[proofplan] The proof has two parts. First, stationarity and the innovation normalization $\mathbb E[\eta_t^2]=1$ turn the GARCH recursion into a scalar linear equation for the common second moment $\mathbb E[\sigma_t^2]$, whose solution gives the unconditional variance. Second, conditioning the same recursion on $\mathcal F_t$ gives a first-order affine recursion for the forecast sequence $v_t(h)$. Solving that recursion yields the displayed mean-reversion formula toward the unconditional variance. [/proofplan] [step:Compute the unconditional second moment from stationarity] Define the constant \begin{align*} \gamma:=\alpha+\beta. \end{align*} By hypothesis $0\leq \gamma<1$. The updated theorem statement assumes that $\sigma_t$ is nonnegative and $\mathcal F_{t-1}$-measurable. Since $\eta_t$ is independent of $\mathcal F_{t-1}$ with $\mathbb E[\eta_t]=0$, we have \begin{align*} \mathbb E[\varepsilon_t\mid \mathcal F_{t-1}] =\mathbb E[\sigma_t\eta_t\mid \mathcal F_{t-1}] =\sigma_t\mathbb E[\eta_t] =0. \end{align*} Taking expectations gives $\mathbb E[\varepsilon_t]=0$. Next, using $\varepsilon_t=\sigma_t\eta_t$, the $\mathcal F_{t-1}$-measurability of $\sigma_t^2$, the independence of $\eta_t$ from $\mathcal F_{t-1}$, and $\mathbb E[\eta_t^2]=1$, we obtain \begin{align*} \mathbb E[\varepsilon_t^2] &=\mathbb E[\sigma_t^2\eta_t^2] \\ &=\mathbb E[\sigma_t^2]\mathbb E[\eta_t^2] \\ &=\mathbb E[\sigma_t^2]. \end{align*} Strict stationarity of $(\varepsilon_t)$ and finite second moments imply that $\mathbb E[\varepsilon_t^2]$ does not depend on $t$. Since we have proved $\mathbb E[\varepsilon_t^2]=\mathbb E[\sigma_t^2]$ for each $t$, the quantity \begin{align*} m:=\mathbb E[\sigma_t^2] \end{align*} also does not depend on $t$. Taking expectations in \begin{align*} \sigma_t^2=\omega+\alpha\varepsilon_{t-1}^2+\beta\sigma_{t-1}^2 \end{align*} therefore gives \begin{align*} m &=\omega+\alpha\mathbb E[\varepsilon_{t-1}^2]+\beta\mathbb E[\sigma_{t-1}^2] \\ &=\omega+\alpha m+\beta m \\ &=\omega+\gamma m. \end{align*} Since $1-\gamma>0$, solving this linear equation yields \begin{align*} m=\frac{\omega}{1-\gamma}=\frac{\omega}{1-\alpha-\beta}. \end{align*} Because $\mathbb E[\varepsilon_t]=0$, we have \begin{align*} \operatorname{Var}(\varepsilon_t) =\mathbb E[\varepsilon_t^2] =m =\frac{\omega}{1-\alpha-\beta}. \end{align*} [guided] The unconditional variance is obtained by reducing the GARCH recursion to an equation for one number. Define \begin{align*} \gamma:=\alpha+\beta. \end{align*} The assumptions $\alpha,\beta\geq 0$ and $\alpha+\beta<1$ give $0\leq\gamma<1$. First we verify that $\varepsilon_t$ has mean zero. The updated theorem statement assumes that $\sigma_t$ is nonnegative and $\mathcal F_{t-1}$-measurable. This is the predictability condition needed to pull $\sigma_t$ out of the conditional expectation. Since $\eta_t$ is independent of $\mathcal F_{t-1}$ with $\mathbb E[\eta_t]=0$, \begin{align*} \mathbb E[\varepsilon_t\mid \mathcal F_{t-1}] =\mathbb E[\sigma_t\eta_t\mid \mathcal F_{t-1}] =\sigma_t\mathbb E[\eta_t] =0. \end{align*} Taking expectation of both sides gives $\mathbb