Calculating probability mass functions with constraints from cumulative distribution This is a self-study question. The name of the book is called: Applied Statistics and Probability for Engineers by Montgomery and Runger. This problem is on page 73. It's exercise 3-41. 
The entire problem is listed as the following: 
Given the following cumulative distribution function:
$$
F(x)=\left\{\begin{matrix}
0 &  & x<-10 \\ 
0.25 &  & -10\leq x< 30 \\ 
0.75 &  &30\leq x< 50  \\ 
1 &  & 50 \leq x
\end{matrix}\right.
$$
Determine each of the probabilities:
a) $P(X<50)$ b) $P(0 \leq X < 10)$ c) $P(-10 < X <10) $... etc etc. The question I have is this:
Why does the following probability mass function evaluate to 0?
$$
P(0 \leq X < 10) = 0
$$
Isn't this set of outcomes a subset of $-10\leq x< 30$ and therefore should be evaluated to 0.25?
 A: The cumulative probability distribution function $F_X(x)$ tells us how much
probability mass there is to the left of $x$ or at $x$ for each $x$
on the real line. (The choice of notation, though almost universally used
is truly dreadful for use in a classroom setting! How on earth does one read out aloud $F_X(x)$ or $P\{X\leq x\}$? F-sub-big X of little x? probability that
random variable $X$ is no larger than lower-case x?) Formally, the value of $F_X(x)$ is just $P\{X \leq x\}$.
As Glen_b's comment says, you really should start by sketching the function
$F_X(x)$ at the very least.
When $X$ is a discrete random variable taking on values $x_1, x_2, \ldots$
with probabilities $p_1, p_2, \ldots $ respectively, a little thought
(instead of rote memorization of the definition) reveals that $F_X(x)$
must be what can be described as a staircase function, increasing from
$0$ to $1$ as $x$ increases, with steps of heights $p_1, p_2, \ldots$
at points $x_1, x_2, \ldots$ etc. The function is discontinuous
at each $x_i$, and is constant in each interval $[x_i, x_{i+1})$ (please
be sure to note the $[$ and $)$ in the description of the intervals).
Note that $F_X(x_i)$ includes $p_i$ so that the value of $F_X(x)$ at
the point $x=x_i$ (where the function is discontinuous) is the
value on the right.  Since you are studying from a text intended for
engineers, you might find this written as $F_X(x) = F_X(x^+)$.
Thus,
$$F_X(x) = P\{X \leq x\} = F_X(x^+) ~ \text{and} ~  P\{X < x\} = F_X(x^-).$$
In fact, for any random variable (not necessarily a discrete random variable
or an integer-valued random variable as in Rusan's answer) and for
any real numbers $a$ and $b$ such that $a \leq b$,
$$\begin{align}
P\{a < X \leq b\} &= F_X(b^+) - F_X(a^+) = F_X(b)-F_X(a),\tag{1}\\
P\{a \leq X \leq b\} &= F_X(b^+) - F_X(a^-) = F_X(b) - F_X(a^-),\tag{2}\\
P\{a \leq X < b\} &= F_X(b^-) - F_X(a^-) = F_X(b^-) - F_X(a^-),\tag{3}\\
P\{a < X < b\} &= F_X(b^-) - F_X(a^+) = F_X(b^-)-F_X(a).\tag{4}
\end{align}$$
For the special case when $b = a$, $(2)$ above becomes
$$P\{X=a\} = F_X(a^+)-F_X(a^-),$$ that is, $P\{X=a\}$ is the jump
(if any) in the value of $F_X(x)$ at $x=a$. If $F_X(x)$ is continuous
at $x=a$, then $P\{X=a\}=0$. 
With this as prologue, note that your given $F_X(x)$ is a staircase
function with jumps of $\frac 14, \frac 12, \frac 14$ at $x=-10, 30, 50$
respectively; that is, $X$ takes on values $-10, 30, 50$ with
probabilities  $\frac 14, \frac 12, \frac 14$ respectively, and
once you have that, the answers to the questions asked are easy
to compute directly, or, if you prefer to read the $F_X(0^-)$
etc off the graph that you have drawn as you apply $(1)$-$(4)$,
that is fine too.
A: Since this question relates specifically to a discrete random variable (as your book says...), I will answer it as such. Dilip Sarwate's answer provides the general result for a discrete random variable.
With those health warnings, the cumulative distribution function (CDF), $F$, between $x=-10$ and $x=30$ does not vary. This implies the probability mass strictly between these points is $0$. In particular, the interval $0\leq x <10$ lies strictly within the interval $-10\leq x <30$ so $\mathbb{P}(0\leq X<10)=0$. 
To see why this is, assume that the CFD does change at some $x$. Because $F$ is a non-decreasing function, and $X$ is a discrete random variable, we can find a sufficiently small $\epsilon>0$ so that 
$$F(x-\epsilon)<F(x)$$
holds, and so can be rearranged as
$$0<F(x)-F(x-\epsilon)=\mathbb{P}(X\leq x)-\mathbb{P}(X\leq x-\epsilon).$$
Because we are free to choose $\epsilon$ as small as we like, subject to $0<\epsilon$, we rewrite the inequality as 
$$0<\mathbb{P}(X\leq x)-\mathbb{P}(X<x)=f(x)=\mathbb{P}(X=x).$$
So the points for which the CDF changes (steps) are those that have positive probability mass. By reversing the argument, you can see the converse is also true.
