[proofplan] We first construct the joint measure explicitly from its density and use Tonelli's theorem to compute its total mass and its $x$-marginal. Then we identify the two kernels as the vertical and horizontal conditional laws of this joint measure, except on null sets where the fallback definitions are irrelevant. Finally, instead of appealing abstractly to a Gibbs-sampler theorem, we verify invariance directly by testing the two-step kernel against an arbitrary non-negative measurable function. [/proofplan]

[step:Verify that the slice density defines a probability measure and has the desired marginal]Define the measurable set \begin{align*} D:=\{(x,u)\in E\times(0,\infty):0<u<\gamma(x)\}. \end{align*} The function $\mathbb 1_D:E\times(0,\infty)\to\{0,1\}$ is measurable because $\gamma$ is $\mathcal E$-measurable and $u\mapsto u$ is Borel measurable on $(0,\infty)$. Since $(E,\mathcal E,\mu)$ is $\sigma$-finite by hypothesis and $((0,\infty),\mathcal B((0,\infty)),\mathcal L^1)$ is $\sigma$-finite, Tonelli's theorem applies to the non-negative measurable function $\mathbb 1_D$ and gives \begin{align*} \int_{E\times(0,\infty)}\pi(x,u)\,d(\mu\otimes\mathcal L^1)(x,u)=\frac{1}{Z}\int_E\mathcal L^1((0,\gamma(x)))\,d\mu(x). \end{align*} Since $\gamma(x)<\infty$ for every $x\in E$, $\mathcal L^1((0,\gamma(x)))=\gamma(x)$, and hence \begin{align*} \int_{E\times(0,\infty)}\pi(x,u)\,d(\mu\otimes\mathcal L^1)(x,u)=\frac{1}{Z}\int_E\gamma(x)\,d\mu(x)=1. \end{align*} Thus $\Pi$ is a probability measure. Let $A\in\mathcal E$. The function $\mathbb 1_{A\times(0,\infty)}\mathbb 1_D$ is non-negative and measurable on the product of the same two $\sigma$-finite measure spaces, so Tonelli's theorem gives \begin{align*} \Pi(A\times(0,\infty))=\frac{1}{Z}\int_A\mathcal L^1((0,\gamma(x)))\,d\mu(x). \end{align*} Therefore \begin{align*} \Pi(A\times(0,\infty))=\frac{1}{Z}\int_A\gamma(x)\,d\mu(x)=\Pi_X(A). \end{align*} So the $E$-marginal of $\Pi$ is $\Pi_X$.[/step]

[guided]The first point is that the displayed density really integrates to one. Define \begin{align*} D:=\{(x,u)\in E\times(0,\infty):0<u<\gamma(x)\}. \end{align*} This set is measurable in $\mathcal E\otimes\mathcal B((0,\infty))$: the map $(x,u)\mapsto\gamma(x)$ is measurable from the product space to $[0,\infty)$, the map $(x,u)\mapsto u$ is measurable, and $D$ is the set where $0<u$ and $u<\gamma(x)$. Now $\mathbb 1_D$ is non-negative and measurable. The [measure space](/page/Measure%20Space) $(E,\mathcal E,\mu)$ is $\sigma$-finite by hypothesis, and $((0,\infty),\mathcal B((0,\infty)),\mathcal L^1)$ is $\sigma$-finite because $(0,\infty)=\bigcup_{n=1}^{\infty}(0,n)$. Thus Tonelli's theorem applies without any prior integrability assumption and gives \begin{align*} \int_{E\times(0,\infty)}\pi(x,u)\,d(\mu\otimes\mathcal L^1)(x,u)=\frac{1}{Z}\int_E\int_0^\infty\mathbb 1_{\{0<u<\gamma(x)\}}\,d\mathcal L^1(u)\,d\mu(x). \end{align*} For fixed $x\in E$, the inner integral is the Lebesgue length of the interval $(0,\gamma(x))$, namely $\gamma(x)$. Hence \begin{align*} \int_{E\times(0,\infty)}\pi(x,u)\,d(\mu\otimes\mathcal L^1)(x,u)=\frac{1}{Z}\int_E\gamma(x)\,d\mu(x)=1. \end{align*} Thus the density defines a probability measure. The same Tonelli argument gives the marginal. For $A\in\mathcal E$, the integrand $\mathbb 1_A(x)\mathbb 1_{\{0<u<\gamma(x)\}}$ is non-negative and measurable on the product of the same $\sigma$-finite measure spaces, hence \begin{align*} \Pi(A\times(0,\infty))=\frac{1}{Z}\int_A\int_0^\infty\mathbb 1_{\{0<u<\gamma(x)\}}\,d\mathcal L^1(u)\,d\mu(x). \end{align*} The inner integral is again $\gamma(x)$, so \begin{align*} \Pi(A\times(0,\infty))=\frac{1}{Z}\int_A\gamma(x)\,d\mu(x)=\Pi_X(A). \end{align*} This proves that the $x$-coordinate under the joint slice law has exactly the target distribution.[/guided]

