Estimate of integral of rescaled function

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We consider a vectorvalued function $v$ on $\Omega \subset \mathbb{R}_{x}^n \times \mathbb{R}_{t}$. This function has it's support in $B_{1}(0,0)$ and satisfies the property $$\int_{\Omega} |v(x,t)| \: dx dt \geq \alpha |\tilde{v}(x_{0},t_{0})|$$ Also we consider the rescaled function $v_{r}(x,t) = v(\frac{x-x_{0}}{r},\frac{t-t_{0}}{r})$ which has it's support in $B_{r}(x_{0},t_{0})$. The part that I don't understand is the following: $$\int_{\Omega} |v_{r}(x,t)| \: dx dt \geq \alpha |\tilde{v}(x_{0},t_{0})| |B_{r}(x_{0},t_{0})|$$ This is what I tried: Because of the support of $v_{r}$ we can rewrite the integral $$\int_{\Omega} |v_{r}(x,t)| \: dx dt = \int_{B_{\mu}(x_{0},t_{0})} |v_{r}(x,t)| \: dx dt$$ Substituting by $\phi(x,t)=(\frac{x-x_{0}}{r},\frac{t-t_{0}}{r})$ gave me $$\int_{\Omega} |v_{r}(x,t)| \: dx dt = r^2 \int_{B_{1}(0,0)} |v(x,t)| \: dx dt \geq r^2 \alpha |\tilde{v}(x_{0},t_{0})|$$ But I don't manage to get the asked estimate. It would be great if someone could help me! :)

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This is not the correct way to make a change of variables. Think of it this way, you want to transform the current integral into an integral over another set. So, you are looking for a one-to-one map between the second set and the first set such that there is a bijective vector valued function of new variables $u$ and $m$ from the second set which gives you unique $x$ and $t$ from the first set. So, your substitution should be $$(x,t)=(ur+x_0 , mr+t_0)$$ Note that $ur+x_0$ is a vector and $mr+t_0$ is a number. Calculate the jacobian matrix by expressing the preceding map in terms of vector components and carrying out the appropriate differentiations. You find the jacobian to be $r^{n+1}$. Using the fact that the volumes of $(n+1)$- dimensional balls are proportional to the $n+1$-th powers of their radii, the volume of $B_r(x,y)$ is $|B_1(x,y)|r^{n+1}$, your inequality becomes $$\int_{\Omega} |v_{r}(x,t)| \: dx dt \geq \frac{\alpha}{|B_1(x_0,t_0)|}|\tilde{v}(x_{0},t_{0})| |B_{r}(x_{0},t_{0})|$$