# Prove that $\int_{B_a}f(tx)\,dx=\int_{B_{ta}}f(x)t^{-n}\,dx$.

Let , $t$ and $a$ be two positive real numbers. Define , $B_a=\{x=(x_1,\cdots ,x_n)\in \mathbb R^n|x_1^2+\cdots +x_n^2 \le a^2\}.$ Then for any compactly supported continuous function $f$ on $\mathbb R^n$ which of the followings is(/are) correct ?

(A) $\displaystyle \int_{B_a}f(tx)\,dx=\int_{B_{ta}}f(x)t^{-n}\,dx$.

(B) $\displaystyle \int_{B_a}f(tx)\,dx=\int_{B_{t^na}}f(x)t\,dx$.

(C) $\displaystyle \int_{\mathbb R^n}f(x+y)\,dx=\int_{\mathbb R^n}f(x)\,dx$ , for some $y\in \mathbb R^n$.

(D) $\displaystyle \int_{\mathbb R^n}f(tx)\,dx=\int_{\mathbb R^n}f(x)t^n\,dx$.

My Attempt :

(A) Let , $tx=X$. Then , $x_1^2+\cdots +x_n^2 \le a^2$ transformed into $X_1^2+\cdots +X_n^2 \le a^2t^2$ which is $B_{ta}$. Also , $\,dx=t^{-n}\,dX$. So , this option is correct , and option (B) is FALSE. Similarly option (D) is FALSE.

(C) For $y=0=(0,0,\cdots ,0)\in \mathbb R^n$ given relation holds. So this option is correct. In fact he relation holds for arbitrary $y\in \mathbb R^n$.

Am I right or wrong ? .. Please verify.

• Are you sure (C) isn't supposed to ask whether the equality is true for all $y\in\mathbb{R}^n$? As it stands it's trivially true. – Joey Zou Dec 22 '15 at 6:21
• NO....' NOT for all $y\in \mathbb R$...' – Empty Dec 22 '15 at 7:19
• Question is correct....According to the question Am I right ? – Empty Dec 22 '15 at 7:20
• You may wish to contact the exam officials and notify them of this issue then. (I don't know if that's how it works in India, but at least that's what I did when I found an error while taking the GRE in the U.S.) – Joey Zou Dec 22 '15 at 18:36
• I think you are correct @S.Panja-1729 – Nitin Uniyal Dec 26 '15 at 4:13