Exercise: Define a function $F:\mathbb{R}\to\mathbb{R}$ such that the associated Lebesgue-Stieltjes measure verifies conditions: $\mu_F({1})=\frac{1}{4}$,$\mu_F((-\infty,1))=0$,$\mu_F((1,\frac{3}{2}))=1$ and $\mu_F([2,x))=\frac{x}{2}$,$\forall x>2$
$\mu_F({1})=F(1)=\frac{1}{4}$
$\mu_F((1,\frac{3}{2}))=F(\frac{3}{2})-F(1)=1\implies F(\frac{3}{2})-\frac{1}{4}=1\implies F(\frac{3}{2})=\frac{5}{4} $
However I have no idea on how to deal with $\mu_F([2,x))=\frac{x}{2}$,$\forall x>2$ . I am not understanding the aim of the exercise I tried to draw the function and I can see a straight line from minus infinity to one at the height of $\frac{1}{4}$ and then a point at $1.5$ below the $X$ axis $\frac{3}{4}$.
Question:
Since the function is not assumed to be either continuous or trivial how should I find the aforementioned function $F$?
Thanks in advance!
Note that we cannot determine $F=\mu_F((-\infty,x])$ uniquely using only given information. Specifically, we cannot determine $F$ on $[3/2,2]$. Notice that $$ \lim_{x\to 2}\mu_F([2,x)) =1=\mu_F(2) = F(2)-F(2^{-}). $$ Here, $F(x^{-})$ denotes left limit of $F$ at $x$. This means that $F$ has a jump discontinuity of size $1$ at $x=2$. Also, note that $$ F(x^{-})=\mu_F((-\infty, x)),\quad\forall x\in\mathbb{R}. $$ Hence $\mu_F([2,x))=\frac{x}{2}$ means that $F(x^{-})-F(2^{-}) = F(x^{-})-F(2)+1=\frac{x}{2}.$ Since $F(x^{-}) = \frac{x}{2} +F(2) -1$ is continuous on $x>2$, we have $$ F(x)=F(x^{-}) = \frac{x}{2} +F(2) -1, \quad\forall x>2.$$
To conclude, $F$ has the following form $$ F(x) = \begin{cases}0 ,\quad x<1\\1/4,\quad x=1\\ 5/4,\quad x=(\frac{3}{2})^{-}\\ 5/4+\delta ,\quad x=2^{-}\\9/4+\delta,\quad x=2\\5/4+\delta +\frac{x}{2},\quad x>2, \end{cases} $$where $0\leq\delta<\infty.$