To prove $P(a \leq \textbf{X} \leq b) = f(a) + F(b) - F(a)$ where $f$ and $F$ are pdf and cdf respectively of random variable X. 
Suppose X is a random variable with probability density function $f$ and cumulative density function  $F$. Show that 
  $$P(a \leq \textbf{X} \leq b) = f(a) + F(b) - F(a)$$

$$F(x)=P(\textbf{X} \leq x)=\int\limits_{-\infty}^x f(t)dt$$
$$\begin{align}
P(a \leq \textbf{X} \leq b) & =P(\textbf{X} \leq b)-P(\textbf{X} <a)\\
                            & = F(b) - \lim\limits_{x \to a^-} F(x)
\end{align}$$
So if I can show that $\lim\limits_{x \to a^-} F(x)= F(a)-f(a)$ I am through. But I seem to be stuck here. 
 A: By the definition of $F(x)$ you have $\Pr(X \le a)=F(a)$ and $\Pr(X \le b)=F(b)$
so if $a \le b$ then $$\Pr(a \lt X \le b) =F(b)-F(a) $$ 
which is close to what you want, though excludes the possibility $X=a$.  That is not an issue if  $\Pr(X=a)=0$, for example when $X$ is a continuous random variable, but for completeness you have $$\Pr(a \le X \le b) =\Pr(X=a)+ \Pr(a \lt X \le b)$$ $$\qquad\qquad\qquad=\Pr(X=a) +F(b)-F(a)$$
and if $X$ is a discrete random variable with probability mass function $f(x)=\Pr(X=x)$ then this would become $f(a) +F(b)-F(a)$. Your question says probability density function in which case you would be back in the continuous random variable case and the original statement would be incorrect
A: Ok, the cumulative distribution function is defined as
$F(x_i) = P(X \leq x_i) = f(1) + f(2) + ... + f(x_i)$
But what happens if we want the cumulative distribution function, within a range?
Per example, i have this cumulative distribution function
$F(x_n) = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) ... f(x_n)$
And i want the cumulative distribution function between $(2, 5]$
So i will do this:
$f(1) + f(2) + f(3) + f(4) + f(5) - f(1) - f(2) => f(3) + f(4) + f(5) $ what is he looking for.
See that $f(1) + f(2) + f(3) + f(4) + f(5) = F(5)$ and $f(1) + f(2) = F(2)$
So that is equal to $F(5) - F(2) = P(2 < X \leq 5)$ .
But, what happens if I want to include the lower interval?
Is easy to see that: $F(5) -F(2) = f(3) + f(4) + f(5)$, and i want to add the lower interval that is only one element, so is equal to $f(2)$, that is the probabilty function. So in general to find the cumulative distribution function between $[a,b ]$, i.e $P([a,b]) = F(b)   - F(a) + f(a)$, with $b > a$
