There's something fundamental that I'm missing regarding the Standard Uniform Distribution (Continuous) So if we have the standard uniform distribution
$$X \sim U(0,1)$$
$$f(x) = 1 \text, 0 < x < 1$$
So now I don't understand how the probability of having any point between 0 and 1 is just 1. Are we just saying that no matter what x is, it will happen with 100% certainty?
 A: As the continuous uniform distribution is a continuous probability distribution, it doesn't make sense to talk about the probability of a single value occurring, instead, we have to talk about the probability of a value in a specific range occurring. 
So for instance:
$$\mathbb{P}[x \in [a,b]]=\int_{a}^{b}f(x)\:\mathrm{d}x$$
In this case, we have the probability of $x$ being in the range $[0,1]$ 
$$\mathbb{P}[x \in [a,b]] = \int_{0}^{1}1\:\mathrm{d}x=1$$
As we expect (it is a valid probability distribution). But if we try to determine the probability of a particular point occurring:
$$\mathbb{P}[x = a] = \lim_{\varepsilon\to0}\int_{a-\varepsilon}^{a+\varepsilon}1\:\mathrm{d}x=(a+\varepsilon)-(a-\varepsilon) = 2\varepsilon \to 0$$
A: I think you misunderstand the meaning of a probability density function. For a continuous distribution $f$, $f(x)$ is not the probability that $x$ happens. Actually this probability is $0$. If $X$ is distributed according to a standard uniform distribution, the probability that $X$ is exactly $0.2$, or exactly $0.7$ or exactly any single point between $0$ and $1$, is $0$.
That is why we need to integrate over an interval (or, to be specific, something that has a positive Lebesgue measure), in order to have a non-zero probability, just as Shaktal explained.
