Why is $P(X\in[a,b])=P(X\in[a,b))=P(X\in(a,b])=P(X\in(a,b))$ I saw, for any continuous random variable $X$, $P(X\in[a,b])=P(X\in[a,b))=P(X\in(a,b])=P(X\in(a,b))$, where $a,b\in\mathbb{R}$, in my textbook.
I don't quite understand why the openness/closeness of an interval doesn't matter in continuous random variable?
Is it because prob. density function $f(x)=0$ whenever $x\in\{a,b\}$
Update:
I realize some fundamental mistakes I have made after checking out the comments and answer from all of you.
PDF $f(x)=0$ whenever $x = a, b$ is so wrong. It should be, WLOG, CDF instead of PDF, $ P(X=a) = F(a) = \int_{a}^{a}f(x) = 0$, which is why the equality holds regardless of the openness of interval. 
 A: Note, $P(X\in [a,b])=P(X\in [a,b))+P(X=b)=P(X\in [a,b))=P(X\in (a,b))+P(X=a)=P(X\in (a,b))$.
A: EDIT: This does not answer the question but may help understand more in-depth about integrals and how to calculate using probability distributions of mixed continous and discrete parts.
For $P(X\leq x_0) \neq P(X< x_0)$ to hold, we will need to expand the concept of integral and function to allow for the integral of a point to be something else than 0. One way to do this is with distributions. The dirac $\delta$ is such a distribution defined so that for any test-function $g(x)$: $$\int_{-\infty}^{\infty}\delta(x)g(x)dx = g(0)$$With such a mechanism we can have a function $g$ encode "atomic" probabilities in some set of points, use a sum of dirac delta function together with the linearity of the integral to "select" them and then add them to the ordinary density function $f$. This way we can represent a mixture of continuous probabilities and discrete jumps in probability or atomic probabilities.
As always we need to make sure that the sum of all probabilities equals one. In our case : that the atomic probabilities and the continuous probability equals 1:$$\sum_{\forall k}g(x_k) + \int_{-\infty}^{\infty} f(x)dx = 1$$ and we can keep using the same formalism, treating $$f(x) + g(x)\sum_{\forall k}\delta(x-x_k)$$ as our new density function.
A: Actually, for continuous distributions, the probability of a single point in a segment is zero because there are infinitely points in that segment. So, in such a case, one can add to or remove from an interval, a finite number of points without changing the probability.
