Take for example the function $u(x)=sin(\pi x)$, which is in $H^1_0(0,1)$, and satisfies
$$ u_t -u^{\prime\prime}=f,\quad u_0=sin(\pi x), \mbox{ on } L^2(0,T;H^{-1}(\Omega)) $$ with $f=\pi^2\sin(\pi x)$. The $L^2(0,T;H^1_0(\Omega))$ norm of $u$ is $$ \|u\|_{L^2(0,T;H^1_0(\Omega))}=\int_0^T \int_0^1 \pi^2 \cos^2(\pi x)dxdt= \frac{T \pi^2}{2}. $$ It isn't in $L^2(0,\infty;H^1_0(\Omega))$.

This answer corresponded to a previous version of the question. The key point that you skip in your question is indeed if the norm stays bounded! If your norm is bounded independently of $n$, then you can show directly (by verifying the definition) that the $v$ you have constructed is indeed in $L^2(0,\infty;V)$.

Take for example the function $u(x)=sin(\pi x)$, which is in $H^1_0(0,1)$, and satisfies
$$ u_t -u^{\prime\prime}=f,\quad u_0=sin(\pi x), \mbox{ on } L^2(0,T;H^{-1}(\Omega)) $$ with $f=\pi^2\sin(\pi x)$. The $L^2(0,T;H^1_0(\Omega))$ norm of $u$ is $$ \|u\|_{L^2(0,T;H^1_0(\Omega))}=\int_0^T \int_0^1 \pi^2 \cos^2(\pi x)dxdt= \frac{T \pi^2}{2}. $$ It isn't in $L^2(0,\infty;H^1_0(\Omega))$.

This answer corresponded to a previous version of the question. The key point that you don't pay attention to in your question is that the norm stays bounded! Your question in the end isn't about linear PDE, it is simply: Let $f$ is locally integrable and there is uniform upper bound on its integral which does not depend on the subdomain, is the function integrable, and the answer is yes, but that's more for stack exchange.

Take for example the function $u(x)=sin(\pi x)$, which is in $H^1_0(0,1)$, and satisfies
$$ u_t -u^{\prime\prime}=f,\quad u_0=sin(\pi x), \mbox{ on } L^2(0,T;H^{-1}(\Omega)) $$ with $f=\pi^2\sin(\pi x)$. The $L^2(0,T;H^1_0(\Omega))$ norm of $u$ is $$ \|u\|_{L^2(0,T;H^1_0(\Omega))}=\int_0^T \int_0^1 \pi^2 \cos^2(\pi x)dxdt= \frac{T \pi^2}{2}. $$ It isn't in $L^2(0,\infty;H^1_0(\Omega))$.

Take for example the function $u(x)=sin(\pi x)$, which is in $H^1_0(0,1)$, and satisfies
$$ u_t -u^{\prime\prime}=f,\quad u_0=sin(\pi x), \mbox{ on } L^2(0,T;H^{-1}(\Omega)) $$ with $f=\pi^2\sin(\pi x)$. The $L^2(0,T;H^1_0(\Omega))$ norm of $u$ is $$ \|u\|_{L^2(0,T;H^1_0(\Omega))}=\int_0^T \int_0^1 \pi^2 \cos^2(\pi x)dxdt= \frac{T \pi^2}{2}. $$ It isn't in $L^2(0,\infty;H^1_0(\Omega))$.

This answer corresponded to a previous version of the question. The key point that you skip in your question is indeed if the norm stays bounded! If your norm is bounded independently of $n$, then you can show directly (by verifying the definition) that the $v$ you have constructed is indeed in $L^2(0,\infty;V)$.