Can someone explain the relation between random variables and continuity? For example, if we look at distributions with $P(T<5)$ and $P(T\le5)$, the two notations equal due to continuity. I don't understand how continuity is related to the reason these two notations are equal.
 A: Optional prefatory remarks:
"Continuity" is a bit ambiguous, in that it can mean either of two things:


*

*The cumulative distribution function is continuous; or

*There is a probability density function.


The second point means that
$$\Pr(T\text{ is in some specified set of numbers}) = \int\limits_{\begin{array}{c} \text{that set} \\ \text{of numbers} \end{array}} f(t)\,dt,$$
where $f$ is the density function.
The second condition is sometimes called "absolute continuity".  Whenever it is true, the first condition---that the C.D.F. is continuous---is also true.  But there are some cases where the first condition is true but the second is not.  The so-called Cantor distribution is one of those.
But we really only need the first conditions---that the C.D.F. is continuous---in order to show that $\Pr(T<t)=\Pr(T\le t)$.
The actual answer begins here:
Observe that if $s<t$ then
$$
\begin{align}
\Pr(T \le t) & = \Pr(T\le s\text{ or }s<T\le t) \\[8pt]
& =\Pr(T\le s)+\Pr(s<T\le t),
\end{align}
$$
and therefore
$$
\begin{align}
\Pr(s<T\le t) & = \Pr(T \le t) - \Pr(T\le s) \\[8pt]
& = F(t) - F(s),
\end{align}
$$
where $F$ is the C.D.F.
Continuity of $F$ implies that you can make $F(t)-F(s)$ as close to $0$ as you want by making $s$ close enough to (but still less than) $t$.  Thus you can make $\Pr(s<T\le t)$ as close to $0$ as you want by pushing $s$ toward $t$ from below (or "from the left", if you think of the usual lay-out of the $t$-axis).  Since $\Pr(T=t)\le\Pr(s<T\le t)$, it follows that you can make $\Pr(T=t)$ as close to $0$ as you want by pushing $s$ toward $t$.  But $\Pr(T=t)$ doesn't change as $s$ moves toward $t$.  Consequently $\Pr(T=t)$ must be $0$.  Now notice that
$$
\begin{align}
\Pr(T\le t) & = \Pr(T<t\text{ or }T=t) \\[8pt]
& = \Pr(T<t)+\Pr(T=t).
\end{align}
$$
Since we've shown that $\Pr(T=t)$ is $0$, it follows that
$$
\Pr(T\le t) = \Pr(T<t).
$$
