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Let $\lambda$ be the Lebesgue measure on $\mathbb{R}$, and let $f : \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = |x|$. Describe the measure $\lambda \circ f^{-1}$.

Since $f^{-1}(x)$ is not injective I'm confused. For example, the number 3 has pre-image 3 and -3. Does this mean that for an interval $(a,b)$ we have pre-image $(a,b) \cup (-b,-a)$? Which has length $2(b-a)$.

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$\lambda \circ f^{-1}$ is a measure.

Take $A \subset \mathbb R$.

Then $(\lambda \circ f^{-1})(A)$ gives the Lebesgue measure of the set $f^{-1}(A)$.

Also, your last statement is wrong. The measure of that set is not always $2(b-a)$. It is true if $a\geq 0$ and $b \geq 0$.

Take the example $a = -1, b=1$.

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  • $\begingroup$ If we take $(-1,1)$ then for the negative numbers in the interval this function is not defined so we only look at $[0,1)$ which has pre-image $((-1,0] \cup [0,1))$ which has length 2. Is that what you mean? My formula was only for an interval with positive bounderies. $\endgroup$ – Nescrio Dec 14 '15 at 17:38
  • $\begingroup$ I get that. But if we take $A = (-2,-1)$, what would be the length $(\lambda \circ f^{-1})(A)$? $\endgroup$ – Nescrio Dec 14 '15 at 17:45

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