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I am trying my hands at a problem in Donald Cohn book on measure theory.

Let $\mu$ be a nonzero finite Borel measure on $\mathbb{R}$, and define the function $F(x) := \mu((−\infty, x])$, and $g$ defined on $(0, \, \mu(\mathbb{R}))$ by $g(x) = \inf \, \{t \in \mathbb{R}: F (t) \geq x\}$.

Prove that $\mu(B)=\lambda(g^{-1}(B))$, where $B=(-\infty, b]$, $b\in\mathbb{R}$.

$\lambda(g^{-1}(B))$, $g$ being measurable, is by definition the push-forward measure of $\lambda$ so we can use the change of variable formula $$\lambda(g^{-1}(B))=\int_\mathbb{R}1_B \; d(\lambda g^{-1}) = \int_\mathbb{R}1_B\circ g \;d\lambda = $$ At this point if I can prove that $g=d\mu/d\lambda$ I think I could conclude but I am not sure it is a viable approach since I know nearly nothing about $\mu$...

The question is then: can this approach succeed or a different more low level one is necessary.

(Note: I recognize that we could normalize the measure $\mu$ to recover the probabilistic interpretation in which $F$ is then the CDF of the random variable $X(\omega)=\omega$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mu/\mu(\mathbb{R}))$ but to procede further I would need that $g$ be the density of $X$ which seems to be equivalent to proving that $g=d\mu/d\lambda$)

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    If $g = d\mu / d\lambda$ was true, then we would have $\int_{\mathbb{R}} 1_B \cdot g \; d\lambda = \mu(B)$, where the dot denotes the __product__ of functions. This is different from the equation you wrote, in which you __compose__ the functions. So the Radon-Nykodym derivative won't help you here (it might not even exist!). The idea here is really to consider $B$ of the form $(-\infty, b]$ and unwind the definitions of $F$ and $g$ to guess what $g^{-1}(B)$ looks like. – 2017-01-09
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    Since my question was on the feasibility of my approach, your answer is the most on-point, however I can't close the question on a comment. I will accept the response thereafter posted by chilango. Thanks – 2017-01-10

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$F$ is non-decreasing, positive and vanishes at $-\infty$. This, together with the definition of $g$, implies first, that $g$ is strictly increasing, hence injective, and then that $g^{-1}(B)=(0,g^{-1}(b)]$.

We have $\tag1\lambda (g^{-1}(B))=\lambda((0,g^{-1}(b)])=g^{-1}(b)$.

Now, $a=g^{-1}(b)\Leftrightarrow g(a)=b,$ so $b=\inf \, \{t \in \mathbb{R}: F (t) \geq a\}$ which means that $F(b)=a$ because $F$ is right-continuous.

Thus,

$\tag2g^{-1}(b)=a=F(b)=\mu (B)$.

Combining 1) and 2) gives the result.