Proof. We may assume that $g = \varnothing $ ; otherwise, pass to the subgraph $G\left [V(G) \setminus \mathrm {dom}(g)\right ]$ and replace $\mathcal {L}$ by the list assignment $x \mapsto \mathcal {L}(x) \setminus \left \{g(y) : y \in N_G(x) \cap \mathrm {dom}(g)\right \}$ . Let $\vartheta \colon V(G) \to {\mathbb {N}}$ be a Borel proper colouring. Fix a Borel linear ordering on $\mathcal {C}$ and recursively define partial $\mathcal {L}$ -colourings $f_n \colon \vartheta ^{-1}(n) \rightharpoonup \mathcal {C}$ as follows: set

$$ \begin{align*} \mathcal{L}_n(x) := \mathcal{L}(x) \setminus \left\{\alpha : \text{there is } y \in N_G(x) \text{ with } \vartheta(y) < n \text{ and } f_{\vartheta(y)}(y) = \alpha\right\}, \end{align*} $$

and let $f_n(x)$ be the smallest colour in $\mathcal {L}_n(x)$ if $\mathcal {L}_n(x) \neq \varnothing $ , leaving $f_n(x)$ undefined otherwise. Then the union $f := \bigcup _{n=0}^\infty f_n$ is a Borel inclusion-maximal proper partial $\mathcal {L}$ -colouring of G, as desired.

Proof. This is immediate from Proposition 2.1 (note that $\chi _{\mathrm {B}}(G) \leq \Delta + 1 < \aleph _0$ , by Theorem 1.2).

Proof. Apply Proposition 2.1 with $\mathcal {L} \colon x \mapsto \left \{0\right \}$ and $g \colon J \to \left \{0\right \} \colon x \mapsto 0$ .

Assumptions for Section 3.1

Fix a Borel graph G of finite maximum degree $\Delta $ with vertex set V and edge set E; a finite set of colours $\mathcal {C}$ of size $k \geq \Delta + 1$ ; and $\mu \in \mathsf {Inv}(G)$ .

Proof. Suppose, toward a contradiction, that f is a $\mu $ -measurable proper $\mathcal {C}$ -colouring of G that is $(\mu , \mathsf {RM}_3)$ -perfect but not $\mu $ -equitable. After passing to a $\mu $ -conull G-invariant Borel subset of V, we may assume that in fact no recolouring move $\varphi \in \mathsf {RM}_3$ improves f with respect to $\mu $ .

For $\gamma \in \mathcal {C}$ , let $V_\gamma := f^{-1}(\gamma )$ be the corresponding colour class. Define $a := \min _{\gamma \in \mathcal {C}} \mu \left (V_\gamma \right )$ and

$$ \begin{align*} \mathcal{A} := \left\{\alpha \in \mathcal{C} : \mu(V_\alpha) = a\right\} \qquad \text{and} \qquad \mathcal{B} := \mathcal{C} \setminus \mathcal{A}. \end{align*} $$

Since f is not $\mu $ -equitable, $\mathcal {B} \neq \varnothing $ . Define $b := \lvert \mathcal {B}\rvert ^{-1} \sum _{\beta \in \mathcal {B}} \mu \left (V_\beta \right )$ and notice that $a < 1/k < b$ . Set

$$ \begin{align*} A := f^{-1}(\mathcal{A}) \qquad \text{and} \qquad B := f^{-1}(\mathcal{B}). \end{align*} $$

Figure 2 The recolouring move from Claim 3.1.1.

Given $x \in B$ and $y \in A$ , we say that y is a solo neighbour of x if y is the unique neighbour of x with the colour $f(y)$ . (This notion plays a similarly central role in [Reference Kierstead, Kostochka, Mydlarz and SzemerédiKie+10].) For $y \in A$ , define

$$ \begin{align*} S(y) := \left\{x \in B : y\ \text{is a solo neighbour of}\ x\right\}. \end{align*} $$

Proof. Take any $x \in S(y)$ and define $\beta := f(x)$ . Suppose that y has no neighbour in $V_{\alpha '}$ and consider the recolouring move $\varphi := \left \{(x, \alpha ), (y, \alpha ')\right \}$ (see Figure 3). Since y is not adjacent to any vertex in $V_{\alpha '}$ , and the only neighbour of x that is coloured $\alpha $ is y, $\varphi $ is acceptable for f. But $\mathcal {D}^+(f, \varphi ) = \left \{\alpha '\right \}, \varphi ) = \left \{\beta \right \}$ and $\mu \left (V_{\alpha '}\right ) < \mu \left (V_\beta \right )$ , so $\varphi $ improves f with respect to $\mu $ .

Figure 3 The recolouring move from Claim 3.1.2.

Proof. Suppose, toward a contradiction, that $G[S(y)]$ is not a clique. This means that there are two distinct nonadjacent vertices $x, x' \in S(y)$ . Define $\alpha := f(y)$ , $\beta := f(x)$ and $\beta ' := f(x')$ . Observe that there is a colour $\gamma \neq \alpha $ such that

(3.3) $$ \begin{align} \left(N_G(y) \cap V_\gamma\right) \setminus \left\{x, x'\right\} = \varnothing. \end{align} $$

Otherwise y would have, in addition to x and $x'$ , at least one neighbour in every colour class $V_\gamma $ except for $V_\alpha $ , which would yield $\deg _G(y) \geq 2 + (k-1) = k+1> \Delta $ . Now fix any $\gamma \neq \alpha $ satisfying equation (3.3) and consider the recolouring move $\varphi := \left \{(x, \alpha ),(x', \alpha ),(y, \gamma )\right \}$ (see Figure 4). The choice of x, $x'$ and $\gamma $ implies that $\varphi $ is acceptable for f. Since $\alpha \in \mathcal {D}^+(f, \varphi )$ , $\mathcal {D}^-(f, \varphi ) \subseteq \left \{\beta , \beta '\right \}$ and $\mu (V_\alpha ) < \mu \left (V_\beta \right )$ , $\mu \left (V_{\beta '}\right )$ , we conclude that $\varphi $ improves f with respect to $\mu $ .

Figure 4 The recolouring move from Claim 3.1.3.

