For continuous random variables $X,Y$ with probability density function $f_{X,Y}(x,y)$, the marginal probability distribution function of $X$ is
$$ f_{X}(x) = \int_{\mathbb{R}} f_{X,Y}(x,y) \, dy$$
and similarly for $f_Y(y)$.
My intuition is that $f_X(x)$ gives us the probability that $X$ attains a value $x$ (since we are essentially summing up ALL possible probabilities of $x$ for every single $y$ value possible).
But this contradicts my thinking, that for any continuous random variable, a probability at a certain point i.e. $\mathbb{P}(X = x)$, is zero.
Clearly, the marginal probability distribution would not always have zero probability at a certain point $x$.
Why is this? Am I misunderstanding it?
This question arose from learning the conditional density function
$$f_{Y\mid X}(y\mid x) = \frac{f_{X,Y}(x,y)}{f_X(x)}$$
which it looks like we're dividing by zero here...
The question and comments make clear that the confusion results from assuming that $f_X(x) = \Pr(X=x).$ That is incorrect. What you need instead is this: $$ \Pr(X\in A) = \int_A f_X(x)\,dx $$ or this: $$ \Pr(a\le X\le b) = \int_a^b f_X(x)\, dx. $$ The uniform distribution on the interval from $0$ to $1/2$ has this density: $$ f(x) = \begin{cases} 2 & \text{if } 0<x<1/2, \\ 0 & \text{otherwise.} \end{cases} $$ But it cannot be that $\Pr(X=x)=2$ no matter what number $x$ is. And how about the following fact about the standard normal density $$ \varphi(x) = \frac 1 {\sqrt{2\pi}} e^{-x^2/2}. $$ This does not exceed $1$ anywhere, but if it were true that $\Pr(X=x) = \varphi(x)$ for some random variable $X$, then we would have $$ \Pr\left(X= 0 \text{ or } \frac 1 {1000} \text{ or } \frac 2 {1000} \text{ or } \cdots \text{ or } \frac {1000}{1000} \right) = \sum_{k=0}^{1000} \varphi\left( \frac k {1000} \right) $$ and you may check that that sum is far bigger than $1$, so it cannot be a probability.
And the normal density with expected value $0$ and standard deviation $1/1000$ is $$ x\mapsto 1000 \varphi(1000x). $$ The value of this function at $0$ is more than $398$. That cannot be the probability that a random variable with this distribution is equal to $0$.