Confusion Regarding Griffith's Quantum Mechanics From Introduction to Quantum Mechanics by Griffiths:

So far, I have assumed that we are dealing with a discrete variable — that is, one that can take on only certain isolated values (in the example, $j$ had to be an integer, since I gave ages only in years). But it is simple enough to generalize to continuous distributions. If I select a random person off the street, the probability that her age is precisely 16 years, 4 hours, 27 minutes, and 3.333... seconds is zero. The only sensible thing to speak about is the probability that her age lies in some interval — say, between 16 and 17. If the interval is sufﬁciently short, this probability is proportional to the length of the interval. For example, the chance that her age is between 16 and 16 plus two days is presumably twice the probability that it is between 16 and 16 plus one day. (Unless, I suppose, there was some extraordinary baby boom 16 years ago, on exactly that day — in which case we have simply chosen an interval too long for the rule to apply. If the baby boom lasted six hours, we’ll take intervals of a second or less, to be on the safe side. Technically, we’re talking about inﬁnitesimal intervals.) Thus
$$ \text{probability that an individual (chosen at random) lies between}\; x \;\text{and}\; (x + \text{dx}) = \rho(x)\text{d}x$$

I really cannot understand the part in bold. What is Griffiths really saying? Please elaborate.
 A: Griffiths is just describing a hypothetical where a dx of one day is not good enough. The takeaway from this passage is that you want your dx on the LHS of 1.14 to be small relative to the length scale at which your problem changes.
A: If 16 years ago there were two days with a baby-boom (lasting for six hours and assuming that there is an asymmetric distribution of these hours over the two days, i.e., more baby-boom hours on one day than on the other), then 16 years later this is translated into a different chance that when selecting a random girl from the street she will be between 16years and 16years_1day old or between 16years+1day and 16years+2days old. So the chance that the girl is between 16years and 16years+2days will not be twice as big when you make the interval twice as big (from 1day to 2days).
A: The other answers give the correct conceptual explanation. In the interest of giving a different point of view (not necessarily better), I want to give one formal interpretation of what Griffifth's is saying.
Consider the cumulative distribution function, $C(x)$, defined as
\begin{equation}
C(x) \equiv P(X\leq x),
\end{equation}
or in words: $C(x)$ is the probability that the random variable $X$ takes a value less than or equal to $x$. If $X$ can be any real number from $x_{\min}$ to $x_{\rm max}$, then $C(x_{\rm min})=0$ and $C(x_{\rm max})=1$.
It's also useful to define the probability density function (PDF), which we will denote as $\pi(x)$. This PDF tells us the probability to find $X$ taking a value near $x$
\begin{equation}
\pi(x)dx = P(x\leq X \leq x+dx).
\end{equation}
The CDF is related to the PDF via
\begin{equation}
C(x) = \int_{x_{\rm min}}^x \pi(x')dx'
\end{equation}
Now, we "normally" expect that $C(x)$ will be a continuous function. In this case, then amount of probability accumulated between two points that are close together will become arbitrarily small as we bring those points together. In equations:
\begin{equation}
\lim_{\epsilon\rightarrow 0}\left[ C\left(x+\frac{\epsilon}{2}\right)-C\left(x-\frac{\epsilon}{2}\right)\right] = 0, \ \ C(x){\rm \ continuous}
\end{equation}
Another way to say this, is that the PDF is the derivative of the CDF (if the derivative exists). Since the probability to find $X$ near $x$ is equal to $P=\pi(x)\epsilon$, then $\epsilon$, if $\pi(x)$ exists and is finite, then $P$ will vanish as $\epsilon\rightarrow 0$.
However, it can happen that $C(x)$ is discontinuous. For example, suppose there was a huge spike of births at exactly the time $t_0$ over a zero time interval (sorry for the mixed notation), then a non-zero amount of probability for births will accumulate slightly before the time $t_0$ and slightly after. If we call the discontinuity $\lambda$, then in equations we have
\begin{equation}
\lim_{\epsilon\rightarrow 0}\left[ C\left(t_0+\frac{\epsilon}{2}\right)-C\left(t_0-\frac{\epsilon}{2}\right)\right] = \lambda \neq 0, \ \ C(x){\rm \ not\ continuous\ at\ } x=t_0
\end{equation}
This means that a finite amount of probability has accumulated over an infinitesimal interval. This is the situation Griffiths is discussing.
We can interpret this in terms of the PDF if you are happy with the notion of a delta function. In particular, we have that
\begin{equation}
\pi(x) = \pi_{\rm smooth}(x) + \lambda \delta(x-t_0)
\end{equation}
where $\pi_{\rm smooth}(x)$ is a normal, not-singular PDF. Then $\pi(t_0)$ is infinite, so we need to regularize the probability $\pi(x) dx$ near $x=t_0$ to find out what is happening. We will find there is a finite probability that accumulates as $x$ goes from infinitesimally less than $t_0$, to infinitesimally more.
That is a mathematical answer (at least a more mathematical answer, I'm a physicist not a mathematician so I am sure what I wrote is not completely rigorous). However the other answers explain physically what is going on. Usually one of the assumptions of our model is breaking down (such as "accumulated number of births per week" being a continuous function). So often the solution is to revisit the assumptions of the model. This is Griffith's point -- in singular situations, often it is a good idea to step back and check if the model makes sense. Occasionally you actually do want to allow these kind of singular probability distributions though (for example if you have a position eigenstate), so it is useful to know what they are and how to work with them.