Proof. Fix an arbitrary colour $\alpha \in \mathcal {A}$ . Let X be the set of all vertices $x \in B$ that have a solo neighbour in $V_\alpha $ , and let Y be the set of all vertices $y \in V_\alpha $ that have a neighbour in X. By Claim 3.1.2, every vertex $y \in Y$ has at least $\lvert \mathcal {A}\rvert - 1 = k - \lvert \mathcal {B}\rvert - 1$ neighbours in A, and thus

(3.4) $$ \begin{align} \lvert N_G(y) \cap B\rvert \leq \Delta - k + \lvert\mathcal{B}\rvert + 1. \end{align} $$

This immediately yields

(3.5) $$ \begin{align} \int_{V_{\alpha}} \lvert N_G(y) \cap B\rvert \mathrm{d} \mu(y) \leq \Delta \mu(V_\alpha) - (k - \lvert\mathcal{B}\rvert - 1) \mu(Y). \end{align} $$

On the other hand, we have

$$ \begin{align*}\hspace{7.8pc} \mu(X) &= \int_{X} 1 \mathrm{d} \mu(x) \\ &= \int_{X} \lvert N_G(x) \cap Y\rvert \mathrm{d}\mu(x) &\qquad\qquad\qquad {(\text{by the definition of}\ X\ \text{and}\ Y)}\\ &= \int_{Y} \lvert N_G(y) \cap X\rvert \mathrm{d}\mu(y) &\qquad\qquad\qquad {(\mu\ \text{is}\ G\text{-invariant})} \\ &\leq (\Delta - k + \lvert\mathcal{B}\rvert + 1) \mu(Y). &\qquad\qquad\qquad {(\text{by formula (3.4)})} \end{align*} $$

This together with Claim 3.1.1 gives

(3.6) $$ \begin{align} \int_{B} \lvert N_G(x) \cap V_\alpha\rvert \mathrm{d} \mu(x) \geq 2\mu(B) - \mu(X) \geq 2\mu(B) - (\Delta - k + \lvert\mathcal{B}\rvert + 1) \mu(Y). \end{align} $$

Since $\mu $ is G-invariant, we can combine formulas (3.6) and (3.5) to get

$$ \begin{align*} \Delta \mu(V_\alpha) - (k - \lvert\mathcal{B}\rvert - 1) \mu(Y) \geq 2\mu(B) - (\Delta - k + \lvert\mathcal{B}\rvert + 1) \mu(Y). \end{align*} $$

Recalling that $\mu (V_\alpha ) = a$ and $\mu (B) = \lvert \mathcal {B}\rvert b$ , we can rewrite the last inequality as

$$ \begin{align*} \Delta a - (k-\lvert\mathcal{B}\rvert-1)\mu(Y) \geq 2\lvert\mathcal{B}\rvert b - (\Delta-k+\lvert\mathcal{B}\rvert+1)\mu(Y). \end{align*} $$

After moving all the terms to one side and using the fact that $b> a$ , we obtain

$$ \begin{align*} \Delta - 2\lvert\mathcal{B}\rvert + (\Delta -2k + 2 \lvert\mathcal{B}\rvert +2)\frac{\mu(Y)}{b}> 0. \end{align*} $$

Since $0 \leq \mu (Y)/b < \mu (Y)/a \leq 1$ , at least one of the following two inequalities holds:

$$ \begin{align*} \Delta - 2\lvert\mathcal{B}\rvert> 0 \qquad \text{or} \qquad \Delta - 2\lvert\mathcal{B}\rvert + (\Delta -2k + 2 \lvert\mathcal{B}\rvert +2) = 2\Delta -2k +2 > 0. \end{align*} $$

The second of these inequalities necessarily fails, since $2\Delta - 2k + 2 \leq 2\Delta - 2(\Delta + 1) + 2 = 0$ . Hence, we must have $\Delta - 2\lvert \mathcal {B}\rvert> 0$ , and thus $\Delta \geq 2\lvert \mathcal {B}\rvert +1$ , as desired.

We are now ready for the final stage of the argument. At least one of the colour classes $V_\beta $ , $\beta \in \mathcal {B}$ , has measure at least b, so by taking J to be any such colour class and applying Corollary 2.3 to the induced subgraph $G[B]$ , we obtain a Borel maximal G-independent subset $I \subseteq B$ such that $\mu (I) \geq b$ . For each $x \in I$ , let $\Sigma (x)$ be the set of all solo neighbours of x. Then, on the one hand,

$$ \begin{align*} \lvert N_G(x) \cap A\rvert \geq 2\lvert\mathcal{A}\rvert - \lvert\Sigma(x)\rvert = 2k - 2\lvert\mathcal{B}\rvert - \lvert\Sigma(x)\rvert, \end{align*} $$

while on the other hand, $\lvert N_G(x) \cap A\rvert \leq \Delta - \lvert N_G(x) \cap B\rvert $ , which yields

(3.7) $$ \begin{align} \lvert\Sigma(x)\rvert \geq 2k - 2\lvert\mathcal{B}\rvert - \Delta + \lvert N_G(x) \cap B\rvert. \end{align} $$

Since I is G-independent, Claim 3.1.3 implies that the sets $\Sigma (x)$ , $x \in I$ , are pairwise disjoint. Using the fact that $\bigcup _{x \in I} \Sigma (x) \subseteq A$ and the G-invariance of $\mu $ , we conclude that

$$ \begin{align*} \mu(A) \geq \int_{I} \lvert\Sigma(x)\rvert \mathrm{d}\mu(x). \end{align*} $$

From formula (3.7), it follows that

$$ \begin{align*} \int_I \lvert\Sigma(x)\rvert \mathrm{d}\mu(x) \geq (2k - 2\lvert\mathcal{B}\rvert - \Delta) \mu(I) + \int_I \lvert N_G(x) \cap B\rvert \mathrm{d} \mu(x). \end{align*} $$

Since I is a maximal G-independent subset of B, every vertex $z \in B \setminus I$ has a neighbour in I, and thus

$$ \begin{align*} \int_I \lvert N_G(x) \cap B\rvert \mathrm{d} \mu(x) = \int_{B \setminus I} \lvert N_G(z) \cap I\rvert \mathrm{d} \mu(z) \geq \mu(B \setminus I) = \mu(B) - \mu(I). \end{align*} $$

To summarise, we have

(3.8) $$ \begin{align} \mu(A) \geq (2k-2\lvert\mathcal{B}\rvert-\Delta-1) \mu(I) + \mu(B). \end{align} $$

Claim 3.1.4 and the inequality $k \geq \Delta +1$ yield $2k-2\lvert \mathcal {B}\rvert -\Delta - 1> 0$ . Hence, since $\mu (I) \geq b$ , the right-hand side of formula (3.8) could only decrease if we replace $\mu (I)$ by b, so

$$ \begin{align*} \mu(A) \geq (2k-2\lvert\mathcal{B}\rvert-\Delta - 1) b + \mu(B). \end{align*} $$

Using $\mu (B) = \lvert \mathcal {B}\rvert b$ , $\mu (A) = \lvert \mathcal {A}\rvert a = (k - \lvert \mathcal {B}\rvert )a$ and $b> a$ , we obtain

$$ \begin{align*} k - \lvert\mathcal{B}\rvert> (2k-2\lvert\mathcal{B}\rvert-\Delta - 1) + \lvert\mathcal{B}\rvert, \end{align*} $$

which after simplifying becomes $k < \Delta + 1$ . This is a contradiction, and Lemma 3.1 is proved.

