For a continuous random variable, why does $P(a < Z < b) = P(a \leq Z < b) = P(a < Z \leq b) = P(a \leq Z \leq b)$ My textbook puts this in a sidebox with the heading "Note" and doesn't explain why. Could you tell me why this statement holds?
$P(a < Z < b) = P(a \leq Z < b) = P(a < Z \leq b) = P(a \leq Z \leq b)$ 
 A: Nothing formal to add to this, but an analogy that really helped me to understand this came from a calculus text. Imagine you have an iron pipe of a certain length and weight. And you wish to cut it into two pieces. If the pipe is say 1 m long you might want to cut it in half at the 0.5 mark. Now think of the pipe's weight as some constant times the length of the pipe, (we assume that all cross-sections of equal length has the same weight).
Cutting the pipe in half at the 0.5 m mark - how much weight do you lose? Remember that the only cross-section you are removing is the 0.5 m mark itself. So what is the length of this cross-section? Consider that 0.49999999... is not apart of it, and neither is 0.5000000000...1, or any other point that is close to, but not equal to 0.5 - so the length of this cross-section is technically zero. Which means you're not really removing any weight at all.
This would explain why $\leq$ and $<$ are basically the same for continuous variables - including or excluding the endpoint really doesn't change anything - for any point you pick close to the endpoint, there is still an infinite amount of points between them.
Does this make any sense?
A: First I will give the definition of an (absolutely) continuous random variable $Z$.    
(Advanced probability is needed, you many skip it!)    
Let $(\Omega,F, P)$ be a probability space and let $Z:\Omega\rightarrow R^n$ be a random vector. The probability $P_X$ on $B(R^n)$ defined by $P_Z(A)=P\{Z\in A\}$, $A \in B(R^n)$ is called the distribution of $Z$. Now if $P_Z\ll \mu,$ where $\mu$ is Lebesgue measure on $R^n$, (i.e. $P$ is absolutely continuous with respect to $\mu$) then we say that $Z$ is an (absolutely) continuous random vector. Now by using Radon–Nikodym theorem, there exists a function $f: R^n\to[0,+\infty]$ such that $P_Z(A)=\int_{A}fd_{\mu}$ for all $A \in B(R^n)$. We call $f$ the density function of $Z$.
Now define the cumulative distribution function (CDF) of an absolutely continuous random variable $Z$ as: $$F_Z(z)=P(Z\leq z).$$     
Before I give a formal proof, let's have an example of a continuous random varible that is uniformly distributed i.e. with a probability density function of $f(z)=1$  for $0\leq z \leq 1$ and 0 otherwise. Now let's try to find $P(z=0.5)$. We have $$P(z=0.5)\sim P(0.4< z\leq 0.6)=\int_{0.4}^{0.6}f(z)d_z=0.2.$$ We can shrink that interval to get a better approximation as follows: $$P(z=0.5)\sim P(0.49< z\leq 0.51)=\int_{0.49}^{0.51}f(z)d_z=0.02,$$ $$P(z=0.5)\sim P(0.499< z\leq 0.501)=\int_{0.499}^{0.501}f(z)d_z=0.002.$$ As you can see, these probabilities are converging to zero as we shrink the length of the interval. Now let's prove it formally. I am goig to show that for any continuous random variable $Z$, we have: $$P(Z=a)=0,$$ by using the CDF. $$P(Z=a)=\lim_{\epsilon\to 0}P(a-\epsilon< Z\leq a+\epsilon)=\lim_{\epsilon\to 0}F_Z(a+\epsilon)-\lim_{\epsilon\to 0}F_Z(a-\epsilon)\\=F_Z(a)-F_Z(a)=0,$$ since the CDF function, $F$, is a "continuous" function for the continuous random variable $Z$. Similarly $P(Z=b)=0$.
Finally note that $$P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B).$$ So $$P(a\leq Z<b)=P\Big(\{Z=a\}\bigcap \{a<Z<b\}\Big)=P(Z=a)+P(a<Z<b)\\=0+P(a<Z<b)=P(a<Z<b).$$ You can use the same argument for other equalities.
A: Perhaps a more intuitive explanation is that for a continuous variable the contribution of the edges (e.g., $a$ or $b$) to the cumulative probability in the surrounding intervals (or semi-intervals) is negligibly small.
