Show $\mu(E) = 2\lambda(E\cap[0,\infty))$, $\mu = \lambda \circ u^{-1}$ (image measure) Let $\lambda$ be the Lebesgue measure on $\mathbb{R}$, and $u:\mathbb{R}\to\mathbb{R}, u(x)=|x|$. Consider the image measure $\mu = \lambda \circ u^{-1}$. Show $\mu(E) = 2\lambda(E\cap[0,\infty))$.

Proof
$$
u^{-1}(E) = \{x\in\mathbb{R}:u(x)\in E\} =\\ \{x\in\mathbb{R}:u(x)\in E\cap(-\infty,0)\} \bigcup \{x\in\mathbb{R}:u(x)\in E\cap[0,\infty)\} = \\ 
u^{-1}(E\cap(-\infty,0)) \cup u^{-1} (E\cap[0,\infty))
$$
Notice that $u^{-1}(E\cap(-\infty,0)) $ and $u^{-1} (E\cap[0,\infty))$ are disjoint.
The Lebesgue measure will therefore be
$$
\lambda(u^{-1}(E)) = \lambda(u^{-1}(E\cap(-\infty,0))) + \lambda(u^{-1} (E\cap[0,\infty)))
$$
Now I'm very close! I'm almost 100 % certain that above is the same as $2\lambda(E\cap[0,\infty))$ but I can not figure out what the formal argument will be?
 A: But $u^{-1}(E\cap(-\infty,0))=\emptyset.$ Maybe this works better:
$\displaystyle \mu(E)=\int_{\mathbb R^+}1_E(y)d\mu(y) =\int_{\mathbb R}(1_E\circ u)(x)d\lambda(x)=\int_{\mathbb R}1_E(|x|)d\lambda(x)=\int_{\mathbb R^+}1_E(x)d\lambda(x)+\int_{\mathbb R^-}1_E(-x)d\lambda(x)=\int_{E\cap \mathbb R^+}d\lambda(x)+\int_{E\cap \mathbb R^+}d\lambda(x)=2\lambda (E\cap\mathbb R^+)$
because $1_E(|x|)=\begin{cases}
       1& |x|\in E \\
      0& |x|\notin E 
    \end{cases}=\begin{cases}
       1& x\in E\quad  \text{or}\ -x\in E\\
      0& \text{otherwise} 
    \end{cases}$
A: Your answer is correct. Here is a little simplification.
The function $u(x)=|x|$ induces a measure on $([0,\infty),\mathscr{B}([0,\infty))$. By monotone class arguments it is enough to see how the push forward of $\lambda$ under $u$ (i.e.$\lambda\circ u^{-1}$)  acts on sets of the form $[0,a]$, $a>0$.
$f^{-1}([0,a])=\{x:|x|\leq a\}=[-a,a]$ and so
$\lambda(f^{-1}([0,a])=\lambda([-a,a])=2\lambda([0,a])$
This $\lambda\circ u^{-1}=2\lambda$ restricted to $([0,\infty),\mathscr{B}([0,\infty))$.