Assumptions for Section 3.2

Fix a finite set of colours $\mathcal {C}$ of size $k \geq 1$ .

Proof. For each distribution $\omega $ on $\mathcal {C}$ , let $\omega ^\ast \colon \left \{1, \dotsc , k\right \} \to [0,1]$ denote the mapping obtained by putting the values of $\omega $ in nondecreasing order; that is, the lists

$$ \begin{align*} \omega(\alpha), \alpha \in \mathcal{C}, \qquad \text{and} \qquad \omega^\ast(1), \omega^\ast(2), \dotsc, \omega^\ast(k) \end{align*} $$

contain the same elements, and also $\omega ^\ast (1) \leq \omega ^\ast (2) \leq \dotsb \leq \omega ^\ast (k)$ .

Proof. For each $\alpha \in \mathcal {C}$ , let $I_\alpha $ be the open interval with endpoints $\omega (\alpha )$ and $\eta (\alpha )$ . Similarly, for each $1 \leq i \leq k$ , let $J_i$ be the open interval with endpoints $\omega ^\ast (i)$ and $\eta ^\ast (i)$ . Then

(3.9) $$ \begin{align} \lVert\omega - \eta\rVert_1 = \sum_{\alpha \in \mathcal{C}} \text{length}(I_\alpha) \qquad \text{and} \qquad \lVert\omega^\ast - \eta^\ast\rVert_1 = \sum_{i=1}^k \text{length}(J_i). \end{align} $$

Define $S := \left \{\omega (\alpha ), \ \eta (\alpha ) : \alpha \in \mathcal {C}\right \}$ . Note that the set S is finite. We shall argue that for each $x \in \mathbb {R}\setminus S$ , the number of colours $\alpha $ with $x \in I_\alpha $ is not less than the number of indices i with $x \in J_i$ ; in view of equation (3.9), this immediately yields the claim. Take any $x \in \mathbb {R}\setminus S$ and define

$$ \begin{align*} \tau(x) := \lvert\left\{\alpha \in \mathcal{C} : \omega(\alpha) < x\right\}\rvert \qquad \text{and} \qquad \sigma(x) := \lvert\left\{\alpha \in \mathcal{C} : \eta(\alpha) < x\right\}\rvert. \end{align*} $$

Then x is in precisely $\lvert \tau (x) - \sigma (x)\rvert $ of the intervals $J_i$ , $1 \leq i \leq k$ . On the other hand, if $\tau (x) \geq \sigma (x)$ , then there are at least $\tau (x) - \sigma (x)$ colours $\alpha $ with $\omega (\alpha ) < x < \eta (\alpha )$ , whereas if $\sigma (x) \geq \tau (x)$ , then there are at least $\sigma (x) - \tau (x)$ colours $\alpha $ with $\eta (\alpha ) < x < \omega (\alpha )$ . In either case, x belongs to at least $\lvert \tau (x) - \sigma (x)\rvert $ of the intervals $I_\alpha $ , $\alpha \in \mathcal {C}$ , and hence we are done.

The proof of Lemma 3.2 rests on the following key observation:

Claim 3.2.2. Suppose that $\omega $ and $\eta $ are distributions on $\mathcal {C}$ and $\omega \vartriangleleft \eta $ . Then there is an index $\ell $ such that $\omega ^\ast (i) \leq \eta ^\ast (i)$ for all $1 \leq i \leq \ell $ , and there is a colour $\alpha \in \mathcal {D}^+(\omega , \eta )$ with

(3.10) $$ \begin{align} \sum_{i=1}^\ell \eta^\ast(i) - \sum_{i=1}^\ell \omega^\ast(i) \geq \eta(\alpha) - \omega(\alpha). \end{align} $$

Proof. Let $\alpha \in \mathcal {D}^+(\omega , \eta )$ be such that for all $\beta \in \mathcal {D}^-(\omega , \eta )$ , we have $\eta (\alpha ) \leq \eta (\beta )$ ; and let $\ell $ be the least index such that $\eta ^\ast (\ell ) = \eta (\alpha )$ . We claim that this choice of $\ell $ and $\alpha $ works.

To begin with, fix a bijection $\left \{1, \dotsc , k\right \} \to \mathcal {C} \colon i \mapsto \alpha _i$ such that $\eta ^\ast (i) = \eta (\alpha _i)$ for all $1 \leq i \leq k$ , and $\alpha _\ell = \alpha $ . Consider any $1 \leq i \leq \ell $ . We have $\eta (\alpha _i) = \eta ^\ast (i) \leq \eta ^\ast (\ell )$ , where equality holds if and only if $i = \ell $ . By the choice of $\alpha $ , this yields $\alpha _i \not \in \mathcal {D}^-(\omega , \eta )$ , and therefore $\omega (\alpha _i) \leq \eta (\alpha _i) = \eta ^\ast (i)$ . Thus, there are at least i colours – namely $\alpha _1, \dotsc , \alpha _i$ – for which the value of $\omega $ does not exceed $\eta ^\ast (i)$ . But this precisely means that $\eta ^\ast (i) \geq \omega ^\ast (i)$ , as desired.

Next we prove formula (3.10). Let $1 \leq t \leq \ell $ be the largest index such that $\omega ^\ast (t) \leq \omega (\alpha )$ (such a t exists, as $\omega ^\ast (1) \leq \omega (\alpha )$ ). We claim that if $t < s \leq \ell $ , then $\omega ^\ast (s) \leq \eta ^\ast (s-1)$ . Indeed, suppose, toward a contradiction, that $\omega ^\ast (s)> \eta ^\ast (s-1)$ . Then $\omega ^\ast (s-1) \leq \eta ^\ast (s-1) < \omega ^\ast (s)$ , meaning that the only colours for which the value of $\omega $ is at most $\eta ^\ast (s-1)$ are $\alpha _1, \dotsc , \alpha _{s-1}$ . Since $\alpha = \alpha _\ell $ is not among them, $\omega (\alpha ) \geq \omega ^\ast (s)$ , contradicting the choice of t and the fact that $s> t$ . Now we can write

$$ \begin{align*}\hspace{1.5pc} \sum_{i=1}^\ell \eta^\ast(i) - \sum_{i=1}^\ell \omega^\ast(i) &= \eta^\ast(\ell) + \sum_{s=t+1}^\ell \left(\eta^\ast(s-1) - \omega^\ast(s)\right) - \omega^\ast(t) \,+\, \sum_{i = 1}^{t-1} \left(\eta^\ast(i) - \omega^\ast(i)\right) \\ &\geq \eta^\ast(\ell) - \omega^\ast(t) \qquad\qquad\qquad\qquad \ \ \, ({\text{the other summands are nonnegative}}) \\[4pt] &\geq \eta(\alpha) - \omega(\alpha), \qquad\qquad\qquad\qquad({\text{since}\ \eta^\ast(\ell) = \eta(\alpha)\ \text{and}\ \omega^\ast(t) \leq \omega(\alpha)}) \end{align*} $$

as desired.

