probability distribution proof $P(a < X \leq b) = F(b) - F(a)$

Question

Let F be the distribution function of the probability $\mathbb{P}$ on $\mathbb{R}$ (induced by some random variable $X$). Prove: $\mathbb{P}((a,b]) =\mathbb{P}(a < X \leq b) = F(b) - F(a)$

This is how far I got, but i cant figure out the next step, as the proof must hold for pdf and pmf.

\begin{align*} \mathbb{P}((a,b]) &= \mathbb{P}(a < X \leq b)\\ &= \mathbb{P}(\{X > a\} \cup \{X \leq b\})\\ &= \mathbb{P}(\{X > a\} \cup (-\infty, b])\\ &= F(b) - \mathbb{P}(\{X > a\})\\ &= \dots\\ \end{align*}

On my notes, the next step should look something like this:

$\mathbb{P}(\{X > a\}) = \mathbb{P}(a \leq X) = F(a)$

I am not sure whether this is true, neither do I understand it...

Answer 1

You certainly have an error as $\mathbb{P}(a < X \leq b) = \mathbb{P}(\{X > a\} \cap \{X \leq b\})$ with an intersection rather than a union, so perhaps (assuming $a \le b$ and so $\{X \leq a\} \subset \{X \le b\}$ ):

\begin{align*} \mathbb{P}((a,b]) &= \mathbb{P}(a < X \leq b)\\ &= \mathbb{P}(\{X > a\} \cap \{X \leq b\})\\ &= \mathbb{P}(\{X \leq b\} \,\backslash\, \{X \le a\}) \\ &= F(b) - F(a)\\ \end{align*}

Answer 2

You have to use continuity at right of $F$. Let $a_n\to a^+$. I denote $\mathbb P_X(A)=\mathbb P\{X\in A\}$. Then

$$\mathbb P_X(a,b]=\mathbb P_X\left(\bigcup_{n\in\mathbb N}[a_n,b]\right)=\lim_{n\to \infty }\mathbb P_X([a_n,b])=\lim_{n\to \infty }(F(b)-F(a_n))=F(b)-F(a).$$