Probability that a continuous random variable assumes a fixed point I'm in an introductory statistics class in which the probability density function for continuous random variables has been defined as $P\left\{X\in B\right\}=\int_B f\left(x\right)dx$.  I understand that the integral of $\int\limits_a^af(x)dx=0$ but I can't rectify this with my intuition of a continuous random variable.
Say X is the random variable equal to the number of minutes from time t that the train arrives.  How do I calculate the probability that the train arrives exactly 5 minutes from now?  How can this probability be zero?  Is it not possible?  What if the train does arrive exactly 5 minutes from now, how could it occur if it had probability 0?
 A: You may be falling into the trap of regarding 'five minutes from now' as lasting some finite period of time (which would have a nonzero probability). 
"Five minutes from now" in the continuous variable sense is truly instantaneous. 
Imagine that the arrival of the next train is uniformly distributed between 8:00 and 8:15. Further imagine we define the arrival of a train as occurring at the instant the front of the train passes a particular point on the station (perhaps the midpoint of the platform if there's no better landmark). Consider the following sequence of probabilities: 
a) the probability a train arrives between 8:05 and 8:10
b) the probability a train arrives between 8:05 and 8:06
c) the probability a train arrives between 8:05:00 and 8:05:01
d) the probability a train arrives between 8:05:00 and 8:05:00.01 (i.e. in the space of one hundredth of a second
e) the probability a train arrives between 8:05 and one billionth of a second later
f) the probability a train arrives between 8:05 and one quadrillionth of a second later
... and so on
The probability that it arrives precisely at 8:05 is the limiting value of a sequence of probabilities like that. The probability is smaller than every $\epsilon>0$.
A: 
What if the train does arrive exactly 5 minutes from now, how could it
  occur if it had probability 0?

A probabilistic statement is not a statement about the possibility/feasibility of an event. It only reflects our attempt to quantify our uncertainty about it happening.  So when a phenomenon is continuous (or is modeled as one), then our tools and current state of knowledge do not permit us to make a probabilistic statement about it taking a specific value. We can only make such a statement related to a range of values. Of course the usual trick here is to discretize the support, to consider"small" intervals of values rather than single values. Since continuous random variables bring great benefits and flexibility compared to discrete random variables, this has been found to be a rather small price to pay, perhaps as small as the intervals we are forced to consider.
A: To give you some intuition for the above, try the following (thought) experiment:
Draw a real line around zero with a ruler. Now take a sharp dart and let it fall from above randomly on the line(let's assume you will always hit the line and only the lateral positioning matters for the sake of the argument).
However many times you let the dart fall randomly on the line, you will never hit the point zero. Why? Think what is the point zero, think what is its width. And after you recognise that its width is 0, do you still think you can hit it?
Will you be able to hit point 1, or -2? Or any other point you pick on the line for that matter?
To get back to maths, this is the difference between the physical world, and a mathematical concept such as real numbers (represented by the real line in my example). Probability theory has quite a bit more complicated definition of probability than you will see in your lecture. To quantify the probability of events and any combination of their outcomes, you need a probability measure. Both the Borel measure and Lebesgue measure are defined for an interval [a, b] on the real line as:
$$\mu([a,b])=b-a$$
from this definition you can see what happens with the probability if you reduce the interval to a number (setting a = b).
The bottom line is that based on our current definition of probability theory (dating back to Kolmogorov) the fact that an event has 0 probability does not mean it cannot occur.
And as far as your example with the train goes, if you will have an infinitely precise watch, your train will never arrive exactly on time.
A: A probability distribution has to have an area of unity. If the measure is continuous then there is an infinite number of values that it can take (i.e. an infinite number of values along the x-axis of the distribution). The only way that the total area of the probability distribution can be finite is for the value at each of the infinite number of values to be zero. One divided by infinity.
In 'real life' there can be no measures that take an infinite number of values (by several different philosophical arguments that don't matter much here) so no value need take a probability of exactly zero. A useful practical argument is based on the finite precision of real-world measurements. If you use a stopwatch that measures to one tenth of a second, the train will have one tenth of a second in which to arrive in 'exactly' five minutes.
A: Other people have answered why the probability is zero (if you approximate time as being continuous, which it is effectively not, but anyway...) so I will just echo it briefly. To answer the last question that the OP asked---"how could it occur if it had probability 0?"---lots and lots of things can occur if they have probability zero. All a set of probability zero $A$ means is that, in the space of possible things that could happen, the set $A$ takes up no space. That is all. It is not more meaningful than this. 
I am writing this to hopefully address something else that the OP said in the comments:

You say "you will never hit the point zero", but what can you say of the point that I hit in my first dart throw? Let  be the point that I hit. Before throwing my dart, you would have said "you will never hit the point ", but I've just hit it. Now what?

This is a very good question and one that, when I began to learn about probability, I struggled with. Here is the answer: it isn't equivalent to the question that you originally asked! What you have done is bring time into the analysis, and that means that the underlying probability structure changes to become much more intricate. Here is what you need to know. A probability space $(\Omega, A, \mu)$ consists of three things:
 an underlying space $\Omega$, such as $\mathbb{R}$ or $\mathbb{Z}$; a set of all possible outcomes on this space, such as the set of all half-open intervals on $\mathbb{R}$, and a measure $\mu$ that satisfies $\mu(\Omega) = 1$.
Your original problem lives in the space 
$([a,b], \text{all half open intervals on $[a,b]$}, \nu)$ where 
$\nu$ is Lebesgue measure (this means that $\nu( [c,d) ) = \frac{1}{d - c}$). In this space, the probability that you hit any single point $x \in [a,b]$ is zero for the reasons discussed above---I think we have this cleared up. But now, when you say things like the quoted passage above, you are defining something called a filtration, which we will write as $\mathcal{F} = \{\mathcal{F}_t\}_{t \geq 0}$. A filtration in general is a collection of subsets of $A$ that satisfy $\mathcal{F}_t \subseteq \mathcal{F}_s$ for all $t < s$.
In your case, we can define the filtration 
$$
\mathcal{F}_t = \{x \in [a,b]: \text{dart hit $x$ at time $t' < t$} \}.
$$
Now, in this new subset of your outcome space, guess what---you're right! You have hit it and, after your first throw, your probability of having hit that point when restricted to the filtration $\mathcal{F}_1$ is 1. 