Now let $(\omega _n)_{n=0}^\infty $ be as in the statement of Lemma 3.2. Define $S_n(\ell ) := \sum _{i=1}^\ell \omega ^\ast _n(i)$ for each $n \in {\mathbb {N}}$ and $1 \leq \ell \leq k$ . Define $R := \left \{n \in {\mathbb {N}} : \omega _n \neq \omega _{n+1}\right \}$ . Claim 3.2.2 shows that for every $n \in R$ , there exist an index $\ell _n$ and a colour $\alpha _n \in \mathcal {D}^+(\omega , \eta )$ such that $\omega ^\ast _n(i) \leq \omega ^\ast _{n+1}(i)$ for all $1 \leq i \leq \ell _n$ , and also

(3.11) $$ \begin{align} S_{n+1}(\ell_n) - S_n(\ell_n) \geq \omega_{n+1}(\alpha_n) - \omega_n(\alpha_n) \geq A^{-1} \cdot \lVert\omega_{n+1} - \omega_n\rVert_1. \end{align} $$

For each $\ell $ , define $R_\ell := \left \{n \in R : \ell _n = \ell \right \}$ and $R_{<\ell } := \left \{n \in R : \ell _n < \ell \right \} = \bigcup _{i=1}^{\ell -1} R_i$ . We will show that

(3.12) $$ \begin{align} \sum_{n \in R_\ell} \left(S_{n+1}(\ell) - S_n(\ell)\right) \leq \frac{(1+A)^\ell - 1}{A} \cdot {\mathrm{disc}}(\omega_0). \end{align} $$

Once inequality (3.12) is proved, we obtain, using formula (3.11), that

$$ \begin{align*} \sum_{n=0}^\infty \lVert\omega_{n+1} - \omega_n\rVert_1 &\leq A \cdot \sum_{n\in R} \left(S_{n+1}(\ell_n) - S_n(\ell_n)\right) \\ &= A \cdot \sum_{\ell=1}^k \sum_{n \in R_\ell} \left(S_{n+1}(\ell) - S_n(\ell)\right) \\ &\leq \sum_{\ell=1}^k \left((1+A)^\ell - 1\right) \cdot {\mathrm{disc}}(\omega_0) = \frac{(1+A)^{k+1} - (k+1)A - 1}{A}\cdot {\mathrm{disc}}(\omega_0)\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \, ({\text{by formula (3.12)}}) \\ &\leq \frac{(1+A)^{k+1}}{A} \cdot {\mathrm{disc}}(\omega_0), \end{align*} $$

as desired. To prove formula (3.12), we use induction on $\ell $ , so suppose that the inequality holds with $\ell $ replaced by any $1 \leq i < \ell $ . Notice that if $\ell _n \geq \ell $ , then $S_{n+1}(\ell ) - S_n(\ell ) \geq 0$ . Hence, for each $m \in {\mathbb {N}}$ ,

$$ \begin{align*} \ell \cdot {\mathrm{disc}}(\omega_0) \geq \frac{\ell}{k} - S_0(\ell) &\geq S_{m+1}(\ell) - S_0(\ell) = \sum_{n = 0}^{m} \left(S_{n+1}(\ell) - S_n(\ell)\right) \\ &\geq \sum_{\substack{n \in R_\ell \\ n \leq m}} \left(S_{n+1}(\ell) - S_n(\ell)\right) - \sum_{\substack{n \in R_{< \ell} \\ n \leq m}} \left\lvert S_{n+1}(\ell) - S_n(\ell)\right\rvert. \end{align*} $$

As m here is arbitrary, we conclude that

$$ \begin{align*} \sum_{n \in R_\ell} \left(S_{n+1}(\ell) - S_n(\ell)\right) \leq \ell \cdot {\mathrm{disc}}(\omega_0) + \sum_{n \in R_{< \ell}} \left\lvert S_{n+1}(\ell) - S_n(\ell)\right\rvert. \end{align*} $$

We can now write the following chain of inequalities:

$$ \begin{align*} \hspace{3.2pc} \sum_{n \in R_{< \ell}} \left\lvert S_{n+1}(\ell) - S_n(\ell)\right\rvert &\leq \sum_{n \in R_{< \ell}} \left\lVert\omega^\ast_{n+1} - \omega^\ast_n \right\rVert_1 \\ &\leq \sum_{n \in R_{< \ell}} \left\lVert\omega_{n+1} - \omega_n \right\rVert_1 & \qquad ({\text{by Claim 3.2.1}}) \\ &\leq A \cdot \sum_{i=1}^{\ell-1} \sum_{n \in R_i} \left(S_{n+1}(i) - S_n(i)\right) & \qquad ({\text{by formula (3.11)}} )\\ &\leq \sum_{i = 1}^{\ell-1} \left((1+A)^i - 1\right) \cdot {\mathrm{disc}}(\omega_0). & \qquad ({\text{by the inductive hypothesis}}) \end{align*} $$

Therefore,

$$ \begin{align*} \sum_{n \in R_\ell} \left(S_{n+1}(\ell) - S_n(\ell)\right) \leq \ell \cdot {\mathrm{disc}}(\omega_0)+ \sum_{i = 1}^{\ell-1} \left((1+A)^i - 1\right) \cdot {\mathrm{disc}}(\omega_0) = \frac{(1+A)^{\ell}-1}{A} \cdot {\mathrm{disc}}(\omega_0), \end{align*} $$

and the proof is complete.

Assumptions for Section 3.3

Fix an aperiodic Borel graph G of finite maximum degree $\Delta $ with vertex set V and edge set E and a finite set of colours $\mathcal {C}$ of size $k \geq \Delta + 1$ .

