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Exercise. Find a non-decreasing and right-continuous function $F\colon\Bbb{R}\to\Bbb{R}$ whose Lebesgue-Stieltjes associated $\mu_F$ satisfies the following conditions at same time

  • $\mu_F(\{0\})=\frac{1}{2}.$

  • $\mu_F([0,\frac{1}{2}])=2.$

  • $\mu_F((-\infty,-\frac{1}{2}])=0.$

  • $\mu_F([1,x))=2x,\forall x>1.$

I was thinking if I can use the following Theorem to solve this question. We denotes $B_{\Bbb{R}}$ the $\sigma$-algebra of Borel.

Theorem. Let $F\colon\Bbb{R}\to\Bbb{R}$ increasing and right-continuous function. So there is a unique measure $\mu_F$ defined in $B_\Bbb{R}$ such that $$\mu_F((a,b])=F(b)-F(a).$$

  • If $G\colon\Bbb{R}\to\Bbb{R}$ is another increasing and right-continuous function, then $$\mu_F=\mu_G\Longleftrightarrow F-G=c,\ \text{where}\ c\in\Bbb{R}\ \text{constant}.$$

  • Conversely, if $\mu$ is a measure defined in $B_\Bbb{R}$ that is finite for any bounded set, then $F\colon\Bbb{R}\to\Bbb{R}$ given by $$F(x)=\left\lbrace\begin{array}{c}\mu((0,x]),x>0\\ 0,x=0\\ -\mu((x,0]),x<0\end{array}\right.$$ is increasing and right-continuous and $\mu_F=\mu$.

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Take $F(x)=0$ for $x<0$, $\frac 1 2$ for $0\leq x <\frac 1 4$, $\frac 3 2$ for $\frac 1 4 \leq x <1$ and $2x$ for $x \geq 1$. Apply the theorem to get $\mu_F$.

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