on the circle. Translation preserves Lebesgue measure $\lambda$ on $\mathbb{R}/\mathbb{Z}$, and $S_f^{-1}(B)$ is Borel, so by Tonelli’s theorem

The same argument with vertical slices gives $\mu(R_g^{-1}(B)) = \mu(B)$. Hence $\mu(T^{-1}B) = \mu\bigl(S_f^{-1}(R_g^{-1}B)\bigr) = \mu(R_g^{-1}B) = \mu(B)$. No smallness of $f$ or $g$ and no differentiability is needed.

(b) Uniform clouds stay uniform

Let $\Phi_n = T_n\circ\cdots\circ T_1$. By (a) applied repeatedly, $\Phi_n$ is a Borel bijection with $\mu(\Phi_n^{-1}B) = \mu(B)$ for every Borel $B$. If $P$ is uniform then $\Pr(\Phi_n(P)\in B) = \Pr(P\in\Phi_n^{-1}B) = \mu(\Phi_n^{-1}B) = \mu(B)$, so $\Phi_n(P)$ is uniform. The maps are deterministic, so for Borel $B_1,\dots,B_M$, using independence of the $P_i$,

So the images are independent and uniform.

(c) Forward Euler does not

The Jacobian matrix of $E$ is $I + hM$ with $M = \nabla v = \begin{pmatrix}\psi_{xy} & \psi_{yy} \\ -\psi_{xx} & -\psi_{xy}\end{pmatrix}$. For any 2 × 2 matrix, $\det(I + hM) = 1 + h\operatorname{tr}M + h^2\det M$. Here $\operatorname{tr}M = 0$ and $\det M = \psi_{xx}\psi_{yy} - \psi_{xy}^2 = K$, so $\det DE = 1 + h^2K$.

Injectivity. Lift $E$ to $\mathbb{R}^2$ by the same formula; $v$ is $\mathbb{Z}^2$-periodic. For $p\neq p'$, $|E(p) - E(p')| \ge |p - p'| - h|v(p) - v(p')| \ge (1 - h\,\mathrm{Lip}(v))\,|p - p'| > 0$. If $E(p) = E(p') + n$ with $n\in\mathbb{Z}^2$, then $E(p) = E(p' + n)$ by periodicity of $v$, so $p = p' + n$. Hence $E$ is injective on $\mathbb{T}^2$.

Positive Jacobian. For every $p$ and unit vector $w$, $|Mw| = |Dv(p)\,w| \le \mathrm{Lip}(v)$, so $\|M\| \le \mathrm{Lip}(v)$ and $h\|M\| < 1$. Since $\operatorname{tr}M = 0$, Cayley–Hamilton gives $M^2 = -(\det M)\,I = -K\,I$, so $|K| \le \|M\|^2$. Therefore $1 + h^2K \ge 1 - h^2\|M\|^2 > 0$ everywhere.

Area is not preserved. Suppose $K(p_0)\neq 0$. By continuity $K$ has a constant nonzero sign on an open disc $\tilde\Omega\subset\mathbb{R}^2$ of diameter less than 1 around a lift of $p_0$; its projection $\Omega$ is an isometric copy, so $\mu(\Omega) = \operatorname{area}(\tilde\Omega)$. The lifted $E$ is $C^1$ and injective on $\tilde\Omega$ with Jacobian $1 + h^2K > 0$, so by the change-of-variables formula $\operatorname{area}(E(\tilde\Omega)) = \int_{\tilde\Omega}(1 + h^2K)\,dA$, which is strictly larger or strictly smaller than $\operatorname{area}(\tilde\Omega)$ according to the sign of $K$. By the injectivity on the torus, the projection $\mathbb{R}^2\to\mathbb{T}^2$ is injective on $E(\tilde\Omega)$ (if $E(p)\equiv E(p')$ with $p, p'\in\tilde\Omega$ then $p\equiv p'$, and $p = p'$ because $\tilde\Omega$ has diameter less than 1), so $\mu(E(\Omega)) = \operatorname{area}(E(\tilde\Omega)) \neq \mu(\Omega)$. $E(\Omega)$ is open by the inverse function theorem, hence Borel. Take $B = E(\Omega)$: injectivity gives $E^{-1}(B) = \Omega$, so $\mu(E^{-1}B) = \mu(\Omega)\neq\mu(B)$, and $E$ does not preserve area in the sense of (a). $\blacksquare$

For the demo’s field, $v = (f(y), g(x))$ with $f, g$ of mean zero is $(\psi_y, -\psi_x)$ for $\psi = F(y) - G(x)$ with $F' = f$, $G' = g$ periodic. Then $\psi_{xx} = -g'(x)$, $\psi_{yy} = f'(y)$, $\psi_{xy} = 0$, so $K = -f'(y)\,g'(x)$, which is not identically zero for any non-constant $f$ and $g$.