Proof. For $\alpha , \beta \in \mathcal {C}$ , define $V_{\alpha \beta } := f^{-1}(\alpha ) \cap g^{-1}(\beta )$ . Then, for each fixed $\alpha \in \mathcal {C}$ ,

$$ \begin{align*} \left\lvert\mu\left(f^{-1}(\alpha)\right) - \mu\left(g^{-1}(\alpha)\right)\right\rvert = \left\lvert\sum_{\beta \in \mathcal{C}} \mu\left(V_{\alpha \beta}\right) - \sum_{\beta \in \mathcal{C}} \mu\left(V_{\beta \alpha}\right) \right\rvert \leq \sum_{\substack{\beta \in \mathcal{C}\\\beta \neq \alpha}} \left(\mu\left(V_{\alpha \beta}\right) + \mu\left(V_{\beta \alpha}\right)\right) \end{align*} $$

and therefore,

$$ \begin{align*} \hspace{-16pt}\lVert f_\ast(\mu)-g_\ast(\mu)\rVert_1 \leq \sum_{\alpha \in \mathcal{C}} \sum_{\substack{\beta \in \mathcal{C}\\\beta \neq \alpha}} \left(\mu\left(V_{\alpha \beta}\right) + \mu\left(V_{\beta \alpha}\right)\right) = 2 \cdot \sum_{\substack{\alpha, \beta \in \mathcal{C}\\\alpha \neq \beta}} \mu\left(V_{\alpha \beta}\right) = 2 \cdot {\mathrm{dist}}_\mu(f,g).\\[-55pt] \end{align*} $$

Proof. Since each $\varphi \in M$ has finite domain, there is a Borel function $p \colon M \to V$ such that $p(\varphi ) \in \mathrm {dom}(\varphi )$ for every $\varphi \in M$ (see Section 2.0.1). Set $P := \mathrm {im}(p)$ . Note that P is Borel, as it is the image of a Borel set under an injective Borel function [Reference KechrisKec95, Corollary 15.2]. Since for each $\varphi \in M$ we have $\lvert \mathrm {dom}(\varphi ) \cap P\rvert = 1$ and $\lvert \mathrm {dom}(\varphi )\rvert \leq m$ , and since $\mu $ is G-invariant, we conclude that $\mu (P) \geq \mu (M)/m$ . For each $x \in P$ , let $\varphi _x \in M$ be the unique recolouring move such that $p(\varphi _x) = x$ . The G-invariance of $\mu $ yields that for each $\alpha \in \mathcal {C}$ ,

where

is defined in expression (3.1). Crucially, $M \subseteq \mathsf {RM}_m\left (f; \mathcal {D}^+, \mathcal {D}^-\right )$ , which means that

for all $\alpha \in \mathcal {D}^+$ and $\varphi \in M$ . Hence, for any $\alpha \in \mathcal {D}^+$ , we can write

$$ \begin{align*} \lVert f_\ast(\mu)-(f \oplus M)_\ast(\mu)\rVert_1 &\geq \mu\left((f \oplus M)^{-1}(\alpha)\right) - \mu\left(f^{-1}(\alpha)\right) \\ &\geq \int_P 1 \mathrm{d}\mu(x) = \mu(P) \geq \frac{\mu(M)}{m} \\ &\geq \frac{{\mathrm{dist}}_\mu(f, f \oplus M)}{m}\\ &\geq \frac{1}{2m}\cdot \lVert f_\ast(\mu)-(f \oplus M)_\ast(\mu)\rVert_1.\qquad\qquad\qquad ({\text{by Lemma 3.3}}) &\\[-42pt]\end{align*} $$

Proof. Since G is aperiodic, every G-invariant probability measure is atomless, so if M is countable, then we can simply take $g = f$ . Hence, we may assume that M is uncountable. Due to the Borel isomorphism theorem [Reference KechrisKec95, Theorem 15.6], we can fix a Borel bijection $\mathbb {R}_{\geq 0} \to M \colon t \mapsto \varphi _t$ . For $t \in \mathbb {R}_{\geq 0} \cup \left \{\infty \right \}$ , define $M_t := \left \{\varphi _s : 0 \leq s < t\right \}$ . In particular, $M_0 = \varnothing $ and $M_\infty = M$ . Define $f_t := f \oplus M_t$ .

Now let $\mu $ be a G-invariant probability measure. Observe the following facts:

• • For each $\alpha \in \mathcal {D}^+$ , the function $t \mapsto \mu \left (f_t^{-1}(\alpha )\right )$ is nondecreasing.

• • For each $\beta \in \mathcal {D}^-$ , the function $t \mapsto \mu \left (f_t^{-1}(\beta )\right )$ is nonincreasing.

• • For each $\gamma \in \mathcal {C} \setminus \left (\mathcal {D}^+ \cup \mathcal {D}^-\right )$ , the function $t \mapsto \mu \left (f_t^{-1}(\gamma )\right )$ is constant.

Furthermore, since $\mu $ is atomless, we have the following additional fact:

• • For each $\gamma \in \mathcal {C}$ , the function $t \mapsto \mu \left (f^{-1}_t(\gamma )\right )$ is continuous.

By definition, $f_\ast (\mu ) \trianglelefteq (f_t)_\ast (\mu )$ if and only if either $f_\ast (\mu ) = (f_t)_\ast (\mu )$ or there is an $\alpha \in \mathcal {D}^+$ such that

$$ \begin{align*} \mu\left(f_t^{-1}(\alpha)\right) \leq \mu\left(f_t^{-1}(\beta)\right) \qquad \text{for all } \beta \in \mathcal{D}^-. \end{align*} $$

This shows that the set of all $t \in \mathbb {R}_{\geq 0} \cup \left \{\infty \right \}$ such that $f_\ast (\mu ) \trianglelefteq (f_t)_\ast (\mu )$ is closed. Hence, if we define

$$ \begin{align*} T(\mu) := \mathop{\mathrm{sup}}\limits \left\{t \in \mathbb{R}_{\geq 0} \cup \left\{\infty\right\} : f_\ast(\mu) \trianglelefteq (f_t)_\ast(\mu)\right\} \qquad \text{and} \qquad g_\mu := f_{T(\mu)}, \end{align*} $$

then $f_\ast (\mu ) \trianglelefteq (g_\mu )_\ast (\mu )$ . Also, by Lemma 3.4, statements (P1) and (P1) hold with $g_\mu $ in place of g.