E[\varepsilon_t]=0$. Now we compare $\mathbb E[\varepsilon_t^2]$ and $\mathbb E[\sigma_t^2]$. The identity $\varepsilon_t=\sigma_t\eta_t$ gives $\varepsilon_t^2=\sigma_t^2\eta_t^2$. Because $\sigma_t^2$ is determined by the past $\mathcal F_{t-1}$ and $\eta_t$ is independent of that past, \begin{align*} \mathbb E[\varepsilon_t^2] &=\mathbb E[\sigma_t^2\eta_t^2] \\ &=\mathbb E[\sigma_t^2]\mathbb E[\eta_t^2] \\ &=\mathbb E[\sigma_t^2]. \end{align*} The normalization $\mathbb E[\eta_t^2]=1$ is exactly what makes the two second moments equal. Strict stationarity of $(\varepsilon_t)$ gives a time-independent value of $\mathbb E[\varepsilon_t^2]$. The identity already proved above, $\mathbb E[\varepsilon_t^2]=\mathbb E[\sigma_t^2]$, transfers this time-independence to $\mathbb E[\sigma_t^2]$. Thus we may denote the common value by \begin{align*} m:=\mathbb E[\sigma_t^2]. \end{align*} Taking expectations in the defining recursion for $\sigma_t^2$ gives \begin{align*} m &=\omega+\alpha\mathbb E[\varepsilon_{t-1}^2]+\beta\mathbb E[\sigma_{t-1}^2] \\ &=\omega+\alpha m+\beta m \\ &=\omega+\gamma m. \end{align*} Since $1-\gamma>0$, this equation has the unique solution \begin{align*} m=\frac{\omega}{1-\gamma}=\frac{\omega}{1-\alpha-\beta}. \end{align*} Finally, because $\mathbb E[\varepsilon_t]=0$, \begin{align*} \operatorname{Var}(\varepsilon_t) =\mathbb E[\varepsilon_t^2] =m =\frac{\omega}{1-\alpha-\beta}. \end{align*} [/guided] [/step] [step:Derive the conditional variance recursion] For fixed $t\in\mathbb Z$ and $h\in\mathbb N$, define the $\mathcal F_t$-measurable nonnegative random variable \begin{align*} v_t(h):=\mathbb E[\sigma_{t+h}^2\mid \mathcal F_t]. \end{align*} For $h=1$, the recursion gives \begin{align*} \sigma_{t+1}^2=\omega+\alpha\varepsilon_t^2+\beta\sigma_t^2. \end{align*} By the adaptedness hypothesis, $\varepsilon_t$ is $\mathcal F_t$-measurable, so $\varepsilon_t^2$ is $\mathcal F_t$-measurable. Also $\sigma_t^2$ is $\mathcal F_{t-1}$-measurable by hypothesis and hence $\mathcal F_t$-measurable because $\mathcal F_{t-1}\subseteq\mathcal F_t$. Therefore \begin{align*} v_t(1) =\mathbb E[\sigma_{t+1}^2\mid \mathcal F_t] =\omega+\alpha\varepsilon_t^2+\beta\sigma_t^2. \end{align*} Let $h\geq 2$. Then \begin{align*} \sigma_{t+h}^2=\omega+\alpha\varepsilon_{t+h-1}^2+\beta\sigma_{t+h-1}^2. \end{align*} Since $\varepsilon_{t+h-1}=\sigma_{t+h-1}\eta_{t+h-1}$, since $\sigma_{t+h-1}^2$ is $\mathcal F_{t+h-2}$-measurable, and since $\eta_{t+h-1}$ is independent of $\mathcal F_{t+h-2}$ with $\mathbb E[\eta_{t+h-1}^2]=1$, \begin{align*} \mathbb E[\varepsilon_{t+h-1}^2\mid \mathcal F_{t+h-2}] =\sigma_{t+h-1}^2. \end{align*} Because $\mathcal F_t\subseteq \mathcal F_{t+h-2}$, the tower property gives \begin{align*} \mathbb E[\varepsilon_{t+h-1}^2\mid \mathcal F_t] &=\mathbb E[\mathbb E[\varepsilon_{t+h-1}^2\mid \mathcal F_{t+h-2}]\mid \mathcal F_t] \\ &=\mathbb E[\sigma_{t+h-1}^2\mid \mathcal F_t] \\ &=v_t(h-1). \end{align*} Taking conditional expectations in the recursion for $\sigma_{t+h}^2$ therefore yields \begin{align*} v_t(h) &=\omega+\alpha v_t(h-1)+\beta v_t(h-1) \\ &=\omega+\gamma v_t(h-1). \end{align*} [guided] Fix $t\in\mathbb Z$. For each horizon $h\in\mathbb N$, we define the forecast as the $\mathcal F_t$-measurable nonnegative random variable \begin{align*} v_t(h):=\mathbb E[\sigma_{t+h}^2\mid \mathcal F_t]. \end{align*} The one-step forecast is immediate from the recursion: \begin{align*} \sigma_{t+1}^2=\omega+\alpha\varepsilon_t^2+\beta\sigma_t^2. \end{align*} We now verify precisely what is known at time $t$. The theorem assumes $\varepsilon_t$ is $\mathcal F_t$-measurable, hence $\varepsilon_t^2$ is $\mathcal F_t$-measurable. It also assumes $\sigma_t$ is $\mathcal F_{t-1}$-measurable; therefore $\sigma_t^2$ is $\mathcal F_{t-1}$-measurable. Since filtrations are increasing, $\mathcal F_{t-1}\subseteq\mathcal F_t$, so $\sigma_t^2$ is $\mathcal F_t$-measurable. Therefore conditioning on $\mathcal F_t$ leaves both terms unchanged: \begin{align*} v_t(1) =\mathbb E[\sigma_{t+1}^2\mid \mathcal F_t] =\omega+\alpha\varepsilon_t^2+\beta\sigma_t^2. \end{align*} Now take $h\geq 2$. The recursion one period before $t+h$ is \begin{align*} \sigma_{t+h}^2=\omega+\alpha\varepsilon_{t+h-1}^2+\beta\sigma_{t+h-1}^2. \end{align*} The only point requiring care is the term $\varepsilon_{t+h-1}^2$. We rewrite it as \begin{align*} \varepsilon_{t+h-1}^2=\sigma_{t+h-1}^2\eta_{t+h-1}^2. \end{align*} Here $\sigma_{t+h-1}^2$ is $\mathcal F_{t+h-2}$-measurable, while $\eta_{t+h-1}$ is independent of $\mathcal F_{t+h-2}$ and has second moment $1$. Hence \begin{align*} \mathbb E[\varepsilon_{t+h-1}^2\mid \mathcal F_{t+h-2}] =\sigma_{t+h-1}^2\mathbb E[\eta_{t+h-1}^2] =\sigma_{t+h-1}^2. \end{align*} Since $h\geq 2$, the filtration relation $\mathcal F_t\subseteq \mathcal F_{t+h-2}$ holds. Applying the tower property of conditional expectation gives \begin{align*} \mathbb E[\varepsilon_{t+h-1}^2\mid \mathcal F_t] &=\mathbb E[\mathbb E[\varepsilon_{t+h-1}^2\mid \mathcal F_{t+h-2}]\mid \mathcal F_t] \\ &=\mathbb E[\sigma_{t+h-1}^2\mid \mathcal F_t] \\ &=v_t(h-1). \end{align*} Conditioning the GARCH recursion on $\mathcal F_t$ now produces the forecast recursion \begin{align*} v_t(h) &=\omega+\alpha\mathbb E[\varepsilon_{t+h-1}^2\mid \mathcal F_t] +\beta\mathbb E[\sigma_{t+h-1}^2\mid \mathcal F_t] \\ &=\omega+\alpha v_t(h-1)+\beta v_t(h-1) \\ &=\omega+\gamma v_t(h-1). \end{align*} [/guided] [/step] [step:Solve the affine forecast recursion] Define the long-run variance level \begin{align*} \bar v:=\frac{\omega}{1-\gamma}. \end{align*} Since $\bar v=\omega+\gamma\bar v$, the recursion $v_t(h)=\omega+\gamma v_t(h-1)$ for $h\geq 2$ implies \begin{align*} v_t(h)-\bar v =\gamma\bigl(v_t(h-1)-\bar v\bigr). \end{align*} Iterating this identity from $h$ down to $1$ gives \begin{align*} v_t(h)-\bar v =\gamma^{h-1}\bigl(v_t(1)-\bar v\bigr). \end{align*} Substituting $\gamma=\alpha+\beta$ and $\bar v=\omega/(1-\alpha-\beta)$ yields \begin{align*} v_t(h)=\frac{\omega}{1-\alpha-\beta} +(\alpha+\beta)^{h-1} \left(v_t(1)-\frac{\omega}{1-\alpha-\beta}\right). \end{align*} Together with the formula for $v_t(1)$, this proves the stated conditional variance forecast formula. [/step]