Expression of the probability of a single point. In my notes I have that the probability of a single point is 
$P(X=a)=P(\{a\})$=$$\lim_{\epsilon \to \ 0^+}P(a- \epsilon<X\leq a)=$$
$$\lim_{\epsilon \to \ 0^+}(F_X(a)-F_X(a- \epsilon))=$$ 
\begin{cases}
0 & \text{ if } F_X \:is\:continuous \\ 
jump\:\:of\:F_X & \text{ otherwise } 
\end{cases}
My question is about the first inequality in $P(\{a\})=\lim_{\epsilon \to -\ 0^+}P(a- \epsilon<X\leq a)$
Does this expression hold for the continuous case? Because the limit of $(a- \epsilon)$ when $\epsilon \to \ 0^+$ is $a$.
I would express it as $P(\{a\})=\lim_{\epsilon \to -\ 0^+}P(a- \epsilon\leq X\leq a)$ though I don't know if it is right from a probabilistic point of view. 
 A: Since
$$
P\left(a-\epsilon< X\le a\right)\le P\left(a-\epsilon\le X\le a\right)\le P\left(a-2\epsilon< X\le a\right)\ ,
$$
it follows from the sandwich theorem that
$$
\lim_{\epsilon\rightarrow0^+}P\left(a-\epsilon\le X\le a\right)=\lim_{\epsilon\rightarrow0^+}P\left(a-\epsilon< X\le a\right)\ ,
$$
so the limit is the same, regardless of whether the left inequality is strict or not.  However if $\ F_X\ $ isn't continuous at $\ a-\epsilon\ $, then
$$
P\left(a-\epsilon\le X\le a\right)\ne F_X(a)-F_X(a-\epsilon)\ ,
$$
whereas the identity
$$
P\left(a-\epsilon< X\le a\right) = F_X(a)-F_X(a-\epsilon)
$$
always holds. Thus, the identity
$$
\lim_{\epsilon\rightarrow0^+}P\left(a-\epsilon< X\le a\right) = \lim_{\epsilon\rightarrow0^+}\left(F_X(a)-F_X(a-\epsilon)\right)
$$
is a trivial consequence of the previous one, but
$$
\lim_{\epsilon\rightarrow0^+}P\left(a-\epsilon\le X\le a\right) = \lim_{\epsilon\rightarrow0^+}\left(F_X(a)-F_X(a-\epsilon)\right)\ ,
$$
while still true, isn't such a trivial consequence.  Thus, in the derivation you give, use of the strict inequality is far preferable to use of the non-strict one.