Proof. We will show that in fact there is no recolouring move $\varphi \in M$ that improves $g_\mu $ . Indeed, suppose that $\varphi \in M$ improves $g_\mu $ . Then $\varphi \not \in M_{T\left (\mu \right )}$ , as otherwise we would have $\varphi \subseteq g_\mu $ . In particular, this means that $T(\mu ) < \infty $ . Since $\varphi $ improves $g_\mu $ , we can fix $\alpha \in \mathcal {D}^+$ such that

$$ \begin{align*} s\mu\left(g_\mu^{-1}(\alpha)\right) < \mu\left(g_\mu^{-1}(\beta)\right) \quad \text{for all } \beta \in \mathcal{D}^-. \end{align*} $$

The functions $t \mapsto \mu \left (f_t^{-1}(\gamma )\right )$ , $\gamma \in \mathcal {C}$ , are continuous, so there is some $\varepsilon> 0$ satisfying

$$ \begin{align*} \mu\left(f_{T\left(\mu\right) + \varepsilon}^{-1}(\alpha)\right) < \mu\left(f_{T\left(\mu\right) + \varepsilon}^{-1}(\beta)\right) \quad \text{for all } \beta \in \mathcal{D}^-. \end{align*} $$

This is a contradiction of the choice of $T(\mu )$ .

The foregoing observations show that with respect to the measure $\mu $ , statements (P1)–(P1) hold with $g_\mu $ in place of g, which yields part (a) of the lemma. To prove part (b), we use the uniform ergodic decomposition theorem. The statement is vacuous if $\mathsf {Erg}(G) = \varnothing $ , so assume that $\mathsf {Erg}(G) \neq \varnothing $ and let $V \to \mathsf {Erg}(G) \colon x \mapsto \mu _x$ be a G-invariant Borel surjection given by Theorem 2.4. Notice that due to [Reference KechrisKec95, Theorem 17.25], the function

$$ \begin{align*} \mathsf{Erg}(G) \to \mathbb{R}_{\geq 0}\cup \left\{\infty\right\} \colon \mu \to T(\mu) \end{align*} $$

is Borel. Hence, if we set $T(x) := T(\mu _x)$ for each $x \in V$ and define

$$ \begin{align*} g \colon V \to \mathcal{C} \colon x \mapsto f_{T(x)}(x), \end{align*} $$

then g is a Borel $\mathcal {C}$ -colouring of G, and this colouring g has all the desired properties. Indeed, for each $\mu \in \mathsf {Erg}(G)$ , g agrees with $g_\mu $ on the $\mu $ -conull G-invariant set $\left \{x \in V : \mu _x = \mu \right \}$ , and thus inherits the properties pertaining to $\mu $ from $g_\mu $ .

Assumptions for Section 3.4

Fix an aperiodic Borel graph G of finite maximum degree $\Delta $ with vertex set V and edge set E and a finite set of colours $\mathcal {C}$ of size $k \geq \Delta + 1$ .

Proof. The proofs of parts (a) and (b) of the lemma are virtually the same, except that the former relies on Lemma 3.5(a) and the latter on Lemma 3.5(b). We give the proof of part (a) and highlight the only place where it needs to be modified to prove part (b).

We start by partitioning $\mathsf {RM}$ into countably many Borel G-separated sets.

Now let $\mu $ be a G-invariant probability measure. We recursively construct a sequence of Borel proper $\mathcal {C}$ -colourings $f_n$ , $n \in {\mathbb {N}}$ , as follows. Fix a Borel function $c \colon \mathsf {RM} \to {\mathbb {N}}$ as in Claim 3.6.1. Let $\left (r_n, \mathcal {D}^+_n, \mathcal {D}^-_n\right )_{n \in {\mathbb {N}}}$ be a sequence of triples such that

1. (R1) for all $n\in {\mathbb {N}}$ , $r_n \in {\mathbb {N}}$ and $\mathcal {D}^+_n, \mathcal {D}^-_n$ are disjoint nonempty subsets of $\mathcal {C}$ ; and

2. (R2) every triple $\left (r, \mathcal {D}^+, \mathcal {D}^-\right )$ as in (R1) appears in the sequence $\left (r_n, \mathcal {D}^+_n, \mathcal {D}^-_n\right )_{n \in {\mathbb {N}}}$ infinitely often.

Set $f_0 := f$ . Once $f_n$ is defined, set

$$ \begin{align*} M_n := \mathsf{RM}_3\left(f_n; \mathcal{D}^+_n, \mathcal{D}^-_n\right) \cap c^{-1}(r_n). \end{align*} $$

Intersecting with $c^{-1}(r_n)$ ensures that $M_n$ is G-separated, so we can apply Lemma 3.5(a) with

$$ \begin{align*} f_n, 3, \mathcal{D}^+_n, \mathcal{D}^-_n \text{ and } M_n \qquad \text{in place of} \qquad f, m, \mathcal{D}^+, \mathcal{D}^- \text{ and } M, \end{align*} $$

in order to obtain a Borel proper $\mathcal {C}$ -colouring $f_{n+1}$ such that:

1. (P1) ${\mathrm {dist}}_\mu (f_n, f_{n+1}) \leq 3 \cdot \lVert (f_n)_\ast (\mu )-(f_{n+1})_\ast (\mu )\rVert _1$ ;

2. (P2) for all $\alpha \in \mathcal {D}^+((f_n)_\ast (\mu ), (f_{n+1})_\ast (\mu ))$ , we have

   $$ \begin{align*} \lVert(f_n)_\ast(\mu) - (f_{n+1})_\ast(\mu)\rVert_1 \leq 6 \cdot \left(\mu\left(f_{n+1}^{-1}(\alpha)\right) - \mu\left(f_n^{-1}(\alpha)\right)\right); \end{align*} $$

3. (P3) $(f_n)_\ast (\mu ) \trianglelefteq (f_{n+1})_\ast (\mu )$ ; and

4. (P4) $f_{n+1}$ is $(\mu , M_n)$ -perfect.

To establish part (b) of the lemma, we instead apply Lemma 3.5(b) to obtain a Borel proper $\mathcal {C}$ -colouring $f_{n+1}$ that satisfies (P1)–(P4) for every $\mu \in \mathsf {Erg}(G)$ . When the sequence $f_n$ , $n \in {\mathbb {N}}$ , is constructed, we define a Borel partial $\mathcal {C}$ -colouring $f_\infty \colon V \rightharpoonup \mathcal {C}$ via the pointwise limit

$$ \begin{align*} f_\infty(x) := \lim_{n \to \infty} f_n(x). \end{align*} $$

Since each $f_n$ is a proper colouring, $f_\infty $ is also proper, and since $k \geq \Delta + 1$ , we can extend $f_\infty $ to a Borel proper $\mathcal {C}$ -colouring g using Corollary 2.2. We claim that this g is as desired.

