Is $P(a \le X \le b) = P(a \lt X \lt b)$ for continuous distributions? 
Does the inclusion/exclusion of the boundary matter in the
      probability of continuous distributions?

Example:
$\displaystyle P(a \le X \le b)=\int_{a}^{b}P(X \in dx)$
$\displaystyle P(a \lt X \lt b)=\int_{a}^{b}P(X \in dx)$ ?
Thus, $P(a \le X \le b) = P(a \lt X \lt b)$ for continuous distributions?
If so, why is that?
 A: What does "continuous probability distribution" mean? By some accounts, a continuous probability distribution on (Borel subsets of) $\mathbb R$ is one for which the cumulative probability distribution function (c.d.f.) is everywhere continuous. And I think that convention is arguably the best.
The c.d.f. of the distribution of a random variable $X$ is
$$
F_X(x) = \Pr(X\le x) \quad \text{ for } x\in \mathbb R.
$$
By additivity of probability measures, one has for $a<b,$
$$
F_X(b) = \Pr(X\le b) = \Pr(X\le a)+\Pr(a<X\le b),
$$
and so
$$
\Pr(a<X\le b) = F_X(b) - F_X(a).
$$
Then by countable additivity, one can write a somewhat more complicated argument to show that
$$
\Pr(X=a) = F_X(a) - \lim_{x \, \uparrow \, a} F_X(x).
$$
This is $0$ if $F_X$ is continuous.
(By another convention, to say a distribution is continuous means there is a probability density function (p.d.f.) for which for every Borel set $A\subseteq\mathbb R,$
$$
\Pr(X\in A) = \int_A f_X(x)\, dx.
$$
A distribution that is continuous in the latter sense is continuous in the former sense, but not conversely.)
