Let $T: \mathbb{R^2} \longrightarrow \mathbb{R^2}$ a linear map.
If we define the following measure, $\mu(E) = \lambda_2 (T(E))$ where $\lambda_2$ is the Lebesgue measure in $\mathbb{R}^2$.
If $\mu([0,1]^2)=3$, then what is the value of $\mu(B(0,1))$ where $B(0,1)=\{ x^2 + y^2 \leq 1 \}$
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I’ve managed to do the following so far.
Given that $\lambda_2([0,1]^2)=1$ and $3=\mu([0,1]^2) = |T| \lambda_2([0,1]^2)=|T|$. Therefore $\mu(B(0,1)) = |T| \lambda_2(B(0,1))= 3 \cdot \lambda_2(B(0,1))$ but I don’t know how to compute that last operation.
Let $I=[0,1]$. Since $T$ is linear, there is a constant $c$ for which $\mu(A)=c\lambda_2(A)$ for $any$ measurable set $A$. To determine $c$, we have, with $A=I\times I,\ 3=\mu(I\times I)=c\lambda_2(I\times I)=c\cdot 1$ so $c=3$. Then, $\mu (B)=3\lambda_2(B)=3\pi.$