To begin with, notice that conditions (P2) and (P3) enable us to apply Lemma 3.2 to the sequence $((f_n)_\ast (\mu ))_{n \in {\mathbb {N}}}$ with $A = 6$ and conclude, using (P1), that

$$ \begin{align*} \sum_{n=0}^\infty {\mathrm{dist}}_\mu(f_n, f_{n+1}) \leq 3 \cdot \sum_{n=0}^\infty \lVert(f_n)_\ast(\mu) - (f_{n+1})_\ast(\mu)\rVert_1 \leq \frac{7^{k+1}}{2} \cdot {\mathrm{disc}}_\mu(f) < \infty. \end{align*} $$

By the Borel–Cantelli lemma, this implies that $f_\infty $ is defined for $\mu $ -almost every $x \in V$ ; furthermore,

$$ \begin{align*} {\mathrm{dist}}_\mu(f, g) = {\mathrm{dist}}_\mu(f, f_\infty) \leq \frac{7^{k+1}}{2} \cdot {\mathrm{disc}}_\mu(f). \end{align*} $$

(For simplicity, we bound the latter expression by $7^{k+1} \cdot {\mathrm {disc}}_\mu (f)$ in formula (3.13).) It remains to show that g is $\mu $ -equitable. To this end, we pass to a $\mu $ -conull G-invariant Borel subset of V and assume that $g = f_\infty $ . Suppose, toward a contradiction, that g is not $\mu $ -equitable. Then, by Lemma 3.1, g is not $(\mu , \mathsf {RM}_3)$ -perfect – that is,

$$ \begin{align*} \mu\left(\left\{\varphi \in \mathsf{RM}_3 : \varphi\ \text{improves}\ g\ \text{with respect to}\ \mu\right\}\right)> 0. \end{align*} $$

As there are only countably many triples $\left (r, \mathcal {D}^+, \mathcal {D}^-\right )$ with $r \in {\mathbb {N}}$ and $\mathcal {D}^+, \mathcal {D}^- \subseteq \mathcal {C}$ disjoint and nonempty, we can choose $\left (r, \mathcal {D}^+, \mathcal {D}^-\right )$ such that

(3.14) $$ \begin{align} \mu\left(\left\{\varphi \in \mathsf{RM}_3\left(g; \mathcal{D}^+, \mathcal{D}^-\right) \cap c^{-1}(r) : \varphi\ \text{improves}\ g\ \text{with respect to}\ \mu\right\}\right)> 0. \end{align} $$

Claim 3.6.2 and formula (3.14) together show that there is $n_0 \in {\mathbb {N}}$ such that for all $n \geq n_0$ ,

$$ \begin{align*} \mu\left(\left\{\varphi \in \mathsf{RM}_3\left(f_n;\mathcal{D}^+, \mathcal{D}^-\right) \cap c^{-1}(r) : \varphi\ \text{improves}\ f_n\ \text{with respect to} \mu\right\}\right)> 0. \end{align*} $$

This precisely means that for all $n \geq n_0$ ,

$$ \begin{align*} f_n\ \text{is not}\ \left(\mu, \mathsf{RM}_3\left(f_n;\mathcal{D}^+, \mathcal{D}^-\right) \cap c^{-1}(r)\right)\text{-perfect}. \end{align*} $$

But this is a contradiction, as there is $n \geq n_0$ with $\left (r_n, \mathcal {D}^+_n, \mathcal {D}^-_n\right ) = \left (r, \mathcal {D}^+, \mathcal {D}^-\right )$ , so

$$ \begin{align*} \mathsf{RM}_3\left(f_n;\mathcal{D}^+, \mathcal{D}^-\right) \cap c^{-1}(r) = M_n, \end{align*} $$

and $f_{n+1}$ is $(\mu , M_n)$ -perfect by (P4).

Proof. The equivalence (i) $\Longleftrightarrow $ (ii) is proven in [Reference Dougherty, Jackson and KechrisDJK94, Proposition 2.1]; Nadkarni’s theorem [Reference Becker and KechrisBK96, Theorem 4.3.1] gives (i) $\Longleftrightarrow $ (iii); and (iii) $\Longleftrightarrow $ (iv) follows by [Reference Dougherty, Jackson and KechrisDJK94, Proposition 3.2].

Proof. Suppose, toward a contradiction, that $U \not \approx _G V(G)$ . Proposition 3.8 then shows that, since $V(G) = [N_G(U)]_G \subseteq [U]_G$ , the set U is not G-compressible. By Theorem 3.7, there is a G-invariant measure $\mu $ on $V(G)$ with $0 < \mu (U) < \infty $ . But then

$$ \begin{align*} 0 < \mu(U) \leq \mu(V(G)) = \mu(N_G(U)) \leq \Delta \cdot \mu(U) < \infty, \end{align*} $$

contradicting the compressibility of G.

Proof. The implication (ii) $\Longrightarrow $ (i) is clear, and (iii) $\Longrightarrow $ (ii) follows by Proposition 3.8. To prove (i) $\Longrightarrow $ (iii), let f be Borel-equitable. This clearly implies that $\left [f^{-1}(\alpha )\right ]_G = V(G)$ for every colour $\alpha $ , so it remains to show that $f^{-1}(\alpha )$ is G-compressible. Take any nonzero G-invariant measure $\mu $ on $V(G)$ . Since f is Borel-equitable, all the colour classes of f have the same $\mu $ -measure, and hence

$$ \begin{align*} \infty = \mu(V(G)) = k \cdot \mu\left(f^{-1}(\alpha)\right). \end{align*} $$

This shows that $f^{-1}(\alpha )$ is G-compressible, by Theorem 3.7.

Proof. Let $V(G) = V_\prec \sqcup V_\approx \sqcup V_\succ $ be a partition given by Proposition 3.11 applied to A and B. Consider any $\mu \in \mathsf {Erg}(G)$ . Since $\mu $ is ergodic, precisely one of $V_\prec $ , $V_\approx $ or $V_\succ $ must be $\mu $ -conull. Since $\mu (A) = \mu (B)$ by assumption, this implies that $\mu (V_\approx ) = 1$ and $\mu (V_\prec \sqcup V_\succ ) = 0$ . As this holds for all $\mu \in \mathsf {Erg}(G)$ , we conclude using Theorems 2.4 and 3.7 that the set $V_\prec \sqcup V_\succ $ is G-compressible. Hence, we can set $V_0 := V_\approx $ and $V_1 := V_\prec \sqcup V_\succ $ .