[step:Identify the vertical conditional kernel outside a null set] Let \begin{align*} N_X:=\{x\in E:\gamma(x)=0\}. \end{align*} Then \begin{align*} \Pi_X(N_X)=\frac{1}{Z}\int_{N_X}\gamma(x)\,d\mu(x)=0. \end{align*} Thus the fallback definition $K_U(x,\cdot)=\delta_1$ is used only on a $\Pi_X$-null set. For $x\in E\setminus N_X$ and $B\in\mathcal B((0,\infty))$, \begin{align*} K_U(x,B)=\frac{\mathcal L^1(B\cap(0,\gamma(x)))}{\gamma(x)}. \end{align*} This is precisely the normalized restriction of one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) to the vertical section $\{u\in(0,\infty):0<u<\gamma(x)\}$. [/step]

[step:Show that bad horizontal slices have zero joint probability] Define the slice-mass function \begin{align*} m:(0,\infty)\to[0,\infty],\qquad m(u):=\mu(S_u). \end{align*} Since $\mathbb 1_D$ is non-negative and measurable on the product of the $\sigma$-finite measure spaces $(E,\mathcal E,\mu)$ and $((0,\infty),\mathcal B((0,\infty)),\mathcal L^1)$, Tonelli's theorem gives \begin{align*} \int_0^\infty m(u)\,d\mathcal L^1(u)=\int_E\gamma(x)\,d\mu(x)=Z<\infty. \end{align*} Consequently $m(u)<\infty$ for $\mathcal L^1$-almost every $u\in(0,\infty)$. Let \begin{align*} G:=\{u\in(0,\infty):0<m(u)<\infty\}. \end{align*} The same Tonelli argument applied to $\mathbb 1_C(u)\mathbb 1_D(x,u)$ shows that the $u$-marginal of $\Pi$ satisfies \begin{align*} \Pi(E\times C)=\frac{1}{Z}\int_C m(u)\,d\mathcal L^1(u) \end{align*} for every $C\in\mathcal B((0,\infty))$. Therefore \begin{align*} \Pi(E\times G^c)=\frac{1}{Z}\int_{G^c}m(u)\,d\mathcal L^1(u)=0, \end{align*} because $m=0$ on $\{m=0\}$ and $\{m=\infty\}$ has $\mathcal L^1$-measure zero with respect to the finite integral of $m$. Hence the fallback definition $K_X(u,\cdot)=\Pi_X$ is used only on a $\Pi$-null set of slice levels. For $u\in G$ and $A\in\mathcal E$, \begin{align*} K_X(u,A)=\frac{\mu(A\cap S_u)}{m(u)}. \end{align*} Thus $K_X(u,\cdot)$ is the normalized restriction of $\mu$ to the horizontal slice $S_u$. [/step]

