In MLE for continuous rv, why is it ok to evaluate a pdf at a point? In MLE for continuous case, my course notes define the likelihood function to be:
$$
L(\theta) = L(\theta;y) = \prod_{i=1}^n f(y_i;\theta)
$$
Where $f$ is the joint pdf of $y_i$ given $\theta$.
I thought normally it doesn't make sense to evaluate a pdf at a point (eg. $f(X=2) = 0$ for some continuous rv X), and that it only make sense to evaluate for the random variable to be $>$ or $<$ to a value.
But in the case of MLE, why is it OK to evaluate a joint pdf at a point?
 A: Intuitively, @dsaxton's answer provides the correct logic. Let me just "translate" it to math language.
Suppose the sample $Y_1, \ldots, Y_n \text{ i.i.d.} \sim f_\theta(y)$, $\theta \in \Theta$, where $\Theta$ is the parameter space and $f_\theta(\cdot)$ are density functions. After a vector of observations $y = (y_1, \ldots, y_n)'$ has been made, the maximum likelihood principle aims to look for an estimator $\hat{\theta} \in \Theta$ such that for any $\delta > 0$, the probability 
\begin{equation}
P_{\hat{\theta}}[(Y_1, \ldots, Y_n) \in (y_1 - \delta, y_1 + \delta) \times \cdots \times (y_n - \delta, y_n + \delta)] \tag{1}
\end{equation}
is the maximum over $\theta \in \Theta$, which suggests us considering the 
quantity
\begin{equation}
L(\theta, \delta) \equiv P_{\theta}[(Y_1, \ldots, Y_n) \in (y_1 - \delta, y_1 + \delta) \times \cdots \times (y_n - \delta, y_n + \delta)], 
\quad \theta \in \Theta.\tag{2}
\end{equation}
Note that from the statistical inference point of view, the probability in $(2)$ should be viewed as a function of $\theta$. 
We now simplify $(2)$ by invoking the i.i.d. assumption and that $P_\theta$ has density $f_\theta$. Clearly,
$$L(\theta, \delta) = \prod_{i = 1}^n P_\theta[y_i - \delta < Y_i < y_i + \delta] = \prod_{i = 1}^n \int_{y_i - \delta}^{y_i + \delta}f_\theta(y) dy. \tag{3}$$
What we need to show, thus answer your question is: if $\hat{\theta}$ maximizes $L(\theta, \delta)$ with respect to $\theta \in \Theta$ for any $\delta > 0$, then it also maximizes the so-called likelihood function
$$L(\theta) = \prod_{i = 1}^n f_{\theta}(y_i). \tag{4}$$
So let's start with assuming for any $\delta > 0$, 
$$L(\hat{\theta}, \delta) \geq L(\theta, \delta), \quad \forall \theta \in \Theta. \tag{5}$$
Divide both sides of $(5)$ by $(2\delta)^n$, then let $\delta \downarrow 0$ 
gives 
$$\prod_{i = 1}^n f_{\hat{\theta}}(y_i) \geq \prod_{i = 1}^n f_{\theta}(y_i), \quad \forall \theta \in \Theta,$$
which is precisely $L(\hat{\theta}) \geq L(\theta), \forall \theta \in \Theta$. 
In other words, maximizing the so-called likelihood function $(4)$ (which is a product of densities) is a necessary condition of carrying out the maximum likelihood principle. From this point of view, the form of densities multiplication makes sense.
Above is just my own interpretation, any comment or critic is very welcomed.
A: It's not "incorrect" to look at a density function at a specific point.  If it were, what's the point of the function?
What you may have heard is that a density function at a given point is not to be interpreted as a probability, but that doesn't make it unimportant.  The density function tells you how "tightly packed" probability is around a certain point, or how likely a random variable is to be close to the given value, and this idea extends to random samples as well.
In the case of maximum likelihood estimation what we're doing then is finding a value for a parameter that causes the sample to belong to a "neighborhood" of high probability, relative to other regions of the sample space.
