Confused by (cumulative) distribution function question... $P(0&lt;=X&lt;1)$ if $X$ is a random variable having a distribution function:

$F(x)=$ {($0, x&lt;0$), ($1/3, 0&lt;=x&lt;1$), ($2/3, 1&lt;=x&lt;2$), ($1, x&gt;=2)$} (hope that makes sense)

But if $x$ can potentially equal anything above $2$ all the way to infinity then surely the probability that $X=x$ where $x$ is between 0 and 1 is practically 0, so is that the answer?

Also as a side question, what is f(x) equal to in this case? F(x) = integral of f(x) right? But all of F(x) are constants?
$x$ cannot be greater than 2, because the cumulative function does not increase, i.e. the probability of x being less than 50000 is the same of being less than 2, so x will not be between 2 and 50000. For your problem, $P(0\leqslant x&lt;1)=\frac{1}{3}$, because 0, 1 and 2 are given probability $\frac{1}{3}$, but only 0 is in $0\leqslant x&lt;1$.

For discrete distributions, the integral is a sum; that's why 0, 1 and 2 are given probability $\frac{1}{3}$.
You have $P(X &lt; 1) = F_X(1-)$ and $P(X \ge 0) = 1 - P(X &lt; 0) = 1 - F_X(0-)$.