[step:Verify invariance of the two-step kernel by direct integration]Let \begin{align*} \varphi:E\times(0,\infty)\to[0,\infty] \end{align*} be a non-negative measurable [test function](/page/Test%20Function). Define the one-step expectation functional \begin{align*} P\varphi:E\times(0,\infty)\to[0,\infty],\qquad P\varphi(x,u):=\int_{(0,\infty)}\int_E \varphi(y,v)\,dK_X(v,y)\,dK_U(x,v). \end{align*} Since $P\varphi(x,u)$ does not depend on the old value of $u$, the $E$-marginal computation gives \begin{align*} \int_{E\times(0,\infty)}P\varphi(x,u)\,d\Pi(x,u)=\int_E\int_{(0,\infty)}\int_E\varphi(y,v)\,dK_X(v,y)\,dK_U(x,v)\,d\Pi_X(x). \end{align*} The set $\{\gamma=0\}$ has $\Pi_X$-measure zero, so the fallback part of $K_U$ contributes zero. Using the formula for $K_U$ on $\{\gamma>0\}$, \begin{align*} \int_E\int_{(0,\infty)}\int_E\varphi(y,v)\,dK_X(v,y)\,dK_U(x,v)\,d\Pi_X(x)=\frac{1}{Z}\int_E\int_0^{\gamma(x)}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mathcal L^1(v)\,d\mu(x). \end{align*} Tonelli's theorem permits swapping the $x$ and $v$ integrations because the integrand is non-negative and measurable on the product of the $\sigma$-finite measure spaces $(E,\mathcal E,\mu)$ and $((0,\infty),\mathcal B((0,\infty)),\mathcal L^1)$: \begin{align*} \frac{1}{Z}\int_E\int_0^{\gamma(x)}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mathcal L^1(v)\,d\mu(x)=\frac{1}{Z}\int_0^\infty\int_{S_v}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mu(x)\,d\mathcal L^1(v). \end{align*} The set $G^c$ has zero contribution with respect to the measure $m(v)\,d\mathcal L^1(v)$. On $G$, the kernel $K_X(v,\cdot)$ is normalized $\mu$ on $S_v$, so \begin{align*} \int_{S_v}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mu(x)=m(v)\frac{1}{m(v)}\int_{S_v}\varphi(y,v)\,d\mu(y). \end{align*} Hence \begin{align*} \int_{E\times(0,\infty)}P\varphi(x,u)\,d\Pi(x,u)=\frac{1}{Z}\int_0^\infty\int_{S_v}\varphi(y,v)\,d\mu(y)\,d\mathcal L^1(v). \end{align*} Since $y\in S_v$ is equivalent to $0<v<\gamma(y)$, Tonelli's theorem applied to the non-negative measurable function $\varphi(y,v)\mathbb 1_{\{0<v<\gamma(y)\}}$ on the same $\sigma$-finite product space gives \begin{align*} \frac{1}{Z}\int_0^\infty\int_{S_v}\varphi(y,v)\,d\mu(y)\,d\mathcal L^1(v)=\int_{E\times(0,\infty)}\varphi(y,v)\,d\Pi(y,v). \end{align*} Therefore \begin{align*} \int_{E\times(0,\infty)}P\varphi\,d\Pi=\int_{E\times(0,\infty)}\varphi\,d\Pi \end{align*} for every non-negative measurable $\varphi$, so $\Pi$ is invariant for $P$.[/step]

