Changing variables for Lebesgue integral over open ball

integrationlebesgue-integral

How do I conduct a change of variables for a multidimensional Lebesgue integral over an open ball, in order to switch the radius of the ball I am integrating over?

If I am integrating a locally integrable function $f$ over the open ball $B(x,r)$ of radius $r$, and I want to switch my integral to one over an open ball of a different radius $r_0$, how would I go about doing this?

For example, if I would rather integrate over an open ball of radius $r_0$ is there any way I can make a change of variables to have $$\int_{B_{(x,r)}}f(y)dy=\int_{B_{(x,r_0)}}f(y+r)dy$$ by some change of variables(Perhaps different than the one I suggest)? I do not think the change of variables I suggested is correct. Any suggestions?

Best Answer

For the integral \begin{align*} \int_{B(x, r)} f(y)dy \end{align*} to switch the domain of integration to a ball of radius $r_0>0$, say, by taking \begin{align*} y=x+\frac{r}{r_0}(z-x) \end{align*} we observe that \begin{align*} |y-x|=\frac{r}{r_0}|z-x|\leq r\ \Longrightarrow\ |z-x|\leq r_0, \end{align*} hence the new domain will be the ball $B(x, r_0)$. The Jacobian of this transformation is given by $dy=(\frac{r}{r_0})^ndz$, therefor by the change of variables theorem we get \begin{align*} \int_{B(x, r)} f(y)dy=\left(\frac{r}{r_0}\right)^n\int_{B(x, r_0)} f\big(x+\tfrac{r}{r_0}(z-x)\big)dz. \end{align*}

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