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Let $f:X\to\mathbb{R}$ measurable, suppose that $\mu$ is $\sigma$-finite and put $v(A):=\mu(f^{-1}(A))$, $A\in\mathcal{B}(\mathbb{R})$ (borelians). Show that $v$ is a Lebesgue-Stieltjes measure, that is, find a nondecreasing right-continuous map $F:\mathbb{R}\to\mathbb{R}$ satisfying \begin{equation}\label{eq1} v(E)=\inf\left\{\sum_{(a_k, b_k]\in\mathcal{F}}[F(b_k)-F(a_k)]\right\},\end{equation} where the infimum is taken over all countable collections $\mathcal{F}$ os half-open intervals of the form $(a_k, b_k]$ such that $$E\subset \bigcup_{(a_k, b_k]\in\mathcal{F}}(a_k, b_k].$$

My question is: What function $F$ should I take? I tried to put $F(x)=\mu(f^{-1}(-\infty, x])$ but failed to prove the equality above.

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    What exactly is your question? – 2017-01-23
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    @Théophile okay, please read again – 2017-01-23
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    Thank you for adding some context. – 2017-01-23
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    You meant $E=A$, right? It is sufficient to verify the equation for $A=(\infty, a]$. Where do you get stuck? – 2017-01-24
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    @user251257 Showing that the infimum above is $\leq$ than $v(E)$. =( – 2017-01-24
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    @user251257 If you take a cover to $E$, say $\cup (a_k, b_k]$, then it is easy to see that $v(E)\leq \sum\mu(f^{-1}(a_k,b_k]) = \sum[F(b_k)-F(a_k)].$ Since this cover is arbitrary, $v(E)$ is less than the infimum. But how can I see the converse? – 2017-01-24

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