[guided]We prove invariance directly, because this avoids invoking any abstract regular-conditional-probability theorem. Let \begin{align*} \varphi:E\times(0,\infty)\to[0,\infty] \end{align*} be a non-negative measurable function. The transition first draws a new level $v$ from $K_U(x,\cdot)$ and then draws a new point $y$ from $K_X(v,\cdot)$, so define \begin{align*} P\varphi:E\times(0,\infty)\to[0,\infty],\qquad P\varphi(x,u):=\int_{(0,\infty)}\int_E\varphi(y,v)\,dK_X(v,y)\,dK_U(x,v). \end{align*} The old value of $u$ is not used in the update. Therefore, when integrating against $\Pi$, only the $x$-marginal matters: \begin{align*} \int_{E\times(0,\infty)}P\varphi(x,u)\,d\Pi(x,u)=\int_E\int_{(0,\infty)}\int_E\varphi(y,v)\,dK_X(v,y)\,dK_U(x,v)\,d\Pi_X(x). \end{align*} The fallback $K_U(x,\cdot)=\delta_1$ occurs only when $\gamma(x)=0$, and this set has $\Pi_X$-measure zero. Thus we may use the uniform formula for $K_U$ inside the integral. Since \begin{align*} d\Pi_X(x)=\frac{\gamma(x)}{Z}\,d\mu(x) \end{align*} and $K_U(x,\cdot)$ has density $1/\gamma(x)$ on $(0,\gamma(x))$ when $\gamma(x)>0$, the factors $\gamma(x)$ and $1/\gamma(x)$ cancel. This gives \begin{align*} \int_E\int_{(0,\infty)}\int_E\varphi(y,v)\,dK_X(v,y)\,dK_U(x,v)\,d\Pi_X(x)=\frac{1}{Z}\int_E\int_0^{\gamma(x)}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mathcal L^1(v)\,d\mu(x). \end{align*} Now we change the order of integration. This is justified by Tonelli's theorem because every term is non-negative and measurable, $(E,\mathcal E,\mu)$ is $\sigma$-finite, and $((0,\infty),\mathcal B((0,\infty)),\mathcal L^1)$ is $\sigma$-finite. The condition $0<v<\gamma(x)$ is exactly the condition $x\in S_v$, so \begin{align*} \frac{1}{Z}\int_E\int_0^{\gamma(x)}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mathcal L^1(v)\,d\mu(x)=\frac{1}{Z}\int_0^\infty\int_{S_v}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mu(x)\,d\mathcal L^1(v). \end{align*} For good levels $v\in G$, the kernel $K_X(v,\cdot)$ is normalized $\mu$ restricted to $S_v$. Therefore \begin{align*} \int_E\varphi(y,v)\,dK_X(v,y)=\frac{1}{m(v)}\int_{S_v}\varphi(y,v)\,d\mu(y). \end{align*} Multiplying by the outer integration over $x\in S_v$ contributes the factor $m(v)=\mu(S_v)$, so \begin{align*} \int_{S_v}\int_E\varphi(y,v)\,dK_X(v,y)\,d\mu(x)=\int_{S_v}\varphi(y,v)\,d\mu(y). \end{align*} Bad levels $v\notin G$ have zero mass under the slice-level marginal, because \begin{align*} \Pi(E\times G^c)=\frac{1}{Z}\int_{G^c}m(v)\,d\mathcal L^1(v)=0. \end{align*} Hence the fallback definition of $K_X$ does not affect the integral. We obtain \begin{align*} \int_{E\times(0,\infty)}P\varphi(x,u)\,d\Pi(x,u)=\frac{1}{Z}\int_0^\infty\int_{S_v}\varphi(y,v)\,d\mu(y)\,d\mathcal L^1(v). \end{align*} Finally, the condition $y\in S_v$ is the same as $0<v<\gamma(y)$. Rewriting the last integral with this indicator is another application of Tonelli's theorem to the non-negative measurable function $\varphi(y,v)\mathbb 1_{\{0<v<\gamma(y)\}}$ on the same $\sigma$-finite product space, and it gives exactly integration against the joint density: \begin{align*} \frac{1}{Z}\int_0^\infty\int_{S_v}\varphi(y,v)\,d\mu(y)\,d\mathcal L^1(v)=\frac{1}{Z}\int_E\int_0^{\gamma(y)}\varphi(y,v)\,d\mathcal L^1(v)\,d\mu(y). \end{align*} Thus \begin{align*} \frac{1}{Z}\int_E\int_0^{\gamma(y)}\varphi(y,v)\,d\mathcal L^1(v)\,d\mu(y)=\int_{E\times(0,\infty)}\varphi(y,v)\,d\Pi(y,v). \end{align*} This proves \begin{align*} \int_{E\times(0,\infty)}P\varphi\,d\Pi=\int_{E\times(0,\infty)}\varphi\,d\Pi \end{align*} for every non-negative measurable test function $\varphi$, which is precisely invariance of $\Pi$ under the two-step kernel.[/guided]

[step:Project the invariant joint law to obtain target invariance] Since $\Pi$ is invariant under $P$, if $(X,U)$ has law $\Pi$ and $(X',U')$ is obtained by one transition of $P$, then $(X',U')$ also has law $\Pi$. Taking the $E$-coordinate marginal of both sides, $X'$ has law equal to the $E$-marginal of $\Pi$, which was shown to be $\Pi_X$. Therefore the $x$-coordinate transition leaves $\Pi_X$ invariant. Since \begin{align*} \Pi_X(A)=\frac{1}{Z}\int_A\gamma(x)\,d\mu(x) \end{align*} for every $A\in\mathcal E$, the invariant $x$-coordinate density with respect to $\mu$ is $\gamma/Z$. [/step]