I want to prove the following:
Let $X$ be a random variable and let us denote with $F_X$ the distribution function. By $a^-$ we denote the left-limit.
Claim:
$P[a \leq X \leq b] = F_X(b) – F_X(a^-)$ if $a \leq b$ and
$P[a < X <b] = F_X(b^-)- F_X(a)$ if $a < b$
Deduce that:
$P[X=b] = F_X(b)- F_X(b^-)$.
Attempt:
To compute: $P[a \leq X \leq b]$ I can use the fact that – given the density function $p$ (can I assume there exists one?) – we have:
$P[a \leq X \leq b] = \int_a^b p(x) dx$.
Now:
$\int_a^b p(x) dx = \int_{-\infty}^bp(x) – \int_{-\infty}^ap(x) = F_X(b) -F_X(a) .$
Why do I need to consider the left limit and why not just write $F_X(a)$ instead? For the first as well as the second point in the claim I get the same result but without considering the left-limit. I don't see where this is necessary.
Thanks for your help!
EDIT
Without assuming that there is a density function:
$P[a \leq X \leq b] = P[X \leq b] – P [X < a] = F_X(b) – F_X(a^-)$
and for the second point
$P[a < x < b] = P[X < b] – P[X \leq a] = F_X(b^-) – F_X(a)$
Best Answer
(Letting $F_X = F$.)
Try showing the following general fact: $$F(a^-) = P[X < a].$$ Everything else follows from that.
To show the above, note that given any increasing sequence $a_n \uparrow a$ (with $a_n < a$ for all $n$), we have $$\{X < a\} = \bigcup_{n \ge 1}\left\{X \le a_n\right\}.$$ Note that the union on the right is an increasing union. Thus, by continuity of probability, you get $$P[X < a] = \lim_{n \to \infty}P\left[X \le a_n\right] = \lim_{n \to \infty} F(a_n).$$ Can you finish it now?