Create combined CDF (Value at Risk) this is my first post and I have probably already made a fool out of myself by the title but I'm desperate to understand this. My question comes from an example so I have the solution but there is a step I want to understand the theory behind. 
Q: $L$ describes percentual damage to a building within next year. We know that $P(L=0)=0.95$. With $p=0.05$ a damage happens and then $L$ has uniform distribution on [50,100]. Find smallest value of m such that $P(L>m)\leq 0.001$ A: 
We know that $P(L=0)=0.95$ and $P(50\leq L \leq 100)=0.05$ and:
\begin{equation}
P(L>m)=1-P(L\leq m)\leq 0.001 \rightarrow P(L\leq m) \geq 0.999
\end{equation}
And we know that a CDF is defined by $F_L(m)=P(L\leq m)$ and we get $F_L$ as: 
\begin{equation}
F_L(m) = \begin{cases} 0 & m<0 \\ 0.95 & 0 \leq m <50 \\ 0.95 + 0.05 \cdot \frac{m-50}{50} & 50 \leq m \leq 100 \\1 & m>100\end{cases}
\end{equation}
My Question: What is the theory behind geting $0.95 + 0.05 \cdot \frac{m-50}{50}$ especially the second term. I do know that the CDF for a uniform r.v. has the form of $\frac{x-a}{b-a}$ but I don't understand how we combine the cases of no damage and damage and then multiply the probability 0.05 with the CDF for a uniform r.v. I do though understand why we add the 0.95. 
I have solved for m and that gives m=99 but there is some kind of essential theory that I don't get. Maybe if I know the correct term I could google it but as my title gives away I'm lost. 
A: Consider the graph of $F_{L}$, which is a non-decreasing function. A point $(x,y)$ indicates that $P(L\leq x)=y$. We know that the graph is constant (horizontal) for $x<0$ (where the constant is $0$), from $0$ to $50$ (where the constant is $.95$), and for $x>100$ (where the constant is $1$). Since what happens in between $50$ and $100$ follows the uniform distribution, the graph must be linear from $x=50$ to $x=100$, and since $F_{L}(50)=.95$ and $F_{L}(100)=1$, the expression follows immediately.
The underlying probability theory is that for $50\leq m\leq 100$, 
$$P(L\leq m)=P(L< 50)+P(50\leq L\leq m) = .95 + .05\cdot\frac{m-50}{50}$$
