Consider the image measure $\gamma=\lambda\circ u^{-1}$. Determine $\gamma([0,a])$, $\gamma([-a,a])$ and $\gamma([a,a+1])$ for all $a\geq 0$

Question

Let $\lambda$ be the Lebesgue measure on $\mathbb{R}$ and let $u:\mathbb{R}\rightarrow \mathbb{R}$, $u(x)=|x|$. Consider the image measure $\gamma=\lambda\circ u^{-1}$. Determine $\gamma([0,a])$, $\gamma([-a,a])$ and $\gamma([a,a+1])$ for all $a\geq 0$

I'm a little uncertain about this, since I'm not sure if I'm getting the notation right. Would the following be correct?

For the first one we have $\gamma([0,a])=\lambda(u^{-1}([0,a]))=\lambda([-a,a])=a-(-a)=2a$

I'm not sure about the second one, since $-a$ is not defined for $u^{-1}$, while $a$ is. But my guess would be it gives the null set, hence $\gamma([-a,a])=\lambda(Ø)=0$?

And for the last one $\gamma([a,a+1])=\lambda(u^{-1}([a,a+1]))=\lambda([-a-1,-a]\cup[a,a+1])=\lambda([-a-1,-a])+\lambda([a,a+1])=-a-(-a-1) +(a+1)-a=2$

Answer 1

The first one is right, and the second and third one are the same: By definition, $$u^{-1}[-a,a]=\left\{x:-a\leq |x|\leq a\right\}$$ since $a\geq 0$, the condition $-a\leq |x|$ is trivial, and thus $$u^{-1}[-a,a]=\left\{x:|x|\leq a\right\}=[-a,a]$$ so $$\gamma[-a,a]=\lambda(u^{-1}[-a,a])=\lambda[-a,a]=2a$$ and the third one is correct: $$\gamma[a,a+1]=\lambda(u^{-1}[a,a+1])=\lambda([-a-1,-a]\cup[a,a+1])=2$$