Proof. Without loss of generality, we may assume that $U_0 \cup U_1 = V(G)$ . From Proposition 3.11 we obtain a partition $V(G) = V_0 \sqcup V_1$ of $V(G)$ into two G-invariant Borel sets such that

$$ \begin{align*} U_1 \cap V_0 \preccurlyeq_G U_0 \cap V_0 \qquad \text{and}\qquad U_0 \cap V_1 \preccurlyeq_G U_1 \cap V_1. \end{align*} $$

We claim that these $V_0, V_1$ are as desired. Suppose that, say, $U_0 \cap V_0 \not \approx _G V_0$ . Since $V(G) = U_0 \cup U_1$ , the relation $U_1 \cap V_0 \preccurlyeq _G U_0 \cap V_0$ implies $[U_0 \cap V_0]_G = V_0$ . By Proposition 3.8, the set $U_0 \cap V_0$ must be not G-compressible, so let $\mu $ be a G-invariant measure on $V(G)$ such that $0 < \mu (U_0 \cap V_0) < \infty $ . Since $U_1 \cap V_0 \preccurlyeq _G U_0 \cap V_0$ , we conclude that

$$ \begin{align*} 0 < \mu(U_0 \cap V_0) \leq \mu(V_0) = \mu(U_0 \cap V_0) + \mu(U_1 \cap V_0) \leq 2\mu(U_0 \cap V_0) < \infty, \end{align*} $$

contradicting the G-compressibility of $V_0$ .

Proof. Applying Corollary 3.13 with $I \setminus N_G(J)$ and $I \cap N_G(J)$ in place of $U_0$ and $U_1$ , we obtain a partition $V(G) = V_0 \sqcup V_1$ of $V(G)$ into two G-invariant Borel sets such that

$$ \begin{align*} (I \setminus N_G(J)) \cap V_0 \approx_G V_0 \qquad \text{and} \qquad (I \cap N_G(J)) \cap V_1 \approx_G V_1. \end{align*} $$

Since we can treat the induced subgraphs $G[V_0]$ and $G[V_1]$ separately, we may assume that either $V_0 = V(G)$ or $V_1 = V(G)$ . Now we consider the two cases.

If $V_0 = V(G)$ and $I \setminus N_G(J) \approx _G V(G)$ , then by Theorem 3.7, the set $I \setminus N_G(J)$ is G-paradoxical, so there is a partition $I \setminus N_G(J) = I_0 \sqcup I_1$ of I into two Borel sets satisfying $I_0 \approx _G I_1 \approx _G V(G)$ . This allows us to set $I' := I \setminus I_1 = (I \cap N_G(J)) \cup I_0$ and $J' := J \cup I_1$ .

On the other hand, if $V_1 = V(G)$ and $I \cap N_G(J) \approx _G V(G)$ , then $J \approx _G V(G)$ by Corollary 3.9, and hence we can simply take $I' := I$ and $J' := J$ .

Proof. Let $\mathcal {C}$ be a k-element set of colours and let $f \colon V(G) \to \mathcal {C}$ be a Borel proper colouring of G. It follows from Corollary 3.13 that there is a partition $V(G) = \bigsqcup _{\alpha \in \mathcal {C}} V_\alpha $ of $V(G)$ into G-invariant Borel sets satisfying $f^{-1}(\alpha ) \cap V_\alpha \approx _G V_\alpha $ for each $\alpha \in \mathcal {C}$ . As we can deal with the induced subgraphs $G[V_\alpha ]$ individually, we may assume that $V(G) = V_\alpha $ for some $\alpha \in \mathcal {C}$ and thus $f^{-1}(\alpha ) \approx _G V(G)$ . Theorem 3.15 then follows through a sequence of $k-1$ applications of Lemma 3.14.

Proof. Suppose, toward a contradiction, that the tuple $(G, \mathcal {L}, g, u)$ forms a counterexample that minimises $\lvert V(G)\rvert $ . Then, clearly, $\lvert V(G)\rvert \geq 2$ . Without loss of generality, we may assume that the partial proper $\mathcal {L}$ -colouring g is inclusion-maximal. This means that for each $x \in V(G) \setminus \mathrm {dom}(g)$ , every colour in $\mathcal {L}(x)$ is assigned by g to some neighbour of x. Since $\lvert \mathcal {L}(x)\rvert \geq \deg _G(x)$ , we conclude that $N_G(x) \subseteq \mathrm {dom}(g)$ , and for each $\alpha \in \mathcal {L}(x)$ , there is exactly one $y \in N_G(x)$ , with $g(y) = \alpha $ .

Consider now an arbitrary vertex $z \in V(G) \setminus \left \{u\right \}$ such that the subgraph $G - z$ is connected (for example, if T is a spanning subtree of G, then any leaf of T distinct from u is such). If $z \not \in \mathrm {dom}(g)$ , then pick any vertex $y \in N_G(z)$ . Since y is the unique neighbour of z with the colour $g(y)$ , we may replace g by the partial colouring $g^\ast $ with domain $\left (\mathrm {dom}(g) \setminus \left \{y\right \}\right ) \cup \left \{z\right \}$ defined by

$$ \begin{align*} g^\ast(x) := \begin{cases} g(x) &\text{if } x \in \mathrm{dom}(g) \setminus \left\{y\right\},\\ g(y) &\text{if } x = z. \end{cases} \end{align*} $$

Therefore, we may assume that $z \in \mathrm {dom}(g)$ . Let $g'$ be the restriction of g onto $\mathrm {dom}(g) \cap \left (V(G) \setminus \left \{z\right \}\right )$ . For each $x \in V(G) \setminus \left \{z\right \}$ , define

$$ \begin{align*} \mathcal{L}'(x) := \begin{cases} \mathcal{L}(x) \setminus \left\{g(z)\right\} &\text{if } x \in N_G(z),\\ \mathcal{L}(x) &\text{otherwise}. \end{cases} \end{align*} $$

Then $\mathcal {L}'$ is a degree-list assignment for $G-z$ and $g'$ is a partial proper $\mathcal {L}'$ -colouring. By the minimality of $\lvert V(G)\rvert $ , $G - z$ has a partial proper $\mathcal {L}'$ -colouring $f'$ such that $\mathrm {dom}(f) \supseteq V(G) \setminus \left \{z,u\right \}$ and $f' \succcurlyeq g'$ . But then the partial colouring $f := f' \cup \left \{(z, g(z))\right \}$ satisfies the conclusion of the lemma.

Proof. Suppose that $x \in V(G)$ satisfies $\lvert \mathcal {L}(x)\rvert \geq \deg _G(x) + 1$ . Due to Lemma 4.1, we may assume that $\mathrm {dom}(g) = V(G) \setminus \left \{x\right \}$ . Then there is a colour in $\mathcal {L}(x)$ that is not assigned by g to any of the neighbours of x, and hence g can be extended to a proper $\mathcal {L}$ -colouring of G – a